Baby Rudin Theorem 2.34.

Theorem 2.34 Compact subsets of a metric space are closed.

Proof.

Suppose $K\subseteq X$, $K$ compact. Let $p\in K^c$, $q\in K$. Let $V_q,W_q$ be neighborhoods of $p$ and $q$ with radius less than $\frac 1 2 d(p,q)$.

Since $K$ is compact, we have $K\subseteq W_{q_1}\cup \cdots\cup W_{q_n}=W$ for some $q_1,...,q_n\in K$.

If V=$V_{q_1}\cap \cdots V_{q_n}$, $V$ is a neighborhood of p which does not intersect $W$, then $V\subseteq K^c$ so $p$ is an interior point of $K^c$. QED.

My understading of this proof is that $V$ is actually $$V= \text {'$\min$'}\left\{V_{q_i}\right\}$$

Where the minimum is to be taken in terms of inclusion.

Also, we have that $V$ doesn't intersect $W$ because $V_{q_i}$ doesn't intersect $W_{q_i}$ for every $i$, right (How do I prove that)?

I'm uploading a picture of my understanding of the situation, am I understanding this correctly?

Solution 1:

Since each $V_{q}$ is a neighborhood of $p$, any finite intersection of them is again a neighborhood of $p$; yes, $V=V_{q_1}\cap\dots\cap V_{q_n}$ is a ball with the radius which is the minimum, but it's mostly irrelevant: the important fact is that it is a neighborhood of $p$.

Now, since $V\subseteq V_{q_i}$, we have by construction that $$ V\cap W_{q_i}\subseteq V_{q_i}\cap W_{q_i}=\emptyset $$ and therefore $$ V\cap W=V\cap(W_{q_1}\cup\dots\cup W_{q_n})= (V\cap W_{q_1})\cup\dots\cup(V\cap W_{q_n})=\emptyset $$ Hence $V\subseteq W^c\subseteq K^c$.

Solution 2:

I believe here you mean $W=W_{q_1}\cup\ldots\cup W_{q_n}$. So, let $x\in V$, for each $1\le k\le n$, $$\begin{align} x\in V & \implies x\in V_{q_k}\\ & \implies x\notin W_{q_k}\\ & \implies x\notin W. \end{align}$$

On the other hand, if $x\in W$, $$\begin{align} x\in W & \implies x\in W_{q_k},\quad\text{for some }1\le k\ne n\\ & \implies x\notin V_{q_k}\\ & \implies x\notin V. \end{align}$$

Solution 3:

$V$ doesn't intersect $W$ because $V$ doesn't intersect $W_{q_i}$ for every $i$. I think that is the same as what you said:

"Also, we have that $V$ doesn't intersect $W$ because $V_{q_i}$ doesn't intersect $W_{q_i}$ for every $i$, right (How do I prove that)?"

Perhaps this follows simply from the fact that $V$ has radius less than $\frac{1}{2}d(p,q_i), \forall i$, and $W$ is at most less than $\frac{1}{2}d(p,q_i)$ away from any $q_i$.

If that does not lead to a simple, direct proof, then I think perhaps you can prove this BWOC (if they did intersect, then some $W_{q_i}$ would need to intersect every $V_{q_j}$, but it can't intersect the $V_{q_i}$ by definition).

Also, I believe you are correct about $V$, the intersection of the neighborhoods, simply being the smallest neighborhood (since they are all neighborhoods of the same point, they different only in radius).

Lastly, your picture seems correct to me.

Solution 4:

Yeah, $ V $ is the smallest set from $V_{q_i}$ in the sense that $ x\in V $ if $ d(x,p) < min\{d(p,q_i)\}$ as $ q_i $ are finite the minimum exist, call it $ d(p,q_m) $. The point here is to show that every $ x\in V $ is not in $ W_{q_i} $ for every $ i={1,...,n} $, that is, if $ x\in V $ then $ d(x,q_i)\geq\frac{1}{2}d(p,q_i) $.

1. $ d(p,q_i)\leq d(p,x)+d(x,q_i) $, by metric space definition.
2. $ d(x,p)<\frac{1}{2}d(p,q_m)\leq \frac{1}{2}d(p,q_i) $.

From (2) $ -\frac{1}{2}d(p,q_i)<-d(x,p) $, then adding this to (1) we get

$ d(p,q_i)-\frac{1}{2}d(p,q_i)< d(p,x)-d(p,x)+d(x,q_i) \rightarrow \frac{1}{2}d(p,q_i)<d(x,q_i)$ for any $i=\{1,...,n\}$.