# Notes II: Sets and Functions: problems and solutions

Let $f: X \rightarrow Y$, let $A \subseteq X$, let $B \subseteq Y$. Then,

$f(A) = \{f(x): x \in A \}$

$f^{-1}(B) = \{ x: f(x) \in B\}$

I. The main properties of the first set mapping are:

1a) $f(\phi) = \phi$, $f(X) \subseteq Y$

1b) $A_{1} \subseteq A_{2} \Longrightarrow f(A_{1}) \subseteq A_{2}$

1c) $f(\bigcup_{i}A_{i}) = \bigcup_{i}f(A_{i})$

1d) $f(\bigcup_{i}A_{i}) \subseteq \bigcap_{i} f(A_{i})$

1a: obvious by definition of f(A) or f.

1b: obvious by definition of f

1c: Let $x_{0} \in \bigcup_{i}A_{i}$ so that for some one i we have $f(x) \in f(A_{i})$. That is, $f(\bigcup_{i}A_{i}) \subseteq \bigcup_{i}f(A_{i})$. Reversing the arguments, we get the other subset relation. So that $f(\bigcup_{i}A_{i}) = \bigcup_{i}f(A_{i})$. The image of the union is the union of images.

1d: Let $x_{0} \in \bigcap_{i}A_{i}$ so that $x_{0}$ belongs to each $A_{i}$. So that LHS is clearly a subset of $\bigcap_{i}f(A_{i})$.

II) Now, we want to verify the following useful/famous relations of the second set mapping:

2a) $f^{-1}{(\phi)} = \phi$, $f^{-1}(Y) = X$

2b) $B_{1} \subseteq B_{2} \Longrightarrow f^{-1}(B_{1}) \subseteq f^{-1}(B_{2})$

2c) $f^{-1}(\bigcup_{i}B_{i}) = \bigcup_{i}f^{-1}(B_{i})$

2d) $f^{-1}(\bigcap_{i}B_{i}) = \bigcap_{i}f^{-1}(B_{i})$

2e) $f^{-1}(B^{'}) = f^{-1}(B)^{'}$

Answer 2c: Let $y_{0}$ belong to at least one $B_{i}$ so that $x_{0} = f^{-1}(y_{0}) = f^{-1}(B_{i})$ for some i. In other words, $f^{-1}(\bigcup_{i}B_{i}) = \bigcup_{i}f^{-1}(B_{i})$

TPT: $f^{-1}(B^{'}) = f^{-1}(B)^{'}$.

Let $y_{0} \in f^{-1}(B^{'})$.

Hence, $f(y_{0}) \in B^{'}$.

Hence, $f(y_{0}) \in U-B$, where U is the universal set

Hence, $f(y_{0}) \in U-f(A)$

Hence, $f(y_{0}) \in U-B$

Hence, $f(y_{0}) \in U-f(A)$

Hence, $f(y_{0}) \in (f(A))^{'}$

Hence, $y_{0} \in f^{-1}(B)^{'}$. Hence, $f^{-1}(B^{'}) \subseteq f^{-1}(B)^{'}$

Reversing the arguments proves the other subset inequality.

Hence, $f^{-1}(B^{'}) = f^{-1}(B)^{'}$. QED.

More Problems :

A) Two mappings $f: X \rightarrow Y$ and $g: X \rightarrow Y$ are said to be equal (and we write them as f=g) if $f(x) = g(x)$ for every x in X. Let f, g, and h be any three mappings of a non-empty set X into itself, and now prove that the multiplication of mappings is associative in the sense that $(fg)h = f(gh)$.

Solution A:

Let $f: X \rightarrow X$, $g: X \rightarrow X$, and $h: X \rightarrow X$ be any three mappings of X into itself. Let $x_{0} \in X$ and let $h(x_{0}) = x_{1} \in X$. Now, we know that $((fg)h)(x)= (fg)(h(x))$. So that now we want to find $(fg)(x_{1}) = f(g(x_{1})) = f(x_{2})$ assuming $g(x_{1}) \in x_{2} \in X$.

Now, on the other hand, $(f(gh))(x)$ means $(f)((gh)(x)) = f(g(h(x_{0})) = f(g(x_{1}))=f(x_{2})$ just as in LHS. QED.

B) Let X be a non-empty set. The identity mapping $i_{X}$ on X is the mapping of X onto itlsef defined by $i_{X}(x)=x$ for every x. Then, $i_{X}$ sends each element of the set X to itself; that is, leaves fixed each element of X. Prove that $fi_{X}= i_{X}f=f$ for any mapping f of X into itself. If f is one-to-one onto, so that its inverse $f^{-1}$ exists, show that $ff^{-1} = f^{-1}f=i_{X}$. Prove further that $f^{-1}$ is the only mapping of X into itself which has this property; that is, show that if g is a mapping of X into itself such that $fg=gf=i_{X}$, then $g=f^{-1}$. Hint: $g=gi_{X}=g(ff^{-1})=(gf)f^{-1} = i_{X}f^{-1}=f^{-1}$, or

$g=i_{X}g=(f^{-1}f)g=f^{-1}(fg) = f^{-1}i_{X}=f^{-1}$

B(i) TPT: $fi_{X}=i_{X}f=f$ for any mapping f of X into itself. Proof: Consider $x_{0}\in X$ and such that $f(x_{0}) = x_{1}$. Hence, $fi_{X}(x_{0}) = f(i_{X}(x_{0}))=f(x_{0}) = x_{1} \in X = f(x_{0})$. Now, on the other $(i_{X}f)(x_{0}) = i_{X}(f(x_{0}))=i_{X}(x_{1})= x_{1}$ also. QED.

B(ii): Let f be one-to-one onto such that $f: X \rightarrow X$. Hence, $f^{-1}$ exists. Also, $f^{-1}: X \rightarrow X$. Let $f(x_{0}) = x_{1} \in X$. Hence, by definition and because $f^{-1}$ is also one-to-one and onto, $f^{-1}(x_{1})=x_{0}$. Hence, clearly, $ff^{-1} = f^{-1}f=i_{X}$. QED.

B(iii) : Prove that such a $f^{-1}$ is unique.

Solution to B(iii): Let there be another function g such that $gf=fg=i_{X}$. Also, we have $ff^{-1}=f^{-1}f=i_{X}$. Use the hint ! 🙂 QED.

Question C:

Let X and Y be non-empty sets and f a mapping of X into Y. Prove the following: (i) f is one-to-one if and only if there exists a mapping g of Y into X such that $gf=i_{X}$ (ii) f is onto if and only if there exists a mapping h of Y into X such that $fh=i_{X}$.

