# Prove Theorem 7.8: Let $X$ be a Hausdorff topological space and $A$ be compact in $X$. Then $A$ is closed in $X$.

My book gave this proof, but I am trying to understand some of the reasoning behind each of the steps. After each step I have numbers in brackets, denoting the places I don't understand the reasoning. I have provided some definitions, but am still trying to understand this proof.

Hausdorff: A topological space $$X$$ is Hausdorff if for every pair of distinct points $$x$$ and $$y$$ in $$X$$, there exists disjoint neighborhoods $$U$$ and $$V$$ of $$x$$ and $$y$$ respectfully.

7.1: Let $$A$$ be a subset of a topological space $$X$$ and let $$O$$ be a collection of subsets of $$X$$.

Cover: The collection $$O$$ is said to cover $$A$$ or to be a cover of $$A$$ if $$A$$ is contained in the union of the the sets in $$O$$

Open Cover: If $$O$$ covers $$A$$, and each set in $$O$$ is open, then we call $$O$$ and open cover of $$A$$.

Subcover: If $$O$$ covers $$A$$, and $$O^\prime$$ is a subcollection of $$O$$ that also covers $$A$$, then $$O^\prime$$ is called a subcover of $$O$$

Compact: A topological space $$X$$ is compact if every open cover of $$X$$ has a finite subcover.

Start of Proof:

Let $$A$$ be compact in the Hausdorff space X. [Given]

Want to show that $$X - A$$ is open [If the complement is open, the space is closed - Definition of closed]

Thus, let $$x \in X-A$$ be arbitrary [if $$x$$ is arbitrary true for any point]

Now, we need to show that there is an open set $$U$$ such that $$x \in U \subset X-A$$ [Theorem 1.4]

Since $$X$$ is Hausdorff, we know that for each $$a \in A$$, there exists disjoint open sets $$U_a$$ and $$V_a$$ such that $$x \in U_a$$ and $$a \in V_a$$. [Definition of Hausdorff]

Then $$O = \{V_a\}_{a \in A}$$ is an open cover of $$A$$. [1]

Because $$A$$ is compact [Given]

There is a finite subcover $$\{V_{a_1},...,V_{a_n}\}$$ of $$O$$. [If A is compact every open cover has a finite subcover by definition]

Let $$V = \bigcup^n_{i=1} V_{a_i}$$ [2]

and $$U = \bigcap^n_i=1 U_{a_i}$$ [3]

Then $$U$$ and $$V$$ are open sets such that $$A \subset V$$ and $$x \in U$$. [4]

Furthermore, since $$U_{a_i}$$ and $$V_{a_i}$$ are disjoint for each $$i$$, it follows that $$U$$ and $$V$$ are disjoint as well. [5]

Thus $$U$$ and $$A$$ are disjoint [6]

and therefore there exists an open set $$U$$ such that $$x \in U \subset X-A$$. [7]

Hence $$X-A$$ is open, implying that $$A$$ is closed.

1. The $$V_a$$ are open for each $$a \in A$$ (that's how it's chosen in the application of Hausdorffness) and each $$a \in A$$ is covered by "its own" $$V_a$$, so it's clear that $$A \subseteq \bigcup \{V_a\mid a \in A\}$$ which is what it means to cover $$A$$.

2. is the just reformulating (using some notations) that $$\{V_a\mid a \in A\}$$ has a finite subcover. So finitely many $$V_a$$ can be found (for diffrent $$a \in A$$ so that still $$A$$ is a subset of their union. The proof just gives a name to the $$a$$ whose $$V_a$$ are used in this finite subcover: there are $$a_1, a_2, \ldots a_n$$ (so $$n\in \Bbb N$$ is the number of sets in the finite subcover) so that $$A \subseteq \bigcup_{i=1}^n V_{a_i} = \bigcup \{V_{a_i}\mid i=1,2,\ldots n\}$$. The name $$V$$ is introduced for this finite union.

3. For those sam $$a_1,a_2,\ldots,a _n$$ we have corresponding $$U_{a_i}$$ which are their disjoint counterparts that all contain $$x$$ by construction. ($$x$$ is the fixed arbitary point outside $$A$$ that we're working with in this part of the proof), and so $$U:= \bigcap_{i=1}^n U_{a_i}$$ is a finite (this is essential) intersection of open sets that all contain $$x$$ and so $$x \in U$$ and $$U$$ is open.

4. That $$A \subseteq V$$ is already part of the 2 and the definition of $$V$$ as the union of the finite subcover. $$x \in U$$ I noted in 3 already. No new info.

5. $$U \cap V = \emptyset$$ is the whole point: If $$p \in U \cap V$$ then $$p \in V_{a_j}$$ for some $$1 \le j \le n$$ (definition of union) and also $$p \in U_{a_j}$$ for that some $$j$$ (as $$p$$ is in the intersection of $$U_{a_i}$$ ,so in all of them). Contradiction, as these sets $$U_{a_j}$$ and $$V){a_j}$$ are the $$U_a$$ and $$V_a$$ sets (for $$a=a_j$$) that were chosen disjointly by Hausdorffness! So no such $$p \in U \cap V$$ can exist.

6. As $$A$$ is a subset of $$V$$, $$U$$ and $$A$$ are certainly disjoint too.

7. So reformulated: $$x \in U \subseteq X-A$$ (two sets are disjoint iff one is a subset of the compelment of the other, a simple logical reformulation).

As each point of $$X-A$$ is an interior point of $$X-A$$, $$X-A$$ is open and is $$A$$ is closed.

Here's some guidance to each of your questions. Hopefully this helps!

1. For all $$a \in A$$, there exist disjoint open sets $$V_{a} \ni a$$ and $$U_{a} \ni x$$. Since $$A = \cup_{a \in A} \{a\} \subseteq \cup_{a \in A} V_{a}$$ is a union of open sets that covers $$A$$, it is an open cover.
2. Since $$A$$ is compact, all open covers (including this one) admit a finite subcover $$\{V_{a_{j}}\}_{j=1}^{n}$$ (that is, $$A \subseteq \cup_{j=1}^{n} V_{a_{j}} = V$$) for some $$\{a_{j} \}_{j=1}^{n} \subseteq A$$.
3. Note that $$U = \cap_{j=1}^{n} U_{a_{j}}$$ is a finite intersection of open sets containing $$\{x \}$$ and so $$x \in U \neq \emptyset$$ is open.
4. Since $$V$$ is an open cover of $$A$$, $$A \subseteq V$$. Since $$x \in U_{a_{j}}$$ for each $$a_{j}$$, it follows that $$x \in U$$.
5. The set $$U$$ is disjoint from $$V$$ since $$V_{a_{j}} \cap U \subseteq V_{a_{j}} \cap U_{a_{j}} = \emptyset$$ for each $$a_{j}$$. Note that $$V \cap U = \cup_{j=1}^{n} (V_{a_{j}} \cap U) = \emptyset$$. This is just de Morgan.
6. Finally $$A \cap U \subseteq V \cap U = \emptyset$$ and so $$A, U$$ are disjoint.
7. For $$x \in X \setminus A$$, $$\exists U \ni x$$ such that $$U \subseteq X \setminus A$$ (or equivalently $$U \cap A = \emptyset$$).
8. Hence, since $$x$$ is arbitrarily chosen from $$X \setminus A$$, it follows that $$X \setminus A$$ is open and $$A$$ is closed in $$X$$.