I was reading Strichartz' The Way of Analysis, there he proves the Uniform Continuity Theorem:

Let $f$ be a function on a compact domain $D$ that is continuous. Then it is uniformly

Proof : Suppose $f$ is not uniformly continuous, that is, there exists $1/m$ such that for all $1/n$ there exists two points $x_n,y_n$ in the domain such that $|x_n - y_n| < 1/n$ but $|f(x_n) - f(y_n)| \ge 1/m$. Since $D$ is assumed compact, the obvious first step is to replace the sequences $x_1,x_2,x_3,\dots$ and $y_1,y_2,y_3,\dots$ by convergent subsequences.The condition $|x_n-y_n|\le 1/n$ implies that both subsequences converge to the same limit, call it $x_0$. Calling the subsequences $x'_1,x'_2,x'_3,\dots$ and $y'_1,y'_2,y'_3,\dots$ we have $|f(x'_n)-f(y'_n)|\ge 1/m$ and yet both $\lim _{n\rightarrow \infty} f(x'_n)=f(x_0) $ and $\lim _{n\rightarrow \infty} f(y'_n)=f(y_0) $ by the continuity of $f$ at the point $x_0$. This is a contradiction; we cannot have $\lim _{n\rightarrow \infty} (f(x'_n)-f(y'_n))=f(x_0)-f(x_0)=0 $ and $|f(x'_n)-f(y'_n)|\ge 1/m$ for all $n$. QED

I cannot understand why convergent subsequences of $x_1,x_2,x_3,\dots$ and $y_1,y_2,y_3,\dots$ will have the same limits: Sequences $\{x_n\}=5,1,0,1,0,1,0,\dots$ and $\{y_n\}=6,1,0,1,0,1,0,\dots$ have the property that $|x_n-y_n|\le 1/n$ for all $n$ but convergent subsequence $1,1,1,\dots$ of $x_n$ and convergent subsequence $0,0,0,\dots$ of $y_n$ have different limits.

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    $\begingroup$ You need to take the corresponding subsequences: $x_n'=x_{\phi(n)}$ and $y_n'=y_{\phi(n)}$. $\endgroup$ – Lord Shark the Unknown Sep 2 '17 at 6:29
  • $\begingroup$ Aha, thank you, I should think that! $\endgroup$ – Silent Sep 2 '17 at 6:32

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