NoteNextra-origin/content/Math4201/Math4201_L29.md

Math4201 Topology I (Lecture 29)

Compact and connected spaces

Compact spaces

Theorem of uncountable compact Hausdorff spaces without isolated points

Any non-empty compact Hausdorff space X without any isolated points is uncountable.

Proof

By contradiction, let X=\{x_n\}_{n\in\mathbb{N}} be a countable set.

We construct inductively a sequence of open non-empty subspaces \{V_i\} of X such that

\overline{V_1}\supseteq \overline{V_2}\supseteq \overline{V_3}\supseteq \dots

where x_i not in \overline{V_i}.

This could imply that for any j\leq i, x_j is not in \overline{V_j} but \overline{V_j}\supseteq \overline{V_i}. This contradicts the fact that x_j not in \overline{V_i}, that is x_1, ..., x_i are not in \overline{V_i}.

This is a contradiction because \{\overline{V_i}\} satisfies the finite intersection property.

\bigcap_{i=1}^\infty \overline{V_i} \neq \emptyset

But in this case, \bigcap_{i=1}^\infty \overline{V_i} = \emptyset because x_i not in \overline{V_i}.

To construct such \{V_i\}, we can start with V_1,\dots, V_{k-1} are constructed, then there is a point y_k in V_{k-1} which isn't same as x_k.

So x_k not in \overline{V_k} because U_k is an open neighborhood of x_k that don't intersect with V_k.

Since X is Hausdorff, there exists an open neighborhood U_k of x_k and U_y of y_k such that U_k\cap U_y=\emptyset.

Let x_k\in U_k and y_k\in W_k\subseteq X that is open, and U_k\cap W_k=\emptyset.

Let V_k=W_k\cap V_{k-1}. Then this is open and is contained in V_{k-1} since \overline{V_k}\subseteq \overline{V_{k-1}}.

Case 1: x_k\notin V_{k-1}. There is such y_k because V_{k-1} is not empty.

Case 2: x_k\in V_{k-1}. Since x_k is not isolated points, any open neighborhood of x_k including V_{k-1} contains another point.

Therefore, X is uncountable.

Definition of limit point compact

A space X is limit point compact if any infinite subset of X has a limit point.

Definition of sequentially compact

A space X is sequentially compact if any sequence has a convergent subsequence. i.e. If \{x_n\}_{n\in\mathbb{N}} is a sequence in X, then there are n_1<n_2<\dots<n_k<\dots such that \{y_i=x_{n_i}\}_{i\in\mathbb{N}} is convergent.

Theorem of limit point compact spaces

If (X,d) is a metric space, then the following are equivalent:

1. X is compact.
2. X is limit point compact.
3. X is sequentially compact.

Example of limit point compact spaces but not compact

Let X'=\{a,b\} with trivial topology, and X=\mathbb{N}\times X' with the product topology where we use the discrete topology on \mathbb{N}.

X isn't compact because \{\{i\}\times X':i\in\mathbb{N}\} is an open cover of X that doesn't have a finite subcover.

because these open sets are disjoint, X is limit point compact.

Let A\subseteq X be an infinite subset of X. In particular, it contains a point of the form (i,a) or (i,b) for i\in\mathbb{N}. Let (i,a)\in A. Then (i,b) is a limit point of A, since any open neighborhood (\{i\}\times X') of (i,b) contains a point of the form (i,a) or (i,b).

X is not sequentially compact because the sequence \{(n,a)\}_{n\in\mathbb{N}} has no convergent subsequence.

Proof

First, we show that 1. implies 2.

Let X be compact and A\subseteq X be an infinite subset of X that doesn't have any limit points.

Then X-A is open because any x\in X-A isn't in the closure of A otherwise it would be a limit point for A, and hence x has an open neighborhood contained in the complement of A.

Next, let x\in A. Since x isn't a limit point of A, there is an open neighborhood U_x of x in X that U_x\cap A=\{x\}. Now consider the open covering of X given as

\{X-A\}\cup \{U_x:x\in A\}

This is an open cover because either x\in X-A or x\in A and in the latter case, x\in U_x since X is compact, this should have a finite subcover. Any such subcover should contain U_x for any x because U_x is the element in the subcover for x.

This implies that our finite cover contains infinite open sets, which is a contradiction.

---

Continue with the proof that 2. implies 3. next time.

Corollary of compact spaces

1. If X is a compact topological space, then it is limit point compact.
2. If X is a sequentially compact topological space, then it is limit point compact.

Proof

Proof of 1. follows from the theorem of limit point compact spaces.