# Other solution to 'compact and continuous imply uniformly continuous'

This is Rudin PMA chapter 4 exercise 10:

Complete details of the following (alternate) proof of this theorem:

Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f$ is uniformly continuous on $X$.

The proof we have to complete is: If $f$ is not uniformly continuous, then for some $\varepsilon>0$ there are sequences $\{p_n\},\{q_n\}$ in $X$ such that $d_X(p_n,q_n)\to 0$ but $d_Y(f(p_n),f(q_n))>\varepsilon$.

I am following Wisconsin Rudin solution which goes like this:

Since $\{p_n\}, \{q_n\}$ are sequences of a compact set, they have convergent subsequences such that $\{p_{n_k}\}\to p$ and $\{q_{n_k}\}\to q$. Since $d_X(p_n,q_n)\to 0$, we see $p=q$. And so on ...

So my problem is with indexing: we get $p=q$ from $d_X(p_n,q_n)\to 0$ when subsequences of $p_n$ and $q_n$ have same index. But, eg, if $\{p_n\}=\{q_n\}=0,1,0,1,0,...$ and subsequence $p_{n_l}=0,0,0,...$ and $q_{n_m}=1,1,1,...$ so even if $d_X(p_n,q_n)\to 0$, we don't get $0=p\ne q=1$. So what guarantees that our subsequences $p_{n_k}$ and $q_{n_k}$ must have same indexes?

• Start with the sequence of pairs $(p_n,q_n)$, and pick a subsequence for which the first components $p$ converge (don't worry about the $q$'s yet). Then pick a subsequence of that subsequence (not merely a subsequence of the original sequence) for which the second components $q$ converge. – Andreas Blass Mar 12 '18 at 17:35
• Thank u so much – Silent Mar 12 '18 at 17:37