Solution C(i) : Given $f: X \rightarrow Y$ such that if $x_{0} \neq x_{1}$, then $f(x_{0}) \neq f(x_{1})$. Now, we want a mapping $g: Y \rightarrow X$ such that $gf=i_{X}$. Let us construct a g such that $g(y_{0})=x_{0}$ and $g(y_{1}) = x_{1}$. Then, $(gf)(x_{0}) = g(f(x_{0}))=g(y_{0})=x_{0}$. Similarly, $gf(x_{1})=x_{1}$. Here, we have assumed that $f(x_{0})=y_{0}$ and $f(x_{1})=y_{1}$. So, knowing f to be one-to-one, we can explicitly construct a mapping g such that $g: Y \rightarrow X$ and $gf=i_{X}$. QED. On the other hand, let it be given that X and Y are two non-empty sets, that $f: X \rightarrow Y$ and $g: Y \rightarrow X$ such that $gf=i_{X}$. That is, if some $x^{'} \in X$, then we have $(gf)(x^{'}) = i_{X}(x^{'})=x^{'}$. That is, $g(f(x^{'})) = x^{'}$. This forces, if $f(x^{'}) = y^{'}$, then $g(y^{'})=x^{'}$. Now, if there be an element $x^{''} \neq x^{'}$, it will force $f(x^{''})=y^{''} \neq x^{'}$. Hence, different inputs give rise to different outputs and so f is one-to-one. QED.

Solution C(ii) : Given $f: X \rightarrow Y$ and f is onto, that is, $f(X) = Y$. We want to show a function $h: Y \rightarrow X$ such that $fh=i_{X}$. Since $f(X)=Y$, with X as domain, Y as codomain equal to range, f is one-to-one, onto. So $h=f^{-1}$. QED.

Question D:

Let X be a non-empty set and f a mapping of X into itself. Show f is one-to-one onto if and only there exists a mapping g of X into itself such that $gf=fg=i_{X}$. If there exists a mapping g with the property, then there is one and only one such mapping. Why?

Solution D: From previous question, C we are done.

Question E:

Let X be a non-empty set, and let f and g be one-to-one mappings of X onto itself. Show that fg is also a one-to-one mapping of X onto itself and that $(fg)^{-1}=g^{-1}f^{-1}$.

HW. Hint: consider the action of the functions on each element.

Question F:

Let X and Y be non-empty sets and f a mapping of X into Y. If A and B are, respectively, subsets of X and Y, prove the following:

(a) $ff^{-1}(B) \subseteq B$, and $ff^{-1}(B)=B$ is true for all B if and only if f is onto.

(b) $A \subseteq f^{-1}f(A)$, and $A = f^{-1}f(A)$ is true for all A if and only if f is one-to-one.

(c) $f(A_{1} \bigcap A_{2}) = f(A_{1}) \bigcap f(A_{2})$ is true for all $A_{1}$ and $A_{2}$ if and only if f is one-to-one.

(d) $f(A)^{'} \subseteq f(A^{'})$ is true for all A if and only if f is onto.

(e) If f is onto — so that, $f(A)^{'} \subseteq f(A^{'})$ is true for all A — then, $f(A)^{'}=f(A^{'})$ is true for all A if and only if f is also one-to-one.

Solution Fa: Put $B=Y$, so that f is onto. To prove the other subset relationship, simply reverse the arguments.

Solution Fb:

We need to apply the definition that different inputs give rise to different outputs.

;

😦 I hope to blog later)

:

Solution Fc:

It certainly implies that $A_{1} \bigcap A_{2} \neq \phi$.

:

(I hope to blog later).

Solution Fd:

Given that $f: X \rightarrow Y$ and $A \subseteq X$ and $B \subseteq Y$. Then, $f(A)^{'}$ is $Y-f(A)$, that is, $Y-B$, that is $B^{'}$, that is, $f(B^{'})$. Put $A=X$, then f is onto. Now, $f(X)^{'} \subseteq f(X^{'})$, that is, $Y-f(X) \subseteq f(\phi)$. This implies $Y=f(X)$. To prove the other subset relationship, simply reverse the arguments. QED.

Regards,

Nalin Pithwa.

# Notes II: Sets and Functions:

Reference I : Topology and Modern Analysis, G F Simmons, Tata McGraw Hill.

III. Functions:

Many kinds of functions occur in topology in a wide variety of situations. In our work, we shall need the full power of the general concept of a function, and since, its modern meaning is much broader and deeper than its elementary meaning, we discuss this concept in considerable detail and develop its main abstract properties.

Let us begin with a brief inspection of some simple examples. Consider the elementary function

$y=x^{2}$

of the real variable x. What do we have in mind when we call this a function and say that y is a function of x? In a nutshell, we are drawing attention to the fact that each real number x has a specific real number y linked to it, which can be calculated according to the rule (or law of correspondence) given by the formula. We have here a process which applied to a real number x does something to it (squares it) to produce another number y (the square of x). Similarly,

$y = x^{2}-3x$ and $y = (x^{2}+1)^{-1}$

are two other simple functions of the real variable x, and each is given by a rule in the form of an algebraic expression which specifies the exact manner in which the value of y depends on the value of x.

The rules for the functions we have just mentioned are expressed by formulas. In general, this is possible for functions of a very simple kind or for those which are sufficiently important to deserve a special symbol of their own. Consider for instance the function of the real variable x defined as follows: for each real number x, write x as an infinite decimal (using the scheme of decimal expansion in which infinite chains of 9s are avoided — in which for example, 1/4 is represented as 0.250000….rather than by 0.24999….); then, let y be the 59th digit after the decimal point. There is of course no standard formula for this but nevertheless it is a perfectly respectable function whose rule is given by a verbal description. On the other hand, the function $y=\sin{x}$ of the real variable x is so important that its rule, though fully as complicated as the one just defined is assigned the special symbol sin. When discussing functions in general, we work to allow for all sorts of rules and to talk about them all at once, so we simply employ non-committal notations like $y=f(x)$, $y=g(z)$, and so on.

Each of the functions mentioned above is defined for all real numbers x. The example $y = \frac{1}{x}$ shows that this restriction is much too severe, for this function is defined only for non zero values of x. Similarly, $y = \log{x}$ is defined only for positive values of x and $y = \arcsin{x}$ only for values of x in the interval $[-1,1]$. Whatever our conception of a function may be, it should certainly be broad enough to include examples like these, which are defined only for some values of the real variable x.

In real analysis, the notion of function is introduced in the following way. Let X be any non-empty set of real numbers. We say that a function $y=f(x)$ is defined on X if the rule f associates a definite real number y with each real number x in X. The specific nature of the rule f is totally irrelevant to the concept of a function. The set X is called the domain of the given function, and the set Y of all the values it assumes is called its range. If we speak of complex numbers here instead of real numbers, we have the notion of function as it is used in complex analysis.

