# Solutions to System of Sets:

Reference: Introductory Real Analysis, Kolmogorov and Fomin. Dover Publications. Translated and edited by Richard A. Silverman:

Problem 1:

Let X be an uncountable set, and let $\mathscr{R}$ be the ring consisting of all finite subsets of X and their complements. Is $\mathscr{R}$ a $\sigma$-ring?

Solution 1:

Recall the definition: A ring of sets is called $\sigma$-ring if it contains the union $S= \bigcup_{n=1}^{\infty}A_{n}$ whenever it contains the sets $A_{1}, A_{2}, \ldots, A_{n}, \ldots$.

There are countably many finite subsets of X (given uncountable set) and their complements (also especially) so that the union of all these is the whole set X. Hence, X is a $\sigma$-ring as it is closed under countably many set unions. QED.

Problem 2:

Are open intervals Borel sets?

Solution 2:

Recall the following:

(1) Borel sets or B-sets are the subsets of the real line belonging to the minimal B-algebra  generated by the set of all closed intervals $[\alpha, \beta]$.

Problem 3:But a B-algebra is also a ring and is closed under differences/complements. Open intervals are complements of closed intervals. And, they belong hence to B-algebra. Note we still have to answer the question whether open sets are Borel sets and so we have to check if they are contained in a minimal B-algebra. As per the following theorem: Given any non-empty system of sets $\mathscr{S}$, there is a unique irreducible (minimal) B-algebra $\mathscr{B}(\mathscr{P})$ generated by the system $\mathscr{S}$ or the Borel closure of $\mathscr{P}$. So, yes indeed open sets are Borel sets. QED.

Problem 3A:

Let $y=f(x)$ be a function defined on a set M and taking values in a set N. Let $\mathscr{M}$ be a system of subsets of M, and let $f(\mathscr{M})$ denote the system of all images $f(A)$ of sets $A \in \mathscr{M}$. Moreover, let $\mathscr{N}$ be a system of subsets of N, and let $f^{-1}(\mathscr{N})$ denote the system of all preimages $f^{-1}(B)$ of sets $B \in \mathscr{N}$. Prove that:

i) If $\mathscr{N}$ is a ring, so is $f^{-1}(\mathscr{N})$.

ii) If $\mathscr{N}$ is an algebra, so is $f^{-1}(\mathscr{N})$

iii) If $\mathscr{N}$ is a B-algebra, so is $f^{-1}(\mathscr{N})$

iv) $\mathscr{R}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{R}(\mathscr{N}))$ where $\mathscr{R}$ stands for a ring.

v) $\mathscr{B}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{B}(\mathscr{N}))$ where $\mathscr{B}$ is irreducible or minimum Borel algebra generated by a non-empty system of sets $\mathscr{S}$.

Problem 3B:

Prove the following:

i) If $\mathscr{M}$ is a ring, so is $f(\mathscr{M})$.

ii) If $\mathscr{M}$ is an algebra, so is $f(\mathscr{M})$.

iii) If $\mathscr{M}$ is Borel-algebra, so is $f(\mathscr{M})$.

iv) $\mathscr{R}(f(\mathscr{M}))=f(\mathscr{R}(\mathscr{M}))$

v) $\mathscr{B}(f(M))=f(\mathscr{B}(\mathscr{M}))$ where $\mathscr{B}$ is irreducible minimum Borel algebra generated by a non-empty system of sets $\mathscr{S}$.

Solution 3A:

Part 1: Given mapping $f: M \rightarrow N$, that is, $y=f(x)$, Let $Y_{1}, Y_{2}, Y_{3}, \ldots$ be subsets of N and let $X_{1}, X_{2}, X_{3}, \ldots$ be subsets of M. Then, as it is given that $\mathscr{N}$ is a ring then :

$Y_{1} \bigcap Y_{2} \in \mathscr{N}$

$Y_{1}\bigcup Y_{2} \in \mathscr{N}$

$Y_{1}-Y_{2} \in \mathscr{N}$

$Y_{1} \triangle Y_{2} = (Y_{1}-Y_{2}) \bigcup (Y_{2}-Y_{1}) \in \mathscr{N}$

The above implies:

$f^{-1}(Y_{1}\bigcap Y_{2}) \in f^{-1}(\mathscr{N})$, that is, $f^{-1}(Y_{1}) \bigcap f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N})$, that is, $X_{1} \bigcap X_{2} \in \mathscr{M}$.

Also, $Y_{1} \bigcup Y_{2} \in \mathscr{N}$ so that $f^{-1}(Y_{1} \bigcup Y_{2}) \in f^{-1}(\mathscr{N})$, that is, $f^{-1}(Y_{1}) \bigcup f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N})$, that is, we conclude $X_{1} \bigcup X_{2} \in \mathscr{M}$.

Also, $Y_{1}-Y_{2} \in \mathscr{N}$ as $\mathscr{N}$ is a ring; so that $f^{-1}(Y_{1}-Y_{2}) \in f^{-1}(\mathscr{N})$ so that for some $X_{0}$, which is $f^{-1}(Y_{1}-Y_{0}) \longrightarrow f^{-1}(Y_{1}) - f^{-1}(Y_{0}) \in \mathscr{M}$

So, we have shown that the system of sets $\mathscr{M}$ is closed under set union, set intersection, set difference, and hence, also set symmetric difference. Hence, it is is a ring. QED.

