Abstract Nonsense

Crushing one theorem at a time

Miscellaneous Rudin Like Topology questions


Problem: Let X be a second countable topological space and U\subseteq X be such that \text{card }U>\aleph_0. Then, \text{card }U\cap D(U)>\aleph_0

Proof: Suppose that D(U)\cap U is countable then U-(U\cap D(U))=U-D(U) is uncountable. But, by definition for each x\in U-D(U) there exists some neighborhood V_x of it such that U\cap V_x=\{x\}. So, clearly \left\{V_x\cap (U-D(U))\right\}_{x\in U-D(U)} is an open cover for U-D(U) and since X is second countable Lindelof’s theorem guarantees that it must have a countable subcover \left\{V_{x_n}\cap (U-D(U))\right\}_{n\in\mathbb{N}}. Thus,

\displaystyle U-D(U)=\bigcup_{n\in\mathbb{N}}\left(V_{x_n}\cap(U-D(U))\right)\subseteq\bigcup_{n\in\mathbb{N}}\left(V_{x_n}\cap U\right)=\left\{x_n:n\in\mathbb{N}\right\}

But this contradicts that U-D(U) is uncountable. The conclusion follows. \blacksquare

Corollary: If f:\mathbb{C}\to\mathbb{C} and \mathcal{S} is the set of all isolated singularities then \text{card }\mathcal{S}\leqslant\aleph_0

Corollary: Let \mathcal{M} be a separable metric space and E\subseteq\mathcal{M} uncountable, then there exists a sequence \{e_n\}_{n\in\mathbb{N}} in E such that e_n\to e\in E.

Problem: Let \mathcal{M} be a metric space and E\subseteq \mathcal{M} perfect and U\subseteq\mathcal{M} open. Then, \overline{E\cap U} is perfect.

Proof: The fact that D(\overline{E\cap U})\subseteq \overline{E\cap U}  follows from the fact that it’s closed. Now, let x\in \overline{E\cap U} and let V be any neighborhood of it. Then, by definition we have that V\cap (E\cap U)\ne \varnothing and so there exists some y\in V\cap E\cap U. Now, since y\in E and V\cap U is a neighborhood of it it follows that there are infinitely many values of E in it. In other words V\cap (U\cap E) is infinite and in particular there exists some y\in V\cap (U\cap E) such that y\ne x. Thus, y\in D(\overline{E\cap U}). The conclusion follows. \blacksquare

Corollary: If \mathcal{M} is a metric space and \varnothing\subsetneq E\subseteq\mathcal{M} is perfect we can choose x\in E and see that \overline{E\cap B_{1}(x)} and  we have a perfect subset of \mathcal{M} which is also bounded.

Problem: Let \mathcal{M} be a compact metric space. Prove that \mathcal{M} is connected if and only if it cannot be written as a union \mathcal{M}=A\cup B with d(A,B)=\inf\left\{d(a,b):a\in A,b\in B\right\}>0

Proof: It is tacitly assumed that A,B\ne\varnothing. So, in any metric space suppose that \mathcal{M}=A\cup B with d(A,B)>0. Remember that \overline{A}=\left\{x\in\mathcal{M}:d(x,A)=0\right\} and thus if x\in B we have that d(x,A)\geqslant d(A,B)>0 and thus x\notin \overline{A}. Thus, \overline{A}\cap B=A\cap\overline{B}=\varnothing (the reverse direction gotten using the exact same logic) from where it follows that \mathcal{M} is disconnected.

Before we continue to prove the converse we prove a nice little lemma

Lemma: Let \mathcal{M} be a compact metric space and E,G\subseteq\mathcal{M} non-empty disjoint closed subspaces of \mathcal{M}. Then, d(E,G)>0.