This point of view towards functions is actually more general than is needed for aims of analysis, but it isn’t nearly general enough for our purposes. The sets X and Y above were taken to be sets of numbers. If we now remove even this restriction and allow X and Y to be completely arbitrary non-empty sets, then we arrive at the most inclusive concept of a function. By way of illustration, suppose that X is the set of all squares in a plane and that Y is the set of all circles in the same plane. We can define a function $y=f(x)$ by requiring that the rule f associate with each square x that circle y which is inscribed in it. In general, there is no need at all for either X or Y to be a set of numbers. All that is really necessary for a function is two non-empty sets X and Y and a rule f which is meaningful and unambiguous in assigning to each element x in X a specific element y in Y.

With these preliminary descriptive remarks, we now turn to the rather abstract but very precise ideas they are intended to motivate.

A function consists of three objects: two non-empty sets X and Y (which may be equal but need not be) and a rule f which assigns to each element x in X a single fully determined element y in Y. The y which corresponds in this way to a given x is usually written f(x), and is called the image of x under the rule f, or the value of f at the element x. (It is fun to draw some figures here). This notation is supposed to be suggestive of the idea that the rule f takes the element x and does something to it to produce the element $y = f(x)$. The rule f is often called a mapping or transformation or operator to amplify this concept of it. We then think of f as mapping x’s to y’s, or transforming x’s to y’s, or operating on x’s to produce y’s. The set X is called the domain of the function, and the set of all f(x)’s for all x’s in X is called its range. A function whose range consists of just one element is called a constant function.

We often denote by $f: X \rightarrow Y$ the function with rule f, domain X and range contained in Y. This notation is useful because the essential parts of the function are displayed in a manner which emphasizes that it is a composite object, the central thing being the rule or mapping f. You can try drawing a figure depicting a convenient way of picturing this function. (these notes don’t have my diagrams from the reference book) On the left, X and Y are different sets, and on the right they are equal — in which case we usually refer to f as a mapping of X into itself. If it is clear that from the context what the sets X and Y are, or if there is no real need to specify them explicitly, it is common practice to identify the function $f: X \rightarrow Y$ with the rule f, and to speak of f alone as if it were the function under consideration (without mentioning the sets X and Y).

It sometimes happens that two perfectly definite sets X and Y are under discussion and that a mapping of X into Y arises which has no natural symbol attached to it. If there is no necessity to invent a symbol for this mapping and if it is quite clear what the mapping is, it is often convenient to designate it by $x \rightarrow y$. Accordingly, the function $y=x^{2}$ mentioned in the beginning of this section can be written as $x \rightarrow x^{2}$ or $x \rightarrow y$ where y is understood to be the square of x.

A function f is called an extension of a function g (and g is called a restriction of f) if the domain of f contains the domain of g and $f(x) = g(x)$ for each x in the domain of y.

Most of mathematical analysis, both classical and modern, deals with functions whose values are real numbers or complex numbers. This is also true of those parts of topology which are concerned with the foundations of analysis. If the range of a function consists of real numbers, we call it a real function, similarly, a complex function is one whose range consists of complex numbers. Obviously, every real function is also complex. We lay very heavy emphasis on real and coomplex functions through out our work.

As a matter of usage, we generally prefer to reserve the term function for real or complex functions and to speak of mappings when dealing with functions whose values are not necessarily numbers.

Consider a mapping $f: X \rightarrow Y$. When we call f a mapping of X into Y, we mean to suggest by this that the elements f(x) — as x varies over all the elements of X — need not fill up Y; but if it definitely does happen that the range of f equals Y, or if we specifically want to assume this, then we call f a mapping of X onto Y. If two different elements in X always have different images under f, then we call f a one=to-one mapping of X into Y. If $f: X \rightarrow Y$ is both onto and one-to-one, then we can define its inverse mapping $f^{-1}: X \rightarrow Y$ as follows: for each y in Y, we find that unique element x in X such that $f(x)=y$ 9 x exists and is unique since f is onto and one-to-one); we then define x to be $f^{-1}(y)$. The equation $x = f^{-1}(y)$ is the result of solving $y=f(x)$ for x in just the same way as $x = \log{y}$ is the result of solving $e^{x}$ for x. Figure 7 illustrates the concept of the inverse of a mapping.

If f is a one-to-one mapping of X onto Y, it will sometimes be convenient to subordinate the conception of f as a mapping sending x’s over to y’s and to emphasize its role as a link between x’s an y’s. Each x has linked to it (or has corresponding to it) exactly one $x = f^{-1}(y)$. When we focus our attention on this aspect of a mapping which is one-to-one onto correspondence between X and Y, and $f^{-1}$ is a one-to-one correspondence between Y and X.

Now, consider an arbitrary mapping $f: X \rightarrow Y$. The mapping f which sends each element of X over to an element of Y induces the following important set mappings. If A is a subset of X, then the image f(A) is the subset of Y defined by

$f^{-1}(B) = \{ x : f(x) \in \hspace{0.1in} B\}$

and the second set mappings pull each B back to its corresponding $f^{-1}(B)$. It is often essential for us to know how these set mappings behave with respect to set inclusion and operations on sets. We develop most of their significant features in the following two paragraphs.

The main properties of the first set mapping are:

$f(\phi) = \phi$

$f(X) \subseteq Y$

$A_{1} \subseteq A_{2} \Longrightarrow f(A_{1}) \subseteq f(A_{2})$

$f(\bigcup_{i})A_{i} = \bigcup_{i}f(A_{i})$

$f(\bigcap_{i}A_{i}) \subseteq \bigcup_{i}f(A_{i})$….call these relations I.

The reader should convince himself of the truth of these statements. For instance, to prove (i) we would have to prove first that $f(\bigcup_{i}A_{i})$ is a subset of $\bigcup_{i}f(A_{i})$, and second that $\bigcup_{i}f(A_{i})$ is a subset of $f(\bigcup_{i}A_{i})$. A proof of the first of these set inclusions might run as follows: an element in $f(\bigcup_{i}A_{i})$ is the image of some element in $\bigcup_{i}A_{i}$, therefore, it is the image of an element in some $A_{i}$, therefore it is some $f(A_{i})$ and so finally it is in $\bigcup_{i}f(A_{i})$. The irregularities and gaps which the reader will notice in the above statements are essential features of this set mapping. For example, the image of an intersection need not equal the intersection of the images, because two disjoint sets can easily have images which are not disjoint. Furthermore, without special assumpitions (see Problem 6), nothing can be said about the relation between $f(A)^{'}$ and $f(A^{'})$.

