Munkres Topology Solution: Chapter 5

§37. Tychonoff Theorem

Ex.37.1

Let $X$ be a space. Let $\mathscr D$ be a collection of subsets of $X$ that is maximal with respect to the finite intersection property.
(a) Show that $x\in \bar D$ for every $D\in\mathscr D$ iff every nbhd of $x$ belongs to $\mathscr D$. Which implication uses maximality of $\mathscr D$?
(b) Let $D\in\mathscr D$. Show that if $A\supset D$, then $A\in\mathscr D$
(c) Show that if $X$ satisfies the $T_1$ axiom, there is at most one point belonging to $\bigcap_{D\in\mathscr D}\bar D$

(a) This is equivalent to saying that $x\in\bigcap_{D\in\mathscr D}\bar D\iff\mathcal N(x)\subset\mathscr D$.

If $x\in\bigcap_{D\in\mathscr D}\bar D$, i.e. $\forall D\in\mathscr D, x\in\bar D$. Then $\forall U\in\mathcal N(x), U\cap D\neq\emptyset$. By maximality of $\mathscr D$, we have $U\in\mathscr D$. Thus we have proven $\mathcal N(x)\subset\mathscr D$.

Conversely, suppose $\mathcal N(x)\subset\mathscr D$, using finite intersection property of $\mathscr D$, we know that $\forall U\in\mathcal N(x)\subset\mathscr D, D\in\mathscr D, U\cap D\neq\emptyset$, which means exactly that $x\in\bar D$.

(b) Let $\mathscr D’=\mathscr D\cup\lbrace A\rbrace$, any finite intersection within $\mathscr D’$ is either a finite intersection within $\mathscr D$ (hence nonempty) or a finite intersection within $\mathscr D$ with $A$: $A\cap D_1\cap…\cap D_n$. Note that $A\cap D_1\cap…\cap D_n\supset D\cap D_1\cap…\cap D_n$, the right side of which is a finite intersection within $\mathscr D$, again nonempty. Therefore $\mathscr D’$ has finite intersection property, and hence equals $\mathcal D$.

(c) THIS IS FALSE. $T_2$ is necessary for (c). A counterexample can be found here.

Suppose there are $x\neq y\in\bigcap_{D\in\mathscr D}\bar D$, by (a) this means $\mathcal N(x),\mathcal N(y)\subset\mathscr D$. Use Hausdorff-ness we get disjoint open sets $U\in\mathcal N(x), V\in\mathcal N(y)$, which contradicts to finite intersection property of $\mathscr D$.

Ex.37.2

A collection $\mathscr A$ of subsets of $X$ has the countable intersection property if every countable intersection of elements of $\mathscr A$ is nonempty. Show that $X$ is a Lindelof space iff for every collection $\mathscr A$ of subsets of $X$ having the countable intersection property, $$\bigcap_{A\in\mathscr A}\bar A$$ is nonempty.

The proof is analogous to the closed set description of compact spaces.

First suppose $X$ is Lindelof, and $\mathscr A$ has countable intersection property. Let $\mathscr B=\lbrace X\setminus \bar A|A\in\mathscr A\rbrace$. This is a family of open sets. Note that the countable intersection property of $\mathscr A$ implies that countable union of $\mathscr B$ cannot be the whole space $X$, i.e. $\mathscr B$ has no countable subcovering. Therefore, provided $X$ is Lindelof, $\mathscr B$ cannot be a covering. $\cup\mathscr B\neq X\iff\bigcap_{A\in\mathscr A}\bar A\neq\emptyset$

Conversely, suppose for every collection $\mathscr A$ of subsets of $X$ having the countable intersection property, $\bigcap_{A\in\mathscr A}\bar A\neq\emptyset$. Let $\mathscr A$ be an open covering of $X$, $\mathscr B=\lbrace X\setminus A|A\in\mathscr A\rbrace$. $\mathscr B$ is a family of closed sets. That $\mathscr A$ is a covering is equivalent to saying $\cap\mathscr B=\emptyset$, thus it cannot have the countable intersection property, i.e. $\exists \lbrace A_n\rbrace_n\subset\mathscr A$, s.t. $\bigcap_n (X\setminus A_n)=\emptyset$. This shows that $\lbrace A_n\rbrace_n$ is a countable subcovering.

Ex.37.3

Consider the three statements:
　　(i) If $X$ is a set and $\mathscr A$ is a collection of subsets of $X$ having th countable intersection property, then there is a collection $\mathscr D$ of subsets of $X$ such that $\mathscr D\supset\mathscr A$ and $\mathscr D$ is maximal w.r.t the countable intersection property.
　　(ii) Suppose $\mathscr D$ is maximal w.r.t. the countable intersection property. Then countable intersections of elements of $\mathscr D$ are in $\mathscr D$. Furthermore, if $A$ is a subset of $X$ that intersects every element of $\mathscr D$, then $A$ is an element of $\mathscr D$.
　　(iii) Products of Lindelof spaces are Lindelof.
(a) Show that (i) and (ii) together imply (iii).
(b) Show that (ii) holds.
(c) Products of Lindelof spaces need not to be Lindelof (see §30). Therefore (i) does not hold. If one attempts to generalize the proof of Lemma 37.1 to the countable intersection property, at what point does the proof break down?

(a) You can prove (a) by directly applying the proof of Theorem 37.3 and change only ‘compact’ to ‘Lindelof’ and ‘finite intersection property’ to ‘countable intersection property’. The fact about compact spaces used there has its corresponding version in Ex.37.2 and Lemma 37.2 has its generalization as (ii), so everything will go through.

(b) This is a direct generalization of Lemma 37.2. Check it yourself.

(c) Recall that $\Bbb R_l$ is Lindelof while $\Bbb R_l^2$ is not (Do you still remember why?). You will break down when you try to find a ‘maximum’ family because this time you have infinite steps to go.

Ex.37.4

Here is another theorem whose proof uses Zorn’s lemma. Recall that if $A$ is a space and if $x,y\in A$, we say that $x,y$ belong to the same quasicomponent of $A$ if there is no separation $A=C\cup D$ of $A$ into two disjoint sets open in $A$ such that $x\in C, y\in D$.
Theorem. Let $X$ be a compact Hausdorff space. Then $x, y$ belong to the same quasicomponent of $X$ iff they belong to the same component of $X$.
(a) Let $\mathscr A$ be a collection of all closed subspaces $A$ of $X$ such that $x, y$ lie in the same quasicomponent of $A$. Let $\mathscr B$ be a subcollection of $\mathscr A$that is simply ordered by proper inclusion. Show that the intersection of the elements of $\mathscr B$ belongs to $A$. [Hint: Compare Exercise 11 of §26.]
(b) Show $\mathscr A$ has a minimal element $D$.
(c) Show $D$ is connected.