I have some difficulties solving the following exercise (Baby Rudin 2.24)

Let $ X $ be a metric space in which every infinite subset has a limit point. Prove that $ X $ is separable.

In order to solve this I try to find a connected set which contains an infinite subset that has no limit point.

So, the interval $ [0 \dots 1] $ is a connected set, but it is a compact one also.

According to Baby Rudin theorem 2.37 any infinte subset $E$ of a compact set $K$ has a limit point in $K$.

It appears that I can't find a subset to contradict exercise 2.24.

I know that there is a hint in the book, I try not to read it.


  • 2
    $\begingroup$ Separability has nothing to do with connectedness. You need to show that $X$ has a countable dense subset. $\endgroup$ – Daniel Fischer Sep 4 '15 at 18:40
  • $\begingroup$ Well' I thought that a set has two choices, being connected or seperated. $\endgroup$ – user251106 Sep 4 '15 at 18:56
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    $\begingroup$ It's a different thing. The words are similar, but mean something completely different. (Also, "separated" often is used as a synonym of Hausdorff.) $\endgroup$ – Daniel Fischer Sep 4 '15 at 18:58
  • $\begingroup$ I don't get it, the definition from the book says "Two subsets $A$ and B$" of a metric space $X$ are said to be separated if both $A \cap \overline{B}$ and $B \cap \overline{A}$ are empty. $\endgroup$ – user251106 Sep 4 '15 at 19:07
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    $\begingroup$ @user251106 Saying that $A,B\subseteq X$ are separated is different than saying that $X$ is separable. The term separable is defined in Exercise 22 just above. $\endgroup$ – Alex Kruckman Sep 4 '15 at 21:05

Fix $\delta > 0$.

Choose $x_1 \in X $.

Choose $x_2$ such that $d(x_1,x_2)>\delta$.

Now choose $x_3$ such that $d(x_1,x_3)>\delta$ and $d(x_2,x_3)>\delta$.

Continue this process...

Basically, we want to cover $X$ with disjoint $\delta$ balls.

Since, {$x_1, x_2, \dots$} forms and infinite set and clearly does not have a limit $\implies$ finitely many $\delta$ balls around $x_i$'s are enough to cover the whole $X$.

Choose $\delta = 1$. Then $\exists$ finitely many points {$x_{11},x_{12}, \dots , x_{1N_1}$} such that X can be covered by $\delta$ balls around them.

Similarly for $\delta = \frac{1}{2}$, let the set be {$x_{21},x_{22}, \dots , x_{2N_2}$}.

Repeating this process for $n \in \mathbb{N}$, $\delta = \frac{1}{n}$, let the set be {$x_{N1},x_{n2}, \dots , x_{nN_n}$}.

Construct a set $A=$ {$x_{11},x_{12},\dots,x_{1N_1}, x_{21},x_{22},\dots x_{2N_2},\dots,x_{n1},\dots,x_{nN_n},\dots$}.

Claim: A is countable and dense in X.

Countable is easy. For denseness, use Archimedean Property.

For any $x \in X$, the $\epsilon$ ball around $x$ contains a point from $A$.

Since A is a countable dense subset of $X \implies X$ is separable.


The given property of $X$ (separated because of $d(x,y)=0 \iff x=y$ sometimes called axiom of separation since ensuring the uniqueness of limits) implies that all sequence contains a convergent subsequence so this space $X$ is sequentially compact then compact (and separated because metric).

NOTE.- For french mathematicians, compact is always separated.

Therefore $X$ is a second-countable space (a.i. X admits a countable basis of open sets). We prouve that this fact implies separability of $X$.

Let {$O_n: n\in \mathbb N$} be a countable basis of X and in each $O_n$ choose a point $a_n$; the set, noted A, of all of these $a_n$ is obviously countable and it remains to prove it is dense . Let $x\in X$ and $O$ an open containing $x$; $O$ is an union of sets $O_n$ and there exist at least un $O_{n_0}$ with $x\in O_{n_0}\subset O$ thus a point $a_n$ belongs to $O$ which shows $A$ is dense.


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