The second set mapping is much better behaved. Its properties are satisfyingly complete, and can be stated as follows:

$f^{-1}(\phi) = \phi$ and $f^{-1}(Y) =X$;

$B_{1} \subseteq B_{2} \Longrightarrow f^{-1}(B_{1}) \subseteq f^{-1}(B_{2})$

$f^{-1}(\bigcup_{i}B_{i}) \Longrightarrow \bigcup_{i}f^{-1}(B_{i})$….(2)

$f^{-1}(\bigcap_{i}B_{i}) = \bigcap_{i}f^{-1}(B_{i})$….(3)

$f^{-1}(B^{'}) = f^{-1}(B)^{'}$….(4)

Again, the reader should verify each of these statements for himself.

We discuss one more concept in this section, that of the multiplication or composition of mappings. If $y= f(x)=x^{2}+1$ and $z = g(y) = \sin{y}$

then these two functions can be put together to form a single function defined by $x = (gf)(x) = g(x^{2}+1)=\sin{(x^{2}+1)}$. One of the most important tools of calculus (the chain rule) explains how to differentiate functions of this kind. This manner of multiplying functions together is of basic importance for us as well, and we formulate it in general as follows. We define the product of these mappings, denoted by $gf: X \rightarrow Z$ by $(gf)(x) = g(f(x))$. In words, an element x in X is taken by f to the element f(x) in Y,and then g maps f(x) to g(f(x)) in Z. Figure 8 is a picture of this process. We observe that the two mappings involved here are not entirely arbitrary, for the set Y which contains the range of the first equals the domain of the second. More generally, the product of two mappings is meaningful whenever the range of the first is contained in the domain of the second. We have regarded f as the first mapping and y as the second, and in forming their product gf, their symbols have gotten turned around. This is a rather unpleasant phenomenon, for which we blame the occasional perversity of mathematical symbols. Perhaps it will help the reader to keep this straight in his mind if he will remember to read the product gf from right to left: first apply f, then g.

Problems:

To be continued next blog.

Regards,

Nalin Pithwa

# Notes I: Sets and Functions

Reference 1: Topology and Modern Analysis, G F Simmons, Tata McGraw Hill Publication.

There are certain logical difficulties which arise in the foundations of set theory (see Problem 1). We avoid these difficulties by assuming that each discussion in which a number of sets are involved takes place in the context of a single fixed set. This set is called the universal set. It is denoted by U in this section and the next, and every set mentioned is assumed to contain elements of U. In later chapters, there will always be on the one hand, a given space within which we work, and this will serve without further comment as our universal set. It is often convenient to have available in U a set containing no elements whatsoever; we call this the empty set and denote it by the symbol $\phi$. A set is said to be finite if it is empty or consists of n elements for some positive integer n; otherwise, it is said to be infinite.

The relation $\subseteq$ is called set inclusion. The following are the properties of set inclusion (it is an equivalence relation):

i) $A \subseteq A$ for every A;

ii) $A \subseteq B$ and $B \subseteq A$ imply $A=B$.

iii) $A \subseteq B$ and $B \subseteq C$ imply $A \subseteq C$.

It is quite important to observe that (i) and (ii) above can be combined into a single statement that $A=B \longrightarrow A \subseteq B$ and $B \subseteq A$. This remark contains a useful principle of proof, namely, that the only way to show that two sets are equal apart from merely inspecting them, is to show that each is a subset of the other.

Problems:

1. Perhaps, the most famous of the logical difficulties referred to in the theory of sets is Russell’s paradox. To explain what this is, we begin by observing that a set can easily have elements which are themselves sets., e.g., $\{ 1, \{ 2,3\}, 4\}$. This means the possibility that a set might well contain itself as one of its elements. We call such a set an abnormal set, and any set which does not have itself as an element we call a normal set. Most sets are notmal,, and if we suspect that abnormal sets are in some way undesirable, we might try to to confine our attention to the set N of all normal sets. Someone is now sure to ask, is N itself normal or abnormal? It is evidently one or the other, as it cannot be both. Show that if N is normal, it must be abnormal. Proof 1: Set N is the set of all normal sets. Clearly, it contains itself, so it is abnormal. Part II: Show that if N is abnormal, then it is normal. Part 2 proof: As N is abnormal, it contains itself as an element. Set N is the set of all normal sets, so it does not contain itself. So, it is normal. We see in this way that each of our two alternatives is self-contradictory, and it seems to be the assumption that such a set N exists which has brought us to this impasse. For further discussion of these matters, refer to Fraenkel’s Abstract Set Theory. Russell’s own account of the discovery can also be found.
2. The symbol we have used for set inclusion is similar to that used for the familiar order relation on the real line: if x and y are real numbers, $x \leq y$ means that $y-x$ is non negative. The order relation on the real line has all the properties mentioned in the text: (a) $x \leq x$ for every x (b) $x \leq y$ and $y \leq x$ means $x=y$, and (c) $x \leq y$ and $y \leq z$ means $x \leq z$. It also has an additional property: (d) For any x and y, either $x \leq y$ or $y \leq x$. Property d says that any two real numbers are comparable with respect to the relation in question, and it leads us to call the order relation on the real line as a total (or linear) order relation. Show by an example that this property is not possessed by set inclusion. It is for this reason that set inclusion is called a partial order relation. Answer: Disjoint sets.
3. (a) Let U be the single element set $\{ 1\}$. There are two subsets, the empty set $\phi$ and $\{1\}$ itself. If A and B are arbitrary subsets of U, there are four possible relations of the form $A \subseteq B$. Count the number of true relations amongst these. Answer: $\phi \subset U$, a true relation; $U \subset \phi$, which is not a true relation; $U \subseteq U$, which is a true relation; $\phi \subseteq \phi$, which is debatable. (3b) Let U be the set $\{ 1,2\}$. There are four subsets, namely, $\phi$, $U$, $\{ 1 \}$ and $\{ 2\}$. If A and B are arbitrary subsets of U, there are 16 possible relations of the form $A \subseteq B$. Count the number of true ones: Answer: $\phi \subseteq U$, $\phi \subseteq \{ 1\}$, $\phi \subseteq \{ 2 \}$, $\{ 1 \} \subseteq U$, $\{ 2\} \subseteq U$, $\{ 1 \} \subseteq \{ 1 \}$, $\{ 2 \} \subseteq \{ 2 \}$. So, there are 7 such true relations out of 16. (3c) Now, let U be the set $\{ 1,2,3\}$. There are 8 subsets, namely, $\phi$, $U$, $\{ 1\}$, $\{ 2\}$, $\{ 3 \}$, $\{1,2\}$, $\{ 1,3\}$, $\{ 2,3\}$. There are 64 possible relations of the form $A \subseteq B$. Count the number of true ones. Answer: $\phi \subseteq U$, $\phi \subseteq \{ 1 \}$, $\phi \subseteq \{ 2 \}$, $\phi \subseteq \{ 3 \}$, $\{ 1 \} \subseteq U$, $\{ 2 \} \subseteq U$, $\{ 3 \} \subseteq U$, $\{ 1, 2\} \subseteq U$, $\{ 1, 3\} \subseteq U$, $\{ 2,3\} \subseteq U$, $\{ 1\} \subseteq \{ 1, 2 \}$, $\{ 1 \} \subseteq \{ 1,3\}$, $\{ 2 \} \subseteq \{ 1,2 \}$, $\{ 2\} \subseteq \{ 2,3\}$, $\{ 3\} \subseteq \{ 1,3\}$, $\{ 3\} \subseteq \{ 2,3\}$. And, also, except $\phi$ and $U$, each is a subset of itself. Thus, there are such 30 true relations. (3d) Let U be the set $\{ 1,2,3,\ldots, n\}$ for an arbitrary positive integer n. How many subsets are there ? Answer: $2^{n}$ subsets. How many possible relations of the form $A \subseteq B$? Answer: $2^{n} \times 2^{n}$, that is, $2^{2n}$ possible pairings. Can you make an informed guess as to how many of these are true? Answer (trial 1): The empty set $\phi$ is paired with each of the remaining $2^{n}-1$ so these give $2^{n}-1$ pairings; each set which is not U and $\phi$ can be paired with U as a superset, so these give $2^{n}-2$ pairings; except $\phi$ and $U$, each subset is a subset of itself, so these are total $2^{n}-2$ possibilities; each two element subset can be paired with 3-element subsets containing the original two elements and one more, and there are $2^{n}-2$ such cases; each two element subset can be paired with 4-element subsets out of which two elements are original and two are chosen out of the remaining $2^{n}-2$, that is, there are ${2^{n}-2} \choose {2}$ pairings; …continuing in this way, we get the following final answer: $2(n-3)+\frac{(n-4)(n-1)}{2} + 2^{n}\times (n-3) + (2^{n-2}-1)(2^{2^{n}-n+2})$

