Understanding Topology Proof

**I-Think** · July 18th 2011, 09:04 AM

Studying on my own currently and having trouble understanding a topology proof.

Theorem: Compact subsets of metric spaces are closed

Proof
Let $K$ be a compact subset of a metric space $X$ . We shall prove that the complement of $K$ is an open subset of $X$
Suppose $p \in{X}$ , $p$ $\notin$ $K$ . If $q\in{K}$ , let $V_q$ and $W_q$ be neighbourhoods of $p$ and $q$ respectively, of radius less than $\frac{1}{2}d(p,q)$
Since $K$ is compact, there are finitely many points $q_1,...,q_n$ in $K$ such that

$K\subset{W_{q_1}\cup...\cup{W_{q_n}}=W$

If $V=V_{q_1}\cap...\cap{V_{q_n}}$ , then $V$ is a neighbourhood of $p$ which does not intersect $W$ . Hence $V\subset{K^c}$ , so that $p$ is an interior point of $K^c$ . The theorem follows.

Question
I am confused as to how $V_q$ and $W_q$ are defined
Is it that
$W_{q_i}:[\forall{p}: d(p,q_i)<\frac{d(p,q)}{2}]$
$V_{q_i}:[\forall{q}: d(p,q_i)<\frac{d(p,q)}{2}]$

Also, how does the chosen radius play a role in the proof?

**hatsoff** · July 18th 2011, 09:35 AM

The idea is that for each point $q\in K$ , there are disjoint (i.e. non-overlapping) neighborhoods $V_q$ and $W_q$ of $p$ and $q$ , respectively. (Notice that $p$ is fixed but $q$ can be any point in $K$ .) In order to ensure they are disjoint, we need to make sure they have a small enough radius. How small? Well if they are no more than half the distance between $p$ and $q$ , that will be small enough. So it doesn't really matter exactly how $V_q,W_q$ are defined as long as they are disjoint neighborhoods. If you like we can define them as you have done. But that's just one possible choice. All we need them to be are disjoint neighborhoods.

And then of course we have $\mathcal{C}=\{W_q:q\in K\}$ an open cover of $K$ . By compactness there is a finite subcover $\mathcal{C}^*$ which is a finite set of the form $\{W_{q_1},\cdots,W_{q_n}\}$ for some $q_1,\cdots,q_n\in K$ . Remember that each $W_{q_i}$ is disjoint from each $V_{q_i}$ . So $V:=\bigcap_{i=1}^n V_{q_i}$ is disjoint from $W:=\bigcup_{i=1}^n W_{q_i}$ (verify this through the algebra of sets), and since $K\subset W$ we have $V\subseteq K^c$ a neighborhood of $p$ . So $p$ is an interior point, and since $p\in K^c$ is arbitrary, all points in $K^c$ are interior---which is equivalent to saying that $K^c$ is open.

EDIT: You have several typos in your OP. For example the theorem should read "compact subsets of metric spaces are closed." And your LaTeX code is off sometimes. Please note that not everyone can see through to what you mean.

**Plato** · July 18th 2011, 09:43 AM

Originally Posted by I-Think

Studying on my own currently and having trouble understanding a topology proof.

Theorem: Compact subsets of metric spaces are compact

There is nothing there to prove. Compact subsets of any space are compact.

There is a theorem that says: Closed subsets of a compact space are compact.

**HallsofIvy** · July 18th 2011, 10:04 AM

There are also "compact subsets of a metric space are bounded" and "compact subsets of a metric space (actually any topological space) are closed." Please go back and reread the problem.

**Opalg** · July 18th 2011, 11:52 AM

The proof obviously shows that compact subsets of metric spaces are closed. So presumably that is what the theorem should be stating.

**I-Think** · July 18th 2011, 05:56 PM

I apologize for typos present. I hope it's more readable now

**Drexel28** · July 18th 2011, 07:25 PM

Originally Posted by HallsofIvy

"compact subsets of a metric space (actually any topological space) are closed."

Well...

.....

**Tinyboss** · July 19th 2011, 07:49 AM

I'd approach this one by contraposition: prove that if a subset S of a metric space is not closed, then S is not compact, by exhibiting a cover with no finite subcover.

Since S isn't closed, there is a point x which is a limit point of S, but not contained in S. So let your open cover be complements of closed 1/n balls around x.

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