Reference: Introductory Real Analysis by Kolmogorov and Fomin.
Reference: Introduction to Analysis by Rosenlicht.
Reference: Analysis by Walter Rudin.
Sec 6.1 Closure of a set. Limit points.
By the open sphere (or open ball) in a metric space R we mean the set of points satisfying the inequality
( is the metric of R.)
Note: Any confusion between “sphere” meant in the sense of spherical surface and “sphere” meant in the sense of a solid sphere (or ball) will always be avoided by judicious use of the adjectives “open” or “closed.”
The fixed point is called the center of the sphere and the number r is called its radius. By the closed sphere (or closed ball) with center and radius r we mean the set of points satisfying the inequality
An open sphere of radius with center will also be called an –neighbourhood of denoted by .
A point is called a contact point of a set if every neighbourhood of x contains at least one point of M. The set of all contact points of a set M is denoted by and is called the closure of M. Obviously, , since every point of M is a contact point of M. By the closure operator in a metric space R, we mean the mapping of R carrying each set into its closure .
The closure operator has the following properties:
- If , then
Property I is obvious.
Proof of property 2:
Let . Then, any given neighbourhood contains a point . Consider the sphere of radius
PS: It helps to draw an illustrative diagram to understand this proof.
Clearly, is contained in . In fact, if then and hence, since , it follows from the triangle inequality that
That is, . Since , there is a point in . But then and hence, , since is an arbitrary neighbourhood of x. Therefore, . But, obviously and hence, , as required.
Proof of property 3:
Let and suppose . Then, and . But then there exist neighbourhoods and such that contains no points of M while contains no points of N. It follows that the neighbourhood , where , contains no points of either M or N, and hence no points of , contrary to the assumption that . Therefore, , and hence,
…call this I
since x is an arbitrary point of . On the other hand, since and , it follows that from property (i) that and . But then,
which together with (i) implies that .
Proof of property 4:
We observe that given any ,
by property (iii).
It follows that . But this is possible for arbitrary M only if . (Alternatively, the set with no elements can have no contact points!). QED.
A point is called a limit point of a set if every neighbourhood of x contains infinitely many points of M. The limit point may or may not belong to M. For example, if M is the set of rational numbers in the interval , then every point of , rational or irrational, is a limit point of M. (It is said that the rationals are dense in R; we will discuss this in detail later).
A point x belonging to a set M is called an isolated point of M if there is a (“sufficiently small”) neighbourhood of x containing no points of M other than x itself.
6.2 Convergence and limits.
A sequence of points in a metric space R is said to converge to a point if every neighbourhood of x contains all points starting from a certain index (more exactly, if, given any , there is an integer such that contains all points with ). The point x is called the limit of the sequence , and we write (as ). Clearly, converges to x if and only if
It is an immediate consequence of the definition of a limit of a sequence that
(1) No sequence can have two distinct limits.
(2) If a sequence converges to a point x, then so does every subsequence of .
(Details are filled in here: I believe it is helpful to write little proofs of little details…I call it playing with small pearls of math…: )
Proof of (1) : Let, if possible, a sequence have two distinct limit points, namely, and . Then, there exists a neigbourhood or open ball which contains all points of the given sequence after some . So also there exists a neighbourhood (again of radius about such that it contains all points of the given sequence after some natural number .
But, we know that . Which in turn means that the two “distinct limit points” lie in the same open ball. Hence, they cannot be distinct. Hence, a sequence can have only one limit point, and it is unique. QED.
Proof of (2):
Definition of subsequence: Given a sequence , where , consider a strictly increasing sequence of positive integers (that is, are positive integers and ) : then the sequence is called a subsequence of the original sequence .
Now, the claim is that: if a given sequence converges to a limit point, so does every subsequence of the given sequence: —-
So, it is given that when for some natural number . Since ( here m is some dummy variable, some appropriate positive integer) for all positive integers m, we have whenever . This means that , which is what we wanted to prove. QED.
A necessary and sufficient condition for a point x to be a contact point of a set M is that there exists a sequence of points of M converging to x.
The condition is necessary since if x is a contact point of M, then every neighbourhood contains at least one point , and these points form a sequence converging to M. The sufficiency is obvious. QED.
A necessary and sufficient condition for a point x to be a limit point of a set M is that there exists a sequence of distinct points of M converging to x.
Clearly, if x is a limit point of M, then the points figuring in the proof of Theorem 2 (just above) can be chosen to be distinct. This proves the necessity and the sufficiency is again obvious. QED.