II. The Algebra of Sets:

In this section, we consider several useful ways in which sets can be combined with one another, and we develop the chief properties of these operations of combination.

As we emphasized above, all the sets we mention in this section are assumed to be subsets of our universal set U. U is the frame of reference, or the universe, for our present discussion. In our later work, the frame of reference in a particular context will naturally depend on what ideas we happen to be considering. If we find ourselves studying sets of real numbers, then U is the set R of real numbers. If we wish to study sets of complex numbers, then we take U to be the set C of all complex numbers. We sometimes want to narrow the frame of reference and to consider (for instance) only subsets of the closed unit interval $[0,1]$, or of the closed unit disc $\{z : |z|\leq 1 \}$ and in these cases we choose U accordingly. Generally speaking, the universal set U is at out disposal, and we are free to select it to fit the needs of the moment. For the present, however, U is to be regarded as a fixed but arbitrary set. This generally allows us to apply the ideas we develop below to any situation which arises in out later work.

It is extremely helpful to the imagination to have a geometric picture available in terms of which we can visualize sets and operations on sets. A convenient way to accomplish this is to represent U by a rectangular area in a plane, and the elements which make up U by the points of this area. Sets can then be pictured by areas within this rectangle, and diagrams can be drawn which illustrate operations on sets and relations between them. For instance, if A and B are sets, then Figure 1 represents the circumstance that A is a subset of B (we think of each set as consisting of all points within the corresponding closed curve). Diagrammatic thought of this kind is admittedly loose and imprecise; nevertheless, the reader will find it invaluable. No mathematics, however, abstract it may appear, is ever carried on without the help of mental images of some kind, and these are often nebulous, personal and difficult to describe.

The first operation we discuss in the algebra of sets is that of forming unions. The union of two sets A and B, written $A \bigcup B$, is defined to be the set of all elements which are in either A or B (including those which are in both). $A \bigcup B$ is formed by lumping together the elements of A and those of B and regarding them as constituting a single set. In Figure 2, $A \bigcup B$ is indicated by the shaded area. The above can also be expressed symbolically:

$A \bigcup B = \{ x : x \in A or x \in B\}$.

The operation of forming union is commutative and associative:

$A \bigcup B = B \bigcup A$ and $A \bigcup (B \bigcup C) = (A \bigcup B) \bigcup C$.

It has the following additional properties: $A \bigcup A = A$, $A \bigcup \phi=A$, and $A \bigcup U = U$.

We also note that $A \subseteq B \Leftrightarrow A \bigcup B = B$ so set inclusion can be expressed in terms of this operation.

Our next operation is that of forming intersections. The intersection of two sets A and B, written $A \bigcap B$ is the set of all elements which are in both A and B. In symbols, $A \bigcap B = \{ x: x \in A and x \in B\}$.

$A \bigcap B$ is the common part of the sets A and B. In Figure 3, $A \bigcap B$ is represented by the shaded area. If $A \bigcap B$ is non-empty, we express this by saying that A intersects B. If, on the other hand, it happens that A and B have no common part, or equivalently, that $A \bigcap B = \phi$, then we say that A does not intersect B, or that A and B are disjoint and a class of sets in which all pairs of distinct sets are disjoint is called a disjoint class of sets. The operation of forming intersection is also commutative and associative:

$A \bigcap B = B \bigcap A$ and $A \bigcap (B \bigcap C) = (A \bigcap B) \bigcap C$.

It has the further properties that

$A \bigcap A = A$, $A \bigcap \phi = \phi$, and $A \bigcap U=A$, and since

$A \subseteq B \Leftrightarrow A \bigcap B = A$,

We see that set inclusion can also be expressed in terms of forming intersections.

We have now defined two of the fundamental operations on sets and we have seen how each is related to set inclusion. The next obvious step is to see how they are related to one another. The facts here are given by the distributive laws:

$A \bigcap (B \bigcup C) = (A \bigcap B) \bigcup (A bigcap C)$ and

$A \bigcup (B \bigcap C) = (A \bigcup B) \bigcap (A \bigcup C)$

These properties depend only on simple logic applied to the meaning of symbols involved. For instance, the first of the two distributive laws says that and element is in A and B or C precisely when it is in A and B or A and C. We can convince ourselves of the validity of these laws by drawing pictures (Venn Diagrams). The second distributive law is illustrated in Figure 4 where $A \bigcup (B \bigcap C$ is formed on the left by shading and $(A \bigcup B) \bigcap (A \bigcup C)$ on the right by cross shading. A moment’s consideration by the reader ought to convince him that one gets the same sets in both cases.

The last of our major operations on sets is the formation of complements. The complement of a set A, denoted by $A^{'}$ is the set of all elements not in A. Since the only elements we are considering are those which make up U, it goes without saying — but it ought to be said that $A^{['}$ consists of all elements of U which are not in A. Symbolically: $A^{'} = \{ x: x \notin A, x \in U\}$

Figure 5 in which $A^{'}$ is shaded indicates this operation. The operation of forming complements has the following obvious properties:

$(A^{'})^{'} = A$

$\phi^{'} = U$

$U^{'}=\phi$

$A \bigcup A^{'} = U$

$A \bigcap A^{'} = \phi$.

