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.

Comments are most welcome.

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