6.3 Dense subsets, Separable spaces:
Let A and B be two subsets of a metric space R. Then, A is said to be dense in B if . In particular, A is said to be everywhere dense (in R) if . A set A is said to be nowhere dense if it is dense in no (open) sphere at all. (Note; it is w.r.t. to the metric function defined in the space R.) (Famous example: The rationals are dense in ).
The set of all rational points is dense in the real line .
We want to prove that . Clearly, as , the set of rationals are dense in rationals themselves. We only need to show that any irrational number is also a limit point of a sequence of rationals. For example, this is crystal clear: consider the sequence of decimal numbers (finitely many decimals) that we get by extracting square root of 2 by long division. The same applies to any other surd or irrational number. Hence, we can always find a rational sequence of numbers converging to an irrational number. Hence, the rationals are dense in the reals. QED.
The set of all points with rational co-ordinates is dense in each of the spaces , , introduced in the following examples (in earlier blog, and presented here again for handy reference):
2a’) the set of all ordered n-tuples of real numbers with distance is a metric space denoted by and called n-dimensional Euclidean space (or, simply Euclidean n-space).
same proof as example 1. QED.
2b) the set of all ordered n-tuples of real numbers with distance is a metric space denoted by .
same proof as example 1. QED.
2c) The set of all ordered n-tuples of real numbers with distance is a metric space denoted by .
once again same proof as example 1 with minor differences. QED.
The set of all points with only finitely many non-zero co-ordinates, each a rational number, is dense in the space introduced in earlier blog (and presented here again for handy reference):
Definition of metric space: Let be the set of all infinite sequences of real numbers satisfying the convergence condition , where distance between points is defined by .
Again, some detailed steps show that the essence of this proof is also the fact that the rationals are dense in the reals.
Example 4: So, also, similarly, the set of all polynomials with rational coefficients is dense in both spaces and . Let me reproduce the definitions of these spaces: The set of all continuous functions defined on the closed interval with distance is a metric space. The set of all functions continuous on the interval with distance function defined by is also a metric space denoted by .
A metric space is said to be separable if it has a countable everywhere dense subset.
The spaces , , , , and are all separable since the sets in Examples 1 to 4 above are all countable. (this can be easily verified).
The “discrete space” M described in Example 1 (reproduced here again) contains a countable everywhere dense subset and hence is separable if and only it is itself a countable set, since clearly in this case. (Put , when and if . Then, this space M is called the discrete metric space).
There is no countable everywhere dense set in the space m of all bounded sequences, introduced in an earlier blog ( To help recall, we present it here again: Consider the set of all bounded infinite sequences of real numbers , and let . This gives a metric space which we denote by m. The fact that it has the three properties of a metric is almost obvious.)
In fact, consider the set E of all sequences consisting exclusively of zeros and ones. Clearly, E has the power of continuum since there is a one-to-one correspondence between E and the set of all subsets of the set (what is the correspondence here? answer:) Now, the distance between any two points of E equals 1. Suppose we surround each point of E by an open sphere of radius , thereby obtaining an uncountably infinite family of pairwise disjoint spheres. Then, if some set M is everywhere dense in m, there must be at least one point of M in each of the spheres. It follows that M cannot be countable and hence, that m cannot be separable.
We say that a subset M of a metric space R is closed if it coincides with its own closure, that is, if . In other words, a set is called closed if it contains all its limit points (we will solve this problem as an exercise).
The empty set and the whole space R are closed sets.
Every closed interval on the real line is a closed set.
Every closed sphere in a metric space is a closed set. In particular, the set of all functions f in the space such that (where K is a constant) is closed.
The set of all functions f in such that (an open sphere) is not closed. The closure of the set is the closed sphere in the preceding example.
Any set consisting of a finite number of points is closed.
The intersection of arbitrary number of closed sets is closed. The union of a finite number of closed sets is closed.
Given arbitrary sets indexed by a parameter , let x be a limit point of the intersection
Then, any neighbourhood contains infinitely many points of F, and hence, infinitely many points of each . Therefore, x is a limit point of each and hence belongs to each , since the sets are all closed. It follows that and hence that F itself is closed.
Next, let be the union of a finite number of closed sets , and suppose x does not belong to F. Then, x does not belong to any of the sets , and hence, x cannot be a limit point of any of them. But then, for every k, there is a neighbourhood containing no more than a finite number of points of . Choosing
we get a neighbourhoood containing no more than a finite number of points of F, so that x cannot be a limit point of F. This proves that a point cannot be a limit point of F. Therefore, F is closed. QED.
6.5 Open sets.
A point x is called an interior point of a set M if x has a neighbourhood , that is, a neighbourhood consisting entirely of points of M. A set is said to be open if all points are interior points (with respect to that metric).
