Context
I am trying to solve Problem 2.25 in Baby Rudin. Here is the problem statement:
(Rudin Problem 2.25)- Prove that every compact metric space has a countable base and that $K$ is therefore separable.
What I want to note in particular is that in Rudin, "countable" means "countably infinite", and "at most countable" is used to refer to "finite or countably infinite".
So in Rudin, a countable base must be countably infinite.
I say all of this because I claim that under Rudin's definitions, the problem as stated is impossible to solve.
This is because if we have a finite compact metric space, then this metric space will have a finite number of subsets and hence a finite number of open sets. So it is impossible for such a compact metric space to have a countable base (remember, Rudin uses "countable" to mean "countably infinite").
Because of this, I am forced to try to interpret what Rudin meant by the problem, and so I have come up with the following interpretation.
Interpretation 1
So the interpretation of the problem I came up with is:
(Interpretation 1)- Prove that every infinite compact metric space has a countable base and that $K$ is therefore separable.
And I should mention that in Rudin, he defines "separable" as:
(Rudin Definition: Separable Metric Space)- A metric space is called "separable" if it contains a countable, dense subset.
So then my problem in this interpretation is that I construct a set $S$ which I believe is countably infinite, but I cannot actually show it. The following section contains my solution to this interpretation of the problem.
My Solution to Interpretation 1
We first make the following definition: For every $n\in\mathbb{N}$, define $\{S_{n}:= N_{\frac{1}{n}}(p) | p\in K \}$.Then $S_{n}$ is an open cover for $K$ and so $S_{n}$ has a finite subcover $\Omega_{n}$ for $K$. Then, let us define $$S:=\bigcup_{n\in\mathbb{N}}\Omega_{n},$$and note that $S$ is a countable union of finite sets, and is therefore at most countable (by corollary to Theorem 2.12).
(I DO NOT KNOW HOW TO SHOW THIS PART): $S$ is countably infinite.
Then we claim that $S$ is a base for $K$. To show this, let $x\in K$ and let $G$ be an open subset of $K$ which contains $x$.
Then since $G$ is open and $x\in G$, there exists some $r>0$ such that $N_{r}(x)\subseteq G$.
By the Archimedean principle there exists some $n\in\mathbb{N}$ such that $\frac{1}{n}< \frac{r}{2}$. Then, since $\Omega_{n}$ covers $K$, let us pick some $N_{\frac{1}{n}}(p)\in \Omega_{n}$ such that $x\in N_{\frac{1}{n}}(p)$. We claim that $N_{\frac{1}{n}}(p)\subseteq N_{r}(x)$ (and hence since $N_{r}(x)\subseteq G$, we have $N_{\frac{1}{n}}(p)\subseteq G$).
Note that if $y\in N_{\frac{1}{n}}(p)$, then $$
d(y,x) \leq d(y,p)+d(p,x)
< \frac{1}{n} + \frac{1}{n}
< \frac{r}{2} + \frac{r}{2}
=r,$$so $y\in N_{r}(x)$ and hence $N_{\frac{1}{n}}(p)\subseteq N_{r}(x)$. Thus, $N_{\frac{1}{n}}(p)\subseteq G$ and so since $N_{\frac{1}{n}}(p)\in\Omega_{n}\subseteq S$, we conclude that $S$ is a base for $K$.
Then we have a lemma (not in Rudin but it's not too hard to show) that if a metric space has a countably infinite base, then it is separable, and so since $S$ is a countably infinite base for $K$ it follows that $K$ is separable. $\square$
Question 1
The only source of confusion I have with my solution to interpretation $1$ is that I cannot figure out how to show that $S$ is countably infinite. It is obviously at most countable, but I cannot figure out how to show it is countably infinite.
It is not enough to say that $S$ is countably infinite just because it is the countable union of finite sets. For instance, $\bigcup_{n\in\mathbb{N}}\{ 1 \}$ is finite.
It is also not enough to say that $S$ is countably infinite just simply because each $\Omega_{n}$ contains neighborhoods of different radii than $\Omega_{m}$ for all $m\neq n$, because of pathological metrics such as $$d(x,y)=\begin{cases}
1, & x\neq y, \\
0, & x=y
\end{cases}$$
So my first question is:
(Question 1)- (Assuming interpretation 1 was the intended interpretation): How can I show that my constructed set $S$ is countably infinite?
Question 2
And then, I just wanted to see if interpretation 1 was actually the correct way to interpret the problem.
(Question 2)- Is interpretation 1 the correct way to interpret the problem? If not, what is the correct way?
Question 3
Finally, my last question is not really about this particular problem, but more generally about learning mathematics. Specifically, working through "Baby Rudin".
I came up with my solution to interpretation 1 pretty quickly but I spent hours trying and failing to show that $S$ is countably infinite, which seems like it should be obviously true. It has been very discouraging to see how slowly I have been progressing through Rudin because I keep getting hung up on little details like this.
(Question 3)- Is it really worth getting hung up on small details like this? If yes, how can I progress faster through Baby Rudin? If no, how do I distinguish between when a detail is worth getting hung up on and when it is best to skip it and move on (surely I shouldn't skip ALL details)?
Best Answer
Regarding Interpretation 1:
For each $n>0$, some $N_n\in\Omega_n$ has more than one point (because the space is infinite). This way we get a sequence $(N_1,N_2,\dots)$ of sets with radii $1/n$, each with at least two points in it. If $S$ is finite, then $(N_n)_{n>0}$ has a subsequence $(N_{n_k})_{k>0}$ such that $N_{n_1}=N_{n_2}=\cdots$. Pick two distinct points $x,y\in N_{n_1}$, and let $k$ be large enough such that $2/n_k<d(x,y)$. Then $N_{n_k}$ cannot contain both $x$ and $y$, a contradiction.
Regarding Question 2:
Very often when Rudin speaks of a countable set $\{x_n\}$, what he really means is a list $(x_1,x_2,\dots)$, such that every member of the set is some $x_n$. He doesn't really care if the list repeats, and usually there's no good reason that it doesn't. Infinite or not, his arguments would go through anyway.
Baby Rudin does have some other minor issues. Sometimes when Rudin says "If every $X$ in $Y$ has property $P$, then...", he assumes that there actually exists an $X$ in $Y$. For example, in Exercise 24 and 26, by "...in which every infinite subset has a limit point..." Rudin assumes that the metric space $X$ being discussed does have an infinite subset.
I myself just define countable as at most countable, and adopt that definition even when reading baby Rudin. I believe this definition is more convenient, hence more commonly accpeted. Most math people now probably find it awkward to exclude finite spaces from separable spaces, etc. Of course, we should still be very careful about when finiteness really makes a difference.
Regarding Question 3:
Getting hung up is part of the training that baby Rudin aims to provide. If you are progressing fast, then either you already know the materials well enough, or you are missing a lot of details that shouldn't be missed. But if you are progressing slowly, then you are definitely doing it right.