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Hi this is a problem from Rudin's Princ. of Mathematics. I was hoping someone could check this part of my proof for the following question, comments would be very appreciated!:

$25.$ Prove that every compact metric space $K$ has a countable base, and that $K$ is therefor separable.

Consider the neighborhoods of radii $\frac{1}{n}$ for every $k_{i} \in K$. Then we have that $K \subset \bigcup\limits_{i=1}^{\infty} N_{1/n}(k_{i})$, and so $\bigcup\limits_{i=1}^{\infty} N_{1/n}(k_{i})$ is an open cover of $K$. Since $K$ is compact, there is a finite sub-collection of this set, let the sequence of these sets be $\{V_{\alpha}\}$. Then we claim that this is a countable base for $K$. Since it is finite, it is clearly countable. Then for any $k \in K$ and open set $G \subset K$ such that $k \in G$, we have that $k \in V_{\alpha}$ for some $\alpha$. Then since these are open sets, there is a neighborhood with a sufficiently small radius that is a subset of an interior neighborhood of $G$.

Ie. since $G$ is open then there is a neighborhood $N_{r}(k) \subset G$. Then let $\delta < \frac{r}{2}$. Since $k \in V_{\alpha}$, we have that there is a neighborhood $N_{\delta}(k)$ for small enough $\delta$ such that $N_{\delta}(k) \in V_{\alpha}$, since $V_{\alpha}$ is open such a neighborhood exists. Then $N_{\delta}(k) \subset N_{r}(k) \subset G$ which is what we wanted: to find a subcollection of $V_{\alpha}$ so that it is a subset of $G$. Then since $G$ and $k$ were arbitrary, $K$ must have a countable base.

To show that $K$ is dense can we say that since $\{V_{\alpha}\}$ is a finite open cover of $K$, then consider $G_{\alpha} = V_{\alpha} \cap K$. Then since $\{V_{\alpha}\}$ is a countable sequence, then so is $\{G_{\alpha}\}$. I think the union of $G_{\alpha}$ for every $\alpha$ is $K$. Then by definition is $K$ dense in $K$, and since it is countable, then $K$ is separable.

Otherwise, we can repeat how we solved problem $24$ just before this. Since $K$ is compact every infinite subset has a limit point in $K$.

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  • $\begingroup$ If you consider a finite sequence $\{V_{\alpha}\}$ such that $K\subset \bigcup_{\alpha} V_{\alpha}$ then $\{V_{\alpha}\}$ is not necessarily a basis for $K.$ Think of $K=[0,1]$ and $V_1=N_1(0)=[0,1)$ and $V_2=N_1(1)=(0,1].$ $\endgroup$
    – mfl
    Jun 9, 2014 at 22:22

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The idea is slightly miscomunicated, at least. For each $n=1,2,\ldots$ we have open covers $$K=\bigcup\limits_{x\in K} B(x,n^{-1})$$

whence we may pick a finite subcover say $$B_1^{(n)}=B(x_1^{(n)},n^{-1}),\ldots,B_{k_n}^{(n)}=B(x_{k_n}^{(n)},n^{-1})$$

Note the number of open sets $k_n$ depends on $n$. Thus the collection of $B_j^{(i)}$ is a countable base and $$D=\{x^{(i)}_{j}:i=1,2,\ldots\; 1\leqslant j\leqslant k_i\}= \{x_1^{(1)},\ldots,x_{k_ 1}^{(1)},x_1^{(2)},\ldots,x_{k_2}^{(2)},\ldots\}$$ is a countable dense subset of $K$.

Can you prove this?

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  • $\begingroup$ Yes, I think what you are recommending is how we solved the problem $24$ just before this. $D$ is the sequence of all finite $\delta$-balls of varying length that covers $K$. Then each of these sets must be finite otherwise we have a contradiction that we will have two points that are closer than the length of $\delta$. Then for any $x$ and $\varepsilon > 0$ we have that $x$ belongs to $N_{r}(q)$ where $q \in K$ and $r < \frac{\varepsilon}{2}$. This shows the density. Then $K$ is separable, is this what you mean? $\endgroup$
    – zzz2991
    Jun 9, 2014 at 23:43
  • $\begingroup$ I don't exactly see why my proof does not show that $\bigcup\limits_{i=1}^{\infty} N_{1/n}(k_{i})$ is a countable base of $K$. The text defines a base as follows: A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if $\forall x \in X$ and every open set $G \subset X$ such that $x \in G$, we have $x \in V_{\alpha} \subset G$ for some $\alpha$ $\endgroup$
    – zzz2991
    Jun 9, 2014 at 23:50
  • $\begingroup$ @DavidJhoo No, it is not necessarily a base, you need the diameters to go to $0$, since if $x\in G$ and $G$ is open, there is a ball $B(x,\varepsilon)\subseteq G$, but there is no guarantee $\varepsilon <1/n$. $\endgroup$
    – Pedro
    Jun 9, 2014 at 23:52
  • $\begingroup$ Thanks for the replies. But if there is $B(x,\varepsilon) \subset G$, we can't continually shrink $\varepsilon$? Given every point in the ball is already in $G$, then won't shrinking the radius still have that it is a subset of $G$, and we can eventually reach some $\varepsilon < 1/n$. I think this is what I am doing in my proof, though I am not sure if it is valid. I thought it OK to shrink balls arbitrarily, regardless of how small it will always have at least the point $K$ which is still a subset of $G$. $\endgroup$
    – zzz2991
    Jun 9, 2014 at 23:57
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    $\begingroup$ @DavidJhoo No, it is the other way around, you want $1/n<\varepsilon$! I switched them above. =) $\endgroup$
    – Pedro
    Jun 10, 2014 at 0:00

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