Abstract Nonsense

Crushing one theorem at a time

Just For Fun(Rudin’s Topology Section) Part IV


28.

Problem: Prove that ever closed set in a separable metric space \mathcal{M}  is the union of a (possibly empty) perfect set and a set which is a set which is countable.

Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if E\subseteq\mathcal{M} is closed we surely have that E\subseteq \mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right). Thus, E=\left(E\cap\mathfrak{C}\right)\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right) but since \mathfrak{C}\subseteq D(E) and D(E)\subseteq E we have then that E=\mathfrak{C}\cup\left(\left(\mathbb{R}-\mathfrak{C}\right)\cap E\right) which is the union of a perfect and countable set respectively. \blacksquare

29.

Problem: Prove that every open set U\subseteq\mathbb{R} may be written as the disjoint union of countably many open intervals.

Proof: We need to prove a quick lemma

Lemma: Let X be a topological space and let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be a class of connected subspace of X such that U_{\alpha_1}\cap U_{\alpha_2}\ne\varnothing,\text{ }\alpha_1,\alpha_2\in\mathcal{A} then \displaystyle \Lambda=\bigcup_{\alpha\in\mathcal{A}} is a connected subspace of X.

Proof: Suppose that \left(E\cap \Lambda\right)\cup\left(G\cap \Lambda\right)=\Lambda is a separation of \Lambda. We may assume WLOG that U_{\alpha_0}\cap E\ne\varnothing for some \alpha_0\in\mathcal{A}. So, now we see that U_{\alpha_0}\subseteq E\cap \Lambda otherwise E\cap U_{\alpha_0},G\cap U_{\alpha_0} would be non-empty disjoint subsets of U_{\alpha_0} whose union is U_{\alpha_0} contradicting that U_{\alpha_0} is connected. Thus, it easily follows that for any  U_\alpha we have that U_\alpha\cap U_{\alpha_0}\ne\varnothing so that U_{\alpha}\cap E\ne\varnothing and thus by a similar reasoning we see that U_\alpha\subseteq E\cap\Lambda. Thus, since \alpha was arbitrary it follows that \Lambda\subseteq E\cap\Lambda contradicting that G\cap\Lambda\ne\varnothing. \blacksquare

So, now for each x\in U define

\mathcal{C}(x)=\left\{V\subseteq U: V\text{ is open in }U,\text{ }V\text{ is connected },\text{ and }x\in V\right\}

And let \displaystyle C(x)=\bigcup_{V\in\mathcal{C}(x)}V and finally we prove that

\Omega=\left\{C(x):x\in U\right\}

is a countable class of disjoint open intervals whose union is U. The fact that each C(x) is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of \mathcal{C}(x) contains x) it is also an open connected subspace of \mathbb{R} (note that each element of \mathcal{C}(x) is open in U but since U is open it is also open in \mathbb{R}. But, it was proven in the book the only connected subspace of \mathbb{R} are intervals and thus C(x) is an interval for each C(x)\in\Omega.

Now, to see that they are disjoint we show that if C(x)\cap C(y)\ne\varnothing then C(x)=C(y) from where the conclusion will follow. So, to see this we first note that if C(x)\cap C(y) is non-empty then C(x)\cup C(y) is an open connected subspace of U containing both x and y and thus C(x)\cup C(y)\subseteq C(x) and C(x)\cup C(y)\subseteq C(y) but this implies that C(y)\subseteq C(x) and C(x)\subseteq C(y) respectively from where the conclusion follows.

Now, to see that \Omega is countable we notice that for each \Omega we have that C(x)\cap\mathbb{Q}\ne\varnothing and so if we let F\left(C(x)\right) be a fixed but arbitrary q\in C(x)\cap\mathbb{Q} then

F:\Omega\to\mathbb{Q}:C(x)\mapsto F\left(C(x)\right)

is an injection since the elements of \Omega are pairwise disjoint. The fact that \Omega is countable follows immediately.

Thus, \Omega is a countable collection of open intervals and

\displaystyle U=\coprod_{C(x)\in\Omega}C(x)

Thus, the conclusion follows. \blacksquare

30.

Problem: Prove that if \displaystyle \mathbb{R}^n=\bigcup_{n\in\mathbb{N}}F_n where each F_n is a closed subset of \mathbb{R}^n then at least one F_n has non-empty interior.

Proof:

Lemma: Let \mathcal{M} be a complete metric space and \left\{K_n\right\}_{n\in\mathbb{N}} a descending sequence of non-empty closed subsets of \mathcal{M} such that \text{diam }K_n\to 0. Then, \displaystyle \bigcap_{n\in\mathbb{N}}K_n contains one point.

Proof: Clearly \displaystyle \bigcap_{n\in\mathbb{N}}K_n does not contain more than one point since \text{diam }K_n\to 0. So, now for each n\in\mathbb{N} choose some x_n\in K_n and let P=\left\{x_n:n\in\mathbb{N}\right\}. Now, if P were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that P is infinite. But, since \text{diam }K_n\to 0 it is evident that \left\{x_n\right\}_{n\in\mathbb{N}} is a Cauchy sequence and thus by assumption it converges to some point x\in\mathcal{M}. We claim that \displaystyle x\in\bigcap_{n\in\mathbb{N}}K_n. To see this we note similarly to problem 26. that since P is infinite it is easy to see that x is a limit point of P and so if x\notin K_{n_0} for some n_0\in\mathbb{N} then \mathcal{M}-K_{n_0} is a neighborhood of x containing only finitely many points of P which clearly contradicts that it is a limit point. The conclusion follows. \blacksquare

So, now suppose each F_n had empty interior (i.e. nowhere dense) . Then, since \mathcal{M} is open and F_1 is nowhere dense there exists an open ball B_1 of radius less than one such that B_1\cap F_1=\varnothing. Let E_1  be the concentric closed ball of B_1 whose radius is half that of B_1. Since F_2 is nowhere dense E_1^{\circ} contains an open ball B_2 of radius less than one-half which is disjoint from F_2.  Let E_2 be the concentric closed ball of B_2 whose radius is one-half that of B_2. Since F_3 is nowhere dense we have that E_2^{\circ} contains an open ball B_3 of radius less than one-fourth which is disjoint from F_3. Let E_3 be the concentric closed ball of B_3 whose radius is half that of E_3. Continuing in this we get a descending sequence of non-empty closed subsets \left\{E_n\right\} for which \text{diam }E_n\to 0. Thus, by our lemma we have that there is some \displaystyle x\in\bigcap_{n\in\mathbb{N}}E_n. This point is clearly not in any of the F_n‘s from where the conclusion follows. \blacksquare

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May 14, 2010 - Posted by | Analysis, Fun Problems, Topology, Uncategorized | , , , , , , , ,

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