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Problem statement:

Prove that if $f$ is uniformly continuous on a bounded set $S$, then $f$ is a bounded function on S.

Proof:

Assume $f$ is unbounded on $S$.

Take sequence $(x_n)$ from $S$ defined so that:

$f(x_2) > f(x_1) > 1$

$f(x_3) > f(x_2) > 2$

$...$

$f(x_{n+1}) > f(x_n) > n$

which is possible since $f$ is unbounded. By the Bolzano-Weierstrass Theorem, there exists a convergent subsequence of $(x_n)$; let it be $(x_{n_k})$ Since $(x_{n_k})$ is convergent, it is a Cauchy sequence. Since $f$ is uniformly continuous on $S$, $f(x_{n_k})$ must be a Cauchy sequence. However, since the sequence is clearly increasing and the function $f$ is unbounded, it cannot be convergent and thus cannot be a Cauchy sequence, which is a contradiction.

Therefore, $f$ is bounded on $S$.

My proof just feels a bit odd. Can someone point out logical holes, if any? Additionally, any proof writing tips would be appreciated.

Dhaivat Pandya
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  • Looks good to me. – Arkady Dec 01 '13 at 06:17
  • There is detail missing. Is $f$ a function between metric spaces? – AnyAD Dec 01 '13 at 06:19
  • I'm pretty sure that the book I'm using at the moment has not mentioned metric spaces; f is a real valued function. – Dhaivat Pandya Dec 01 '13 at 06:25
  • The set $S$ is assumed bounded (not closed). What allows you then to conclude that the subsequence $x_{n_{k}}$ is convergent to a point $x$ (the aim $f(x_{n_{k}})\rightarrow f(x)$ is otherwise good). – AnyAD Dec 01 '13 at 09:14
  • I'm not exactly sure. Is there any way to salvage the proof? As far as I understand Bolzano-Weierstrass, I thought that it implied exactly this. – Dhaivat Pandya Dec 01 '13 at 14:53
  • See http://math.stackexchange.com/questions/88257/uniform-continuity-and-boundedness – AnyAD Dec 02 '13 at 02:41

0 Answers0