# Any continuous mapping on compact set is uniformly continous

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.

• – Exp ikx Jan 27 at 15:37
• @Expikx It will be useful, but I need to prove that via RAA, not directly. – Caenorhabditis Elegans Jan 27 at 15:50