Understanding Topology Proof

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

Theorem: Compact subsets of metric spaces are closed

Proof

Let $\displaystyle K$ be a compact subset of a metric space $\displaystyle X$. We shall prove that the complement of $\displaystyle K$ is an open subset of $\displaystyle X$

Suppose $\displaystyle p \in{X}$, $\displaystyle p$ $\displaystyle \notin$ $\displaystyle K$. If $\displaystyle q\in{K}$, let $\displaystyle V_q$ and $\displaystyle W_q$ be neighbourhoods of $\displaystyle p$ and $\displaystyle q$ respectively, of radius less than $\displaystyle \frac{1}{2}d(p,q)$

Since $\displaystyle K$ is compact, there are finitely many points $\displaystyle q_1,...,q_n$ in $\displaystyle K$ such that

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

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

Question

I am confused as to how $\displaystyle V_q$ and $\displaystyle W_q$ are defined

Is it that

$\displaystyle W_{q_i}:[\forall{p}: d(p,q_i)<\frac{d(p,q)}{2}]$

$\displaystyle V_{q_i}:[\forall{q}: d(p,q_i)<\frac{d(p,q)}{2}]$

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

Re: Understanding Topology Proof

The idea is that for each point $\displaystyle q\in K$, there are disjoint (i.e. non-overlapping) neighborhoods $\displaystyle V_q$ and $\displaystyle W_q$ of $\displaystyle p$ and $\displaystyle q$, respectively. (Notice that $\displaystyle p$ is fixed but $\displaystyle q$ can be any point in $\displaystyle K$.) 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 $\displaystyle p$ and $\displaystyle q$, that will be small enough. So it doesn't really matter exactly how $\displaystyle V_q,W_q$ 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 $\displaystyle \mathcal{C}=\{W_q:q\in K\}$ an open cover of $\displaystyle K$. By compactness there is a finite subcover $\displaystyle \mathcal{C}^*$ which is a finite set of the form $\displaystyle \{W_{q_1},\cdots,W_{q_n}\}$ for some $\displaystyle q_1,\cdots,q_n\in K$. Remember that each $\displaystyle W_{q_i}$ is disjoint from each $\displaystyle V_{q_i}$. So $\displaystyle V:=\bigcap_{i=1}^n V_{q_i}$ is disjoint from $\displaystyle W:=\bigcup_{i=1}^n W_{q_i}$ (verify this through the algebra of sets), and since $\displaystyle K\subset W$ we have $\displaystyle V\subseteq K^c$ a neighborhood of $\displaystyle p$. So $\displaystyle p$ is an interior point, and since $\displaystyle p\in K^c$ is arbitrary, all points in $\displaystyle K^c$ are interior---which is equivalent to saying that $\displaystyle K^c$ 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.

Re: Understanding Topology Proof

Quote:

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.

Re: Understanding Topology Proof

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.

Re: Understanding Topology Proof

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

Re: Understanding Topology Proof

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

Re: Understanding Topology Proof

Quote:

Originally Posted by

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

Well...(Thinking)(Thinking).....(Itwasntme)

Re: Understanding Topology Proof

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.