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

Proof: Let $f$ is not uniformly continuous then for some $\varepsilon>0$ and arbitrary $\delta>0$ exists $x,y\in E$ and $d_X(x,y)<\delta$ but $d_Y(f(x),f(y))>\varepsilon.$ Taking $\delta=\frac{1}{n}$ then exists sequences $\{p_n\},\{q_n\}$ in $X$ such that $d_X(p_n,q_n)\to 0$ but $d_Y(f(p_n),f(q_n))>\varepsilon.$

Since $X$ is compact then $\{p_n\}$ and $\{q_n\}$ has limit point, namely $p$ and $q$. Then exists some subsequence $\{p_{n_k}\}$ that converges to $p$. Using triangle inequality we get that $\{q_{n_k}\}$ also converges to $p$.

Since $f$ is a continuous mapping then $f(p_{n_k})\to f(p)$ and $f(q_{n_k})\to f(p)$. Combine these statements we get $\varepsilon<d_Y(f(p_{n_k}),f(q_{n_k}))<\varepsilon$ and it's a contradiction.

Is this proof correct?

  • 2
    $\begingroup$ This is the standard proof of Heine-Cantor 's Theorem. $\endgroup$ – Crostul Sep 26 '15 at 7:44
  • $\begingroup$ @Crostul, I don't know about this standard proof. Before this I knew only one proof from baby Rudin theorem 4.19 $\endgroup$ – ZFR Sep 26 '15 at 7:46
  • 2
    $\begingroup$ OK. By the way, it is correct. $\endgroup$ – Crostul Sep 26 '15 at 7:46

The proof is correct.

Some of your conjunctions (in the grammatical sense) are a little confusing. For instance, "exists $x, y \in E$ and $d_X(x, y) < \delta$", which should read "exists $x, y \in E$ with $d_X(x, y) < \delta$" or even better "such that". Doesn't affect the mathematics of the proof, but it makes the reader do a bit more work than they should really have to.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.