Compactness and Lebesgue Number proof

Question

Let $X$ be a compact metric space. Then, every open cover of $X$ has a lebesgue number.

Suppose $X$ has an open cover $(U_i)_{i\in I}$, which admits no lebesgue number. In particular, for each $n$, there exists a sequence of non-empty sets, $A_n$, for which, $A_n$ is not contained in any of the $U_i$.Considering AOC, we choose $x_n\in A_n\backslash U_i$. Then, $x_n$ has a convergent subsequence, $(x_{n_k})$ such that $x_{n_k}\rightarrow x$ for some $x\in U_m$. Since $x_{n_k}$ converges, there must exists $K$, for which, $k\geq K$ implies $x_{n_k}\in U_m$. Contradiction.

Is this correct?

Answer 1

Your proof has a logical error in the second sentence. The statement

... there exists a sequence of non-empty sets $A_n$ for which $A_n$ is not contained in any of the $U_i$...

does not imply what follows. That statement is not quite what you want. Instead, what you want is this:

... there exists a sequence of non-empty sets $A_n$ for which $A_n$ is not contained in any of the $U_i$ and such that the diamater of $A_n$ approaches $0$ as $n \to \infty$.

Also, your final contradiction is not a contradiction: there is nothing contradictory about the implication

... $k \ge K$ implies $x_{n_k} \in U_m$.

The contradiction you want is

... $k \ge K$ implies $A_{n_k} \subset U_m$.

The last thing I'll say is: nowhere in your proof have you used the metric in any substantial way. That's a big clue that there are some ideas missing.