I'm trying to prove this theorem:
Let $(X,d)$ be a complete metric space, with no isolated points, prove X is uncountable.
I've heard that this can be proved using Baire's category theorem but we haven't learned that theorem in my course so I'm looking for a proof that does not use it.
I started by assuming this is not true and marking $X = \{x_n|n\in \Bbb{N}\}$. Knowing that X is complete I can use the fact that any Cauchy series converges to some element in X.
Best Answer
I think that an argument as the following might work:
Suppose that $X$ is countable, then you can index its elements by $\mathbb{N}$, as $\{x_n\}_{\mathbb{N}}$.
Consider a ball $B(x_1,r)$, then you have infinitely many $x_i$s in the ball (because in $X$ there are not isolated points). Pick the first $i_1>1$ such that $x_{i_1}\in B(x_1,r)$, and a ball $B(x_{i_1}, r_1)$ contained in $B(x_1,r)$ such that $x_1$ is not in $B(x_{i_1},r_{i_1})$. Inductively you can construct in this way a sequence of $x_{i_n}$s such that $x_{i_k}\in B(x_{i_n},r_n)$ for $k\geq n$, and such that $x_{i_{n+1}}$ is the element with the first index $j$ such that $x_j\in B(x_{i_n},r_n)$. Moreover you can impose that $x_{i_n}$ is not in $B(x_{i_{n+1}},r_{n+1})\subset B(x_{i_n},r_n)$ and choose the $r_n$s in a way that they tend to $0$.
The sequence then is clearly Cauchy (it is contained in balls that get smaller and smaller), but it has no limit. In fact if $x_l$ is the limit of the sequence, then $l$ can't be one of the $i_n$s by the construction of the sequence. Then you have that $i_{j}<l<i_{j+1}$ for some $j$, but since you must have $x_l\in B(x_{i_j},r_j)$ (it is the limit) this is not possible by the construction of the sequence (remember that $i_{j+1}$ is the first index such that the corresponding element is contained in the ball). This proves the claim by contradiction.