Every open interval on the real line is an open set. In fact, if , choose . Then, clearly .
Every open sphere in a metric space is an open set. In fact, implies that . Hence, choosing we have .
Let M be the set of of all functions f in such that , where g is a fixed function in . Then, M is an open subset of .
A subset M of a metric space R is open if and only its complement is closed.
Proof of theorem 4:
If M is open, then every point has a neighbourhood (entirely) contained in M. Therefore, no point can be a contact point of . In other words, if x is a contact point of then , that is, is closed.
Conversely, if is closed, then any point must have a neighbourhood contained in M; since, otherwise every neighbourhood of x would contain points of , that is, x would be a contact point of , not in . Therefore, M is open.
corollary to theorem 4:
The empty set and the whole space R are open sets.
An immediate consequence of Theorem 4 and Example 1 in the previous section.
The union of an arbitrary number of open sets is open. The intersection of a finite number of open sets is open.
Proof of theorem 5:
this is the dual of theorem 3. The proof is an immediate consequence of theorem 4 and DeMorgan laws. QED.
6.6 Open and closed sets on the real line:
The structure of open and closed sets in a given metric space can be quite complicated. This is true even for open and closed sets in a Euclidean space of two or more dimensions. In the one-dimensional case, however, it is an easy matter to give a complete description of all open sets (and, hence, of all closed sets).
Every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals.
Proof of theorem 6:
Let x be an arbitrary point of G. By the definition of an open set, there is at least one open interval containing x and contained in G. Let be the union of all such open intervals. Then, as we now show,
is itself an open interval.
In fact, let and where we allow the cases and .
Then, obviously, ….call this II.
Moreover, let us assume that y is an arbitrary point of distinct from x, where, to be explicit, we assume that . Then, there is a point such that . (HW: think, why!) Hence, G contains an open interval containing the points and . But, then this interval also contains y, that is, . (The case is treated similarly. )
Moreover, , by hypothesis. It hence follows that , and hence, by (II) that . Thus, is itself an open interval, as asserted, in fact the open interval .
Next, we want to show that is a union of countable, pairwise, disjoint open intervals. Let us proceed as follows:
By its very construction, the interval is contained in G and is not a subset of a larger interval contained in G. Moreover, it is clear that two intervals and corresponding to distinct points x and either coincide or else are disjoint (otherwise, and would both be contained in a larger interval . There are no more than countably many such pairwise disjoint intervals . In fact, choosing an arbitrary rational point in each we establish a one-to-one correspondence between the intervals and a subset of the rational numbers. Finally, it is obvious that
COROLLARY TO PROOF 6:
Every closed set on the real line can be obtained by deleting a finite or countable system of pairwise disjoint intervals from the line.
Every closed interval is a closed set (here, a and b are necessarily finite).
Every single element set is closed.
The union of a finite number of closed intervals and single element sets is a closed set.
EXAMPLE 4 (The CANTOR Set):
A more interesting example of a closed set on the real line can be constructed as follows:
Delete the open interval from the closed interval and let denote the remaining closed set, consisting of two closed intervals. Then, delete the open intervals and from and let denote the remaining closed set consisting of four closed intervals. Then, delete the “middle third” from each of these four intervals, getting a new closed set , and so on. Continuing this process indefinitely, we get a sequence of closed sets such that
(such a sequence is said to be decreasing). The intersection
of all these sets is called the CANTOR SET.
Clearly, F is closed, and is obtained from the unit interval by deleting a countable number of open intervals. In fact, at the nth stage of construction, we delete intervals, each of length .
To describe the structure of the set we first note that F contains the points, 0,1, , , , , , , ,….call this III.
that is, the end points of the deleted intervals (together with the points 0 and 1)
However, F contains many other points.
In fact, given any , suppose we write x in ternary notation, representing x as a series :
…call this IV.
where each of the numbers is 0, 1 or 2 in radix-3. Then, it is easy to see that x belongs to F if and only if x has a representation (IV) such that none of the numbers equals 1 (you need to think about this as HW!!)
Remarkably enough, the set F has the power of the continuum, that is, there are as many points in F as in the whole interval , despite the fact that the sum of the lengths of the deleted intervals equals
To see this, we associate a new point
with each point IV, where
, if ; ; if .
In this way, we set up a one-to-one correspondence between F and the whole interval . It follows that F has the power of the continuum, as asserted. Let be the set of points III. Then,
where the set is uncountable, since is countable and F itself is not. The points of are often called “points of F of the first kind,” while those of are called “points of the second kind.”