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.


2 Answers 2

  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 $X$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.