Further, it is related to set inclusion by $A \subseteq B \Longleftrightarrow B^{'} \subseteq A^{'}$, and to the formation of unions and intersections by

$(A \bigcup B)^{'} = A^{'} \bigcap B^{'}$

$(A \bigcap B)^{'} = A^{'} \bigcup B^{'}$.

In words, the above laws, called De Morgan’s Laws, can be beautifully related as: The complement of union of two sets is the intersection of the complements; the complement of the intersection of two sets is the union of the complements.

Or, in yet other words, the first De Morgan law says: an element is not in either of the two sets precisely when it is outside of both; and the second De Morgan law can be expressed as: an element is not in both sets precisely when it is outside of one or the other.

The operations of forming unions and intersections are primarily binary operations, that is, each is a process which applies to a pair of sets and yields a third. We have emphasized this by our use of parentheses to indicate the order in which the operations are to be performed, as in $(A_{1} \bigcup A_{2})\bigcup A_{3}$, where the parentheses direct us first to unite $A_{1}$ and $A_{2}$, and then to unite the result of this with $A_{3}$. Associativity makes it possible to dispense with parentheses in an expression like this and to write $A_{1} \bigcup A_{2} \bigcup A_{3}$ where we understand that these sets are to be united in any order and the order in which we do it is irrelevant (that is, all possible different orders yield the same result). Similar remarks apply to $A_{1} \bigcap A_{2} \bigcap A_{3}$. Furthermore, if $\{ A_{1}, A_{2}, \ldots, A_{n}\}$ is any finite class of sets, then we can form

$A_{1} \bigcup A_{2} \bigcup \ldots \bigcup A_{n}$

$A_{1} \bigcap A_{2} \bigcap \ldots \bigcap A_{n}$

in much the same way without any ambiguity of meaning whatever. In order to shorten the notation, we let $I = \{1, 2, 3, \ldots, n \}$ be the set of subscripts which index the sets under consideration. I is called the index set. We then compress the symbols for the union and intersection just mentioned to $\bigcup_{i \in I}A_{i}$ and $\bigcap_{i \in I}A_{i}$. For the sake of both brevity and clarity, these sets are often written $\bigcup_{i=1}^{n}A_{i}$ and $\bigcap_{i=1}^{n}A_{i}$.

These extensions of our ideas and notation don[‘t reach nearly far enough. It is often necessary to form unions and intersections of large (really large!) classes of sets. Let $\{A_{i} \}$ be an entirely arbitrary class of sets indexed by a set I of subscripts. Then,

$\bigcup_{i \in I}A_{i} = \{x : x \in A \hspace{0.1in} for \hspace{0.1in} at \hspace{0.1in} least \hspace{0.1in} one \hspace{0.1in} i \in I \}$.

And,

$\bigcap_{i \in I}A_{i} = \{ x: x \in A_{i} \hspace{0.1in} for \hspace{0.1in} each \hspace{0.1in} i \in I\}$

define their unions and intersections.

As above, we usually abbreviate these notations in $\bigcup_{i}A_{i}$ and $\bigcap_{i}A_{i}$ and if the class $\{ A_{i}\}$ consists of a sequence of sets, that is, if $\{ A_{i}\} = \{A_{1}, A_{2}, \ldots, A_{n} \}$, then their unions and intersections are often written in the form $\bigcup_{i=1}^{\infty}A_{i}$ and $\bigcap_{i}A_{i}$. Observe that we did not require the class $A_{i}$ to be non-empty. If it does happen that the class is empty, then the above definitions give (remembering that all sets under consideration are subsets of U) $\bigcup_{i}A_{i}$ and $\bigcap_{i}A_{i}=U$. The second of these facts amounts to the following statement: if we require of an element that it belongs to each set of the class, but there be no set present in the class, then every element satisfies this requirement. (We call this as the school of “vacuous logic” in theory of sets.) Note that if we had not made the explicit assumption that all sets under consideration are to be considered as subsets of U, we would not be able to fix a meaning to the intersection of an empty class of sets. A moments reflection makes it clear that the De Morgan equations are valid for arbitrary unions and intersections: (we can write a little proof also; it is easy):

$(\bigcup_{i}A_{i})^{'} = \bigcap_{i}A_{i}^{'}$

Complement of arbitrary union is intersection of complements.

$(\bigcap_{i}A_{i})^{'} = \bigcup_{i}A_{i}^{'}$

Complement of arbitrary intersection is union of complements.

**********************

It is instructive to verify the above general De Morgan laws for the case in which the class $\{ A_{i}\}$ is empty. We present the answers below:

Consider the first general DeMorgan law: $(\bigcup_{i}A_{i})^{'} = \bigcap_{i}A_{i}^{'}$. Now, as $\{ A_{i} \} = \phi$, so $\bigcup_{i}A_{i}=\phi$ and hence, the LHS is $(\bigcup_{i}A_{i})^{'}= (\phi)^{'}=U$ whereas RHS Is $\bigcap_{i}A_{i}^{'}= \bigcap_{i}U=U$ same as LHS. QED.

Now, consider the second generalized De Morgan Law:

$(\bigcap_{i}A_{i})^{'} = \bigcup_{i}A_{i}^{'}$, which we want to verify when the class of sets $\{ A_{i}\}$ is empty. Consider $\bigcap_{i}A_{i}=\phi$ and so $(\bigcap_{i}A_{i})^{'} = (\phi)^{'}=U$, which is our LHS. Now let us see RHS: $\bigcup_{i}A_{i}^{'} = \bigcup_{i}(\phi)^{'}=\bigcup_{i}U=U$ which is same as LHS. QED.

**********************

We conclude our treatment of the general theory of sets with a brief discussion of certain special classes of sets which are of considerable importance in topology, logic, and measure theory. We usually denote classes of sets by capital letters in boldface.

First some general remarks which will be useful both now and later, especially in connection with topological spaces. We shall often have occasion to speak of finite unions and finite intersections, by which we mean unions and intersections of finite classes of sets, and by a finite class of sets we always mean one which is empty or consists of n sets for some positive integer n. If we say that a class of sets A is closed under the formation of finite unions, we mean that A contains the union of each of its finite subclasses; and since the empty subclass qualifies as a finite subclass of A , we see that its union, the empty set, is necessarily an element of A. In the same way, a class of sets which is closed under the formation of finite intersections necessarily contains the universal set.