Proof: Since \mathcal{M} is compact and E,G closed subspaces it follows that they themselves are compact and thus so is E\times G. So, notice that F:E\times G\to\mathbb{R}:(e,g)\mapsto d(e,g) is continuous since it is the distance function (trivially continuous) restricted to E\times G. Thus, by the extreme value theorem it follows that \inf\text{ }F(E\times G)=d(e_0,g_0) for some (e_0,g_0)\in E\times G and since e_0\ne g_0 we see that d(E,G)=d(e_0,g_0)>0. \blacksquare

So, now assume that \mathcal{M} were not connected. Then, \mathcal{M}=E\cup G where E,G are non-empty disjoint closed subsets of \mathcal{M}. But, by the lemma this contradicts the assumption that \mathcal{M} cannot be written as A\cup B with d(A,B)>0. \blacksquare

Problem: Let \mathcal{M} be a separable metric space, then for any subspace \mathcal{N} we have that \mathcal{N} is separable.

Proof: We prove something stronger. But, first we prove a small lemma

Lemma: Let X be a second countable topological space, then any subspace Y is second countable.

Proof: Let \mathfrak{B} be the countable base for X and let \mathfrak{B}_Y=\left\{B\cap Y:B\in\mathfrak{B}\right\}. Clearly \mathfrak{B}_Y is a countable collection of open subsets of Y. Now, let y\in Y be arbitrary and let U be any neighborhood of it. Then, by definition U=V\cap Y for some open set V in X. So, we may find some B\in\mathfrak{B} such that y\in B\subseteq V and thus y\in B\cap Y\subseteq V\cap Y=U and since B\cap Y\in\mathfrak{B}_Y the conclusion follows. \blacksquare

So, if \mathcal{M} is a separable metric space, then as proven earlier it is second countable and thus so is (by the lemma) \mathcal{N} but since every second countable space is trivially separable the conclusion follows. \blacksquare

Problem: Let \mathcal{M} be a compact metric space and \varphi:\mathcal{M}\to\mathcal{M} an isometry. Then, \varphi is surjective.

Proof: Suppose x\in\mathcal{M}-f(\mathcal{M}) and consider \left\{f^n(x)\right\}_{n\in\mathbb{N}}. Since \mathcal{M} is a compact metric space this must have a convergent subsequence \left\{f^{n_m}(x)\right\}_{m\in\mathbb{N}}. In particular, that sequence must be Cauchy. So, noting that \{x\}f(\mathcal{M}) are disjoint non-empty compact subspaces of a compact space it follows by an earlier lemma that d(x,M)=\delta>04. So, since \left\{f^{n_m}(x)\right\}_{m\in\mathbb{N}} is Cauchy there exists M\in\mathbb{N} such that d(f^{n_{M+1}}(x),f^{n_M}(x))<\delta. So,

\delta> d\left(f^{n_{M+1}}(x),f^{n_M}(x)\right)=d\left(f^{n_M}(x),f^{n_{M-1}}(x)\right)=\cdots=d(f(x),x)>\delta

which is a contradiction. \blacksquare

Problem: Let \mathcal{M} be a compact metric space and \varphi:\mathcal{M}\to\mathcal{M} be a function such that

d(\varphi(x),\varphi(y))<d(x,y)

for all distinct x,y\in\mathcal{M}. Prove that \varphi has precisely one fixed point.

Proof: Clearly if there is a fixed point it’s unique because if \varphi(x)=x,\varphi(y)=y,\text{ }x\ne y then d(x,y)=d(\varphi(x),\varphi(y))<d(x,y). Now, to see that it must have a fixed point we note that

\psi:\mathcal{M}\to\mathcal{M}:x\mapsto d(x,\varphi(x))

is continuous since evidently the diagram

commutes. So, since \psi is continuous and \mathcal{M} it follows that \psi obtains a minimum at some x_0\in\mathcal{M}. Now, assume that \varphi(x_0)\ne x_0 then d(\varphi(\varphi(x_0)),\varphi(x_0))<d(\varphi(x_0),x_0) contradicting that \psi achieved it’s minimum at x_0. The conclusion follows. \blacksquare


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

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