Nuances of the “compact set is closed” proof

Question

smart guys! I am studying Mathematical Analysis by Rudin and got stuck with the proof “compact set is closed”. The particular thing that puzzles me is: There is an intersection of a finite number of neighborhoods of q within 1/2d(p,q). I do not get why this intersection exists, or to put it differently, cannot be a null set, why it should necessarily contain elements. Or it does not distort the logic of the proof in general? Thanks for any help in advance.

Answer 1

If $U_1, \dots, U_n$ are neighborhoods of $q$, there exists $r_1, \dots, r_n >0$ such that for all $k$ we have $B(q,r_k) \subset U_k$. Therefore, $$ \bigcap_{k=1}^n B(q,r_k) \subset \bigcap_{k=1}^n U_k $$

but $\bigcap_{k=1}^n B(q,r_k) = B(q,r)$ where $r = \displaystyle \min_{1 \leq k\leq n} r_k > 0$ so $\bigcap_{k=1}^n U_k$ is a neighborhood of $q$.

Does that answer your question?

Answer 2

Rudin is taking it for granted that $U_{q_i}= B_{r_i}(p) = \{x\in X| d(p,x)< r_i\}$ are neighborhoods centered at $p$[1] then any finite intersection $U_{q_1} \cap U_{q_2} \cap ... \cap U_{q_n}= B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)$ will itself be a neighborhood centered at $p$ with a radius equal to the least radius of $\{r_1,..., r_n\}$. That is $ B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)=B_{\min r_i}(p)$ and that $B_{\min r_i}(p)$ is not empty because it contains $p$.

It's easy to prove this claim.

If $\{r_1,....,r_n\}$ are a finite set of strictly positive values then there exist some least element, call it $\min r$.

If $x \in B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)$ then $x \in B_{r_i}(p)$ for all $r_i$ and in particular $x \in B_{\min r}(p)$ so $B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)\subset B_{\min r}(p)$.

And if $x\in B_{\min r}(p)$ then $d(x,p)< \min r \le r_i$ for all $r_i$ so $x\in B_{r_i}(p)$ for all $B_{r_i}(p)$ so $x \in B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)$. So $B_{\min r}(p) \subset B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)$.

So $B_{r_1}(p) \cap B_{r_2} (p)\cap .....\cap B_{r_n} (p)= B_{\min r}(p)$.

[1]Rudin defines "neighborhood" as an "open ball" with a center point $p$ and a radius $r_i$ and so the "neighborhood" at $p$ with radius, $r$, is $\{x\in X|d(x,p) < r\}$.

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Rudin's actual proof is subtle.

Let $K$ be a compact set. He proves $K$ is closed be proving $K^c$ is open.

He lets $p \not \in K$.

Now for any $q\in K$ then $d(q,p) > 0$ and we can find a radius $r_q < \frac 12d(q,p)$ and let $W_q = B_{r_q}(q)$. If we do this for all $q\in K$ then $q \in W_q$ and so $K \subset \cup_{q\in K} W_q$. So $\{W_q|q\in K\}$ is an open cover of $K$.

$K$ is compact so it has a finite subcover $W_{q_1},W_{q_2}...., W_{q_n}$ and $K \subset \cup W_{q_i} = W$.

For these finite number of points $q_i$ we can find $U_{q_i}$ that are neighborhoods of $p$ with radius $r_{q_i} < \frac 12 d(q_i, p)$.

Then the intersection of the $U_{q_i}$ is $B_{\min r_{q_i}} (p)$. If $y\in K \subset W=\subset W_{q_i}$ then $y \in W_{q_j}$ for some $W_{q_j}$ sp $d(q_j,y) < r_j < \frac 12 d(q_j,p)$

But then $d(q_j,y) + d(y,p) \ge d(q_j, p)$ so $d(y,p) \ge d(q_j,p)-d(q_j,y) \ge d(q_j,p)-d(q_j,p)=\frac 12d(d_j,p) > r_j \ge \min r_{q_i}$.

So $y \not \in B_{\min r_{q_i}}(p)$ for any $y \in K$

Co $B_{\min r_{q_i}}(p) \subset K^c$.

So $K^c$ is open.

So $K$ is closed.

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Answer 3

I believe that this is a theorem where using the metric is more confusing than helpful.

The key point is that a metric space is Hausdorff. Suppose $C$ is a compact subset of the Hausdorff topological space $X$ and suppose $x\notin C$.

For each $y\in C$, choose an open set $U_y$ and an open set $V_x$ such that

1. $y\in U_y$;
2. $x\in V_y$;
3. $U_y\cap V_y$.

This is guaranteed by the Hausdorff hypothesis.

Then $(U_y)_{y\in C}$ is an open cover of $C$ and therefore we have $$ C\subseteq \bigcup_{k=1}^n U_{y_k} $$ for some $y_1,y_2,\dots,y_n\in C$. Then $$ V=\bigcap_{k=1}^n V_{y_k} $$ is an open neighborhood of $x$ and $V\cap C=\emptyset$. Therefore $x\notin\bar{C}$ and so $C$ is closed.

What's the path with metric spaces? Our $U_y$ will be of the form $B(y,r_y)$ (open ball centered at $y$, with radius $r_y$) and $V_y=B(x,r_y)$, where you can take $r_y=d(x,y)/2$. By the triangle inequality, $B(y,r_y)\cap B(x,r_y)=\emptyset$.

One can choose the finite set $\{y_1,y_2,\dots,y_n\}$ as before and the point is that $$ \bigcap_{k=1}^n B(x,r_{y_k}) $$ is not necessarily an open ball centered at $x$, but it surely contains one, for instance $B(x,r)$, where $r=\min\{r_{y_1},r_{y_2},\dots,r_{y_n}\}$. Then, as before, $B(x,r)\cap C=\emptyset$.