Now, for the special classes of sets mentioned above. For the remainder of this section, we specifically assume that the universal set is non-empty. A Boolean algebra of sets is a non-empty class A of subsets of U which has the following properties:

(i) $A \hspace{0.1in}and \hspace{0.1in} B \in {\bf {A}} \Longrightarrow A \bigcup B \in \bf{A}$

(ii) $A \hspace{0.1in} and \hspace{0.1in} B \in {\bf{A}} \Longrightarrow A \bigcap B \in \bf{A}$

(iii) $A \in {\bf{A}} \Longrightarrow A^{'} \in \bf{A}$.

Since A is assumed to be non-empty, it must contain at least one set A. Then, property iii above shows that $A^{'} \in {\bf{A}}$ also, so that $A \bigcap A^{'} = \phi \in {\bf{A}}$ also. Moreover, $A^{'} \bigcup A = U \in {\bf{A}}$ also. Since the class consisting of only the empty set and the Universal set is also clearly a boolean algebra of sets, these two distinct sets are the only ones which every Boolean algebra of sets must contain. It is equally clear that the class of all subsets of U is also a Boolean algebra of sets. There are many other less trivial kinds, and their applications are manifold in fields of study as diverse as electronics and statistics.

Let A be a Boolean algebra of sets. It is obvious that if $\{ A_{1}, A_{2}, \ldots, A_{n}\}$ is a non-empty finite sub-class of A then

$A_{1}\bigcup A_{2} \bigcup \ldots \bigcup A_{n}$ and $A_{1}\bigcap A_{2} \bigcap \ldots \bigcap A_{n}$ are both sets in A; and since A contains both the empty set and the universal set, it is easy to see that A is a class of sets which is closed under the formation of finite unions, finite intersections and complements. We now go in the other direction, and let A be a class of sets which is closed under the formation of finite unions, finite intersections, and complements. By these assumptions, A automatically contains the empty set and the universal set, so it is non-empty and is easily seen to be a Boolean algebra of sets. We conclude from these remarks that Boolean algebras of sets can be described alternatively as classes of sets which are closed under the formation of finite unions, finite intersections, and complements. It should be emphasized once again that when discussing Boolean algebras of sets, we always assume that the universal set is non-empty.

One final comment. We speak of Boolean algebras of sets because there are other kinds of Boolean algebras than those which consist of sets, and we wish to preserve this distinction. We explore this topic further in our Appendix on Boolean algebras.

Problems:

1. If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, prove that $\bigcup_{i}A_{i} \subseteq \bigcup_{j}B_{j}$ and $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$.

Answer 1: Recall that a class of sets means a set of sets. By this definition, the following obviously holds: $\bigcup_{i}A_{i} \subseteq \bigcup_{j}B_{j}$. But, the second claim (I have yet to prove it for myself): $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$.

2. The difference between two sets A and B, denoted by $A-B$, is the set of all elements in A and not in B, thus $A-B=A \bigcap B^{'}$. Show that the following holds true:

$A-B = A - (A \bigcap B) = (A \bigcup B) -B$

$(A-B)-C=A-(B \bigcup C)$

$A - (B-C) = (A-B) \bigcup (A \bigcap C)$

$(A \bigcup B) - C = (A-C) \bigcup (B-C)$

$A-(B \bigcup C) = (A-B) \bigcap (A-C)$

3. The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is thus the union of their differences in opposite orders. Prove the following:

$A \triangle (B \triangle C) = (A \triangle B) \triangle C$….that is, symmetric difference is associative

$A \triangle \phi = A$

$A \triangle A = \phi$

$A \triangle B = B \triangle A$…that is, symmetric difference is commutative

Distributive law like relation: prove: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$

4. A ring of sets is a non-empty class A of sets such that if A and B are in A, then $A \triangle B$ and $A \bigcap B$ are also in A. Show that A must also contain the empty set, $A \bigcup B$, and $A-B$. Prove that if a non empty class of sets contains the union and difference of any pair of its sets, then it is a ring of sets. Prove that a Boolean algebra of sets is a ring of sets.

5. Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra of sets.

6. Show that the class of all finite unions of closed open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.

7. Assuming that the universal set U is non-empty, show that Boolean algebras of sets can be described rings of sets which contain U.

Let me host the solutions after a while.

Regards,

Nalin Pithwa

# Mathematical Morsels : IV

The words set and space are often used in loose context to one another. A set is merely an amorphous collection of elements, without coherence or form. When some kind of algebraic or geometric structure is imposed on a set, so that its elements are organized into a systematic whole, then it becomes a space.

# Equivalence relations and partitions: some core basic theorems

Suppose R is an equivalence relation on a set S. For each a in S, let [a] denote the set of elements of S to which a is related under R, that is, $[a] = \{ x: (a,x) \in R\}$

We call [a] the equivalence class of a in S under R. The collection of all such equivalence classes is denoted by S/R, that is, $S/R = \{ [a]: a \in S\}$. It is called the quotient set of S by R.

The fundamental property of an equivalence relation and its quotient set is contained in the following theorem:

Theorem I:

Let R be an equivalence relation on a set S. Then, the quotient set S/R is a partition of S. Specifically,

(i) For each $a \in S$, we have $a \in [a]$.

(ii) $[a]=[b]$ if and only if $(a,b) \in R$.

(iii) If $[a] \neq [b]$, then [a] and [b] are disjoint.

Proof of (i):

Since R is reflexive, $(a,a) \in R$ for every $a \in S$ and therefore $a \in [a]$.

Proof of (ii):

Assume: $(a,b) \in R$.

we want to show that $[a] = [b]$. That is, we got to prove, (i) $[b] \subseteq [a]$ and (ii) $[a] \subseteq [b]$.

Let $x \in [b]$; then, $(b,x) \in R$. But, by hypothesis $(a,b) \in R$ and so, by transitivity, $(a,x) \in R$. Accordingly, $x \in [a]$. Thus, $[b] \subseteq [a]$.

To prove that $[a] \subseteq [b]$, we observe that $(a,b) \in R$ implies, by symmetry, that $(b,a) \in R$. Then, by a similar argument, we obtain $[a] \subseteq [b]$. Consequently, $[a]=[b]$.

Now, assume $[a] = [b]$.

Then by part (i) of this proof that for each $a \in S$, we have $a \in [a]$. So, also, here $b \in [b]=[a]$; hence, $(a,b) \in R$.

Proof of (iii):

Here, we prove the equivalent contrapositive of the statement (iii), that is:

If $[a] \bigcap [b] \neq \emptyset$ then $[a] = [b]$.

if $[a] \bigcap [b] \neq \emptyset$ then there exists an element $x \in A$ with $x \in [a] \bigcap [b]$. Hence, $(a,x) \in R$ and $(b,x) \in R$. By symmetry, $(x,b) \in R$, and, by transitivity, $(a,b) \in R$. Consequently, by proof (ii), $[a] = [b]$.

The converse of the above theorem is also true. That is,

Theorem II:

Suppose $P = \{ A_{i}\}$ is a partition of set S. Then, there is an equivalence relation $\sim$ on S such that the set $S/\sim$ of equivalence classes is the same as the partition $P = \{ A_{i}\}$.

