0

Let $X$ be a compact metric space with no isolated points (so every point is a limit point). Show that $X$ is uncountable.

I think I can show this by showing that $X$ has cardinality at least that of the Cantor Set. Essentially, begin with two disjoint nonempty open sets $U_0$ and $U_1$ of $X$. Then choose disjoint nonempty open sets $U_{00}, U_{01}$ of $U_0$ and $U_{10}, U_{11}$ of $U_1.$ Though I'm not really sure how to show that we can always choose such sets. The axiom of choice is likely involved here.

However, if we can always choose such sets, then after choosing the $2^n$ sets $U_{0\cdots 0},\cdots, U_{1\cdots 1},$ for each of the $2^n$ sets $U_{j_1\cdots j_n}$, we can choose two sets $U_{j_1\cdots j_n0}, U_{j_1\cdots j_n1}$ that are disjoint and nonempty in $U_{j_1\cdots j_n}.$ How would one formally show that this results in a "copy" of the Cantor set in $X$?

Edit: I do not want to use the Baire Category theorem to solve this problem.

Alfred
  • 859

2 Answers2

1

Hint.
Your answer is basically good. But (a) you want closed sets (not open) so that, if $j_1j_2\dots$ is an infinite sequence, you can conclude that $$ \bigcap_{n=1}^\infty U_{j_1j_2\dots j_n} \tag1$$ is nonempty. And (b) you want to arrange that the diameters if $U_{j_1j_2\dots j_n}$ are ${} < 1/n$ [or something going to $0$], so that $(1)$ is a single point.

GEdgar
  • 111,679
  • my answer is very rough. I was looking for some answer that closely sticks with the definitions. In particular, why can we always choose these closed sets? – Alfred Jun 28 '21 at 13:40
  • For instance, a limit point of a set $A$ is a point $a$ such that for all $r > 0, B^(a,r)\cap A\neq \emptyset,$ where $B^(a,r)$ is the punctured ball of radius $r$ centred at $a$. – Alfred Jun 28 '21 at 13:52
  • can you prove a link to a resource that might be useful for proving this? – Alfred Jun 28 '21 at 19:22
  • Try https://math.stackexchange.com/a/1654430/442 – GEdgar Jun 28 '21 at 20:18
0

One way is to use the Baire Category Theorem, which applies to all complete metric spaces, which includes all compact metric spaces.

If $X$ is a non-empty compact metric space with no isolated points then each member of the family $F=\{\,\{x\}:x\in X\}$ is nowhere-dense in $X,$ but $\cup F=X$ so by Baire, $F$ cannot be countable.

DanielWainfleet
  • 57,985