I've got some problems while I'm proving that statement.

I'm trying to show that with using RAA(it is PMA exercise Problem 10, chapter 4).

Here is my idea.

Suppose $f$ is continuous but not uniformly. Then there is some $\varepsilon>0$ which satisfies that for any $\delta>0$, there is $x,y\in X$ such that $$d(x,y)<\delta\land d(f(x),f(y))\ge\varepsilon.$$ In other words, there are sequences $\{p_n\}$ and $\{q_n\}$ in $X$ so that for any $\delta>0$ there is some $n\ge N$ such that $$d(p_n,q_n)<\delta\land d(f(p_n),f(q_n))>\varepsilon.$$ Since $X$ is compact metric space, for any sequence in $X$ has a convergent subsequence. Let $p_{n_k}\to a\in X$ and $q_{n_k}\to b\in X$. To simplify let $p_{n_k}=a_k$ and $q_{n_k}=b_k$. Then $$0\le d_X(b_k,a)\le d_X(b_k,a_k)+d_X(a_k,a).$$ Since both $d_X(b_k,a_k)$ and $d_X(a_k,a)$ converges to $0$ as $k\to \infty$, $d_X(b_k,a)\to0$, i.e., $a=b$. Since $f$ is continuous, $$f(a)=\lim\limits_{x\to a}f(x)=\lim\limits_{k\to\infty}f(a_k)=\lim\limits_{k\to\infty}f(b_k).$$ That is, for fixed $\varepsilon>0$, there are $N_1>0$ and $N_2>0$ such that $$k\ge N_1\Longrightarrow d_Y(f(a),f(a_k))<\frac{\varepsilon}{2}$$ and $$k\ge N_2\Longrightarrow d_Y(f(a),f(b_k))<\frac{\varepsilon}{2}.$$ Hence, if $k\ge\max(N_1,N_2)$, $$d_Y(f(a_k),f(b_k))\le d_Y(f(a_k),f(a))+d_Y(f(a),f(b_k))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

Thus I got the result that $d_Y(f(p_{n_k}),f(q_{n_k}))<\varepsilon$, but I think it is not sufficient. Even though $d_Y(f(p_{n_k}),f(q_{n_k}))<\varepsilon$ for all $k\ge \max(N_1,N_2)$, it does not guaratee that $d_Y(f(p_n),f(q_n))<\varepsilon$ for all $n\ge N$ for some $N$.

Is there something I am missing? I cannot catch it.


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