Specifically, for $a, b \in S$, the equivalence $\sim$ in Theorem I is defined by $a \sim b$ if a and b belong to the same cell in P.

Thus, we see that there is a one-one correspondence between the equivalence relations on a set S and the partitions of S.

Proof of Theorem II:

Let $a, b \in S$, define $a \sim b$ if a and b belong to the same cell $A_{k}$ in P. We need to show that $\sim$ is reflexive, symmetric and transitive.

(i) Let $a \in S$. Since P is a partition there exists some $A_{k}$ in P such that $a \in A_{k}$. Hence, $a \sim a$. Thus, $\sim$ is reflexive.

(ii) Symmetry follows from the fact that if $a, b \in A_{k}$, then $b, a \in A_{k}$.

(iii) Suppose $a \sim b$ and $b \sim c$. Then, $a, b \in A_{i}$ and $b, c \in A_{j}$. Therefore, $b \in A_{i} \bigcap A_{j}$. Since P is a partition, $A_{i} = A_{j}$. Thus, $a, c \in A_{i}$ and so $a \sim c$. Thus, $\sim$ is transitive.

Accordingly, $\sim$ is an equivalence relation on S.

Furthermore,

$[a] = \{ x: a \sim x\}$.

Thus, the equivalence classes under $\sim$ are the same as the cells in the partition P.

More later,

Nalin Pithwa.

# Some more foundation mathematics — notions from set theory: Tutorial problems

It’s a strange way of starting a lecture that I adopt..sometimes…I first give my students quizzes or exams…Here is some foundation mathematics for my deserving students and also, if any of my reader is interested:

Basic Notions from Set Theory:

Reference: Introduction to Analysis, Maxwell Rosenlicht, Dover Publications,

Exercises:

Question 1:

Let $\Re$ be the set of real numbers and let the symbols $<, \leq$ have their conventional meanings:

a) Show that $\{ x \in \Re: 0 \leq x \leq 3\} \bigcap \{x \in \Re: -1

b) List the elements of

$(\{2,3,4 \} \bigcup \{ x \in \Re: x^{2}-4x+3 = 0\}) \bigcap \{x \in \Re: -1 \leq x < 3 \}$

c) Show that

$(\{ x \in \Re: -2 \leq x \leq 0\} \bigcup \{ x \in \Re: 2 < x <4\}) \bigcap \{ x \in \Re: 0 \leq x \leq 3\} = \{ x \in \Re: 2 < x \leq 3\} \bigcup \{ 0\}$

Question 2:

If A is a subset of the set S, prove that :

2a) $(A^{'})^{'}=A$

2b) $A \bigcup A = A \bigcap A = A \bigcup \phi = A$

2c) $A \bigcap \phi = \phi$

2d) $A \times \phi = \phi$

Question 3:

Let A, B, C be elements of a set S. Prove the following statements and illustrate them with diagrams:

(a) $A^{'} \bigcup B^{'} = (A\bigcap B)^{'}$… a De Morgan law. In words, it can be said that the union of two complements is the complement of the intersection of the two.

(b) $A \bigcap (B \bigcup C) = (A \bigcap B) \bigcup (A \bigcap C)$

(c) $A \bigcup (B \bigcap C) = (A \bigcup B) \bigcap (A \bigcup C)$.

Question 4:

If A, B, C are sets, then prove that :

i) $(A-B) \bigcap C = (A \bigcap C)-B$

ii) $(A \bigcup B)-(A \bigcap B) = (A-B) \bigcup (B-A)$

iii) $A-(B-C) = (A-B) \bigcup (A \bigcap B \bigcap C)$

iv) $(A-B) \times C = (A \times C) - (B \times C)$

Question 5:

Let f be a non-empty set and for each $i \in I$, let X be a set. Prove that

(i) for any set B, we have :

$B \bigcap \bigcup_{i \in I}X_{i} = \bigcup_{i \in I}(B \bigcap X_{i})$

(ii) if each $X_{i}$ is a subset of a given set S, then

$(\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}$

Question 6:

Prove that if $f: X \rightarrow Y$, $g: Y \rightarrow Z$, and $h: Z \rightarrow W$ are functions, then

$h \circ (g \circ f) = (h \circ g) \circ f$

Question 7:

Let $f: X \rightarrow Y$ be a function, let A and B be subsets of X, and let C and D be subsets of Y. Prove that:

(i) $f(A \bigcup B) = f(A) \bigcup f(B)$

(ii) $f(A \bigcap B) \subset f(A) \bigcap f(B)$

(iii) $f^{-1}(C \bigcup D) = f^{-1}(C) \bigcup f^{-1}(D)$

(iv) $f^{-1}(C \bigcap D) = f^{-1}(C) \bigcap f^{-1}(D)$

(v) $f^{-1}(f(A)) \supset A$

(vi) $f(f^{-1}(C)) \subset C$

Question 8:

(a) Prove that a function f is one-to-one if and only if $f^{-1}(f(A)) = A$ for all $A \subset X$.

(b) Prove that a function f is onto if and only if $f(f^{-1}(C)) = C$ for all $C \subset Y$.

Cheers,

Nalin Pithwa

PS: These tutorial problems can be used for IIT JEE Maths, Pre RMO, RMO Maths etc. also.

# Some foundation mathematics

Well-Ordering Principle:

Every non-empty set S of non-negative integers contains a least element; that is, there is some integer a in S such that $a \leq b$ for all b’s belonging to S.

Because this principle plays a role in many proofs related to foundations of mathematics, let us use it to show that the set of positive integers has what is known as the Archimedean property.

Archimedean property:

If a and b are any positive integers, then there exists a positive integer n such that $na \geq b$.

Proof:

Assume that the statement of the theorem is not true so that for some a and b, we have $na for every positive integer n. Then, the set $S = \{ b-na : n \in Z^{+}\}$ consists entirely of positive integers. By the Well-Ordering Principle, S will possess a least element, say, $b-ma$. Notice that $b- (m+1)a$ also lies in S; because S contains all integers of this form. Further, we also have $b-(m+1)a=(b-ma)-a contrary to the choice of $b-ma$ as the smallest integer in S. This contradiction arose out of original assumption that the Archimedean property did not hold; hence, the proof. QED.

First Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) Whenever the integer k is in S, the next integer $k+1$ is also in S.

Then, S is the set of all positive integers.

Second Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) If k is a positive integer such that $1,2,\ldots k$ belong to S, then $(k+1)$ must also be in S.

Then, S is the set of all positive integers.

So, in lighter vein, we assume a set of positive integers is given just as Kronecker had observed: “God created the natural numbers, all the rest is man-made.”

More later,

Nalin Pithwa.