Part ii:

We have already shown that $f^{-1}(\mathscr{N})$ is a ring, now $\mathscr{N}$ is also an algebra meaning a ring of sets with a unit element. Let it contain a unit E $\in \mathscr{N}$ such that if $Y_{0}$ is any set belonging to $\mathscr{N}$, then $Y_{0} \bigcap E = Y_{0}$. so that $f^{-1}(Y_{0} \bigcap E) = f^{-1}(Y_{0})$, hence, $f^{-1}(Y_{0}) \bigcap f^{-1}(E) = f^{-1}(Y_{0})$, which in turn implies that $f^{-1}(E)$ is a unit of $f^{-1}(\mathscr{N})$. Hence, $f^{-1}(\mathscr{N})$ is a ring of sets with a unit element, or in other words, an algebra of sets.

Part iii:

To prove that: if $\mathscr{N}$ is a Borel algebra, so is $f^{-1}(\mathscr{N})$. From parts i and ii above, $f^{-1}(\mathscr{N})$ is a ring with a unit or in other words, $f^{-1}(\mathscr{N})$ is an algebra. Moreover, if it contains $\bigcup_{k=1}^{\infty}Y_{k}$ whenever it contains the sets $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}, \ldots$ all of which are members of $\mathscr{N}$.

But for an arbitrary (finite or infinite) collection of sets:

$f^{-1}(\bigcup_{\alpha}A_{\alpha}) = \bigcup_{\alpha}f^{-1}(A_{\alpha})$

Now, $\bigcup_{k=1}^{\infty}Y_{k} \in \mathscr{N}$

hence $f^{-1}(\bigcup_{k=1}^{\infty}Y_{k}) \in f^{-1}(\mathscr{N})$

Consequently, $\bigcup_{k=1}^{\infty}f^{-1}(Y_{k}) \in f^{-1}(\mathscr{N})$, which in turn implies that $\mathscr{N}$ is a Borel algebra also. QED.

Part iv and part v:

Using parts i and iii above, these can be very easily proved.

Solutions 3b:

i) Prove: If $\mathscr{M}$ is a ring, so is $f(\mathscr{M})$. Note that $f: M \rightarrow N$, $y=f(x)$. Proof (i): As $\mathscr{M}$ is a system of subsets of M. and it is a ring, so if $A_{1} \in \mathscr{M}$, and $A_{2} \in \mathscr{M}$, it implies that $A_{1} \bigcup A_{2} \in \mathscr{M}$; $A_{1} \bigcap A_{2} \in \mathscr{M}$; $(A_{1}-A_{2}) \in \mathscr{M}$. We know that $f(A_{1}\bigcup A_{2}) = f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M})$ and $f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$. And, quite obviously, $f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$ so that it is true that if $\mathscr{M}$ is a ring, then $f(\mathscr{M})$ is also a ring.

ii) Check if it is true or false: if $\mathscr{M}$ is an algebra, so is $f(\mathscr{M})$. Argument: As $\mathscr{M}$ is an algebra, it is already a ring. But additionally, $\mathscr{M}$ contains an element E such that $A_{1} \bigcap E= A_{1}$ (whenever $A_{1} \in \mathscr{M}$ but $f(A_{1} \bigcap E) \subset f(A_{1}) \bigcap f(E) \in f(\mathscr{M})$ but it need not be true that $f(E)$ is a unit element of $f(\mathscr{M})$. So, we can say that in case $\mathscr{M}$ is an algebra, then $f(\mathscr{M})$ may or may not be an algebra. QED.

iii) Check if it is true or false: If $\mathscr{M}$ is a Borel algebra, then so also $f(\mathscr{M})$ is also a Borel algebra. Argument: As in previous case, $f(\mathscr{M})$ may not have a unit element. So, $f(\mathscr{M})$ need not be a Borel algebra. QED.

iv) $\mathscr{R}(f(\mathscr{M})) = f(\mathscr{R}(\mathscr{M}))$. Check if this is true or false. Argument: Let $f(A_{1}) \in f(\mathscr{M})$, $f(A_{2}) \in \mathscr{M}$. By definition of a ring, $f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M})$, $f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$, $f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$, hence we get the following: $f(A_{1} \bigcup A_{2}) \subset f(\mathscr{M})$; $f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$ and $f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$. This in turn means that $A_{1} \bigcup A_{2} \in \mathscr{M}$, $A_{1} \bigcap A_{2} \in \mathscr{M}$ and $A_{1} - A_{2} \in \mathscr{M}$. Hence, $LHS \subset RHS$. Similarly, we can show that $RHS \subset LHS$. QED.

v) $\mathscr{B}(f(\mathscr(M))) = f(\mathscr{B}(\mathscr{M}))$. Check if this is true or false. (We will discuss this later; meanwhile, if you wish, please try as homework and let me know).

Cheers,

Nalin Pithwa.