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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?

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    $\begingroup$ This is the standard proof of Heine-Cantor 's Theorem. $\endgroup$
    – Crostul
    Commented Sep 26, 2015 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 $\endgroup$
    – RFZ
    Commented Sep 26, 2015 at 7:46
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    $\begingroup$ OK. By the way, it is correct. $\endgroup$
    – Crostul
    Commented Sep 26, 2015 at 7:46

1 Answer 1

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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.

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