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Let $f:(0,1]\rightarrow \mathbb{R} $ be a continuous function and bounded is it uniformly continuous?

I know this isn't true, but I can't find a good counter example I was thinking something like this

$f(x)=\min\{3,1/x\}$ can someone give a better example.

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  • $\begingroup$ Have some wave that oscillates really fast near the origin, e.g. $\sin(1/x)$ $\endgroup$ – jlammy Jan 24 at 14:40
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You can simply take $f(x)=\sin\left(\frac1x\right)$. It is not uniformly continuous because, for any $\delta>0$, there are $x,y\in(0,1]$ such that $|x-y|<\delta$ and that $\bigl|f(x)-f(y)\bigr|=2$.

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You example $f(x)$ doesn't work, i.e. $f(x)$ is uniformly continuous on $(0,1]$. This is because it extends to a continuous function on $[0,1]$ by defining $f(0)=3$, and all continuous functions on closed and bounded intervals (or compact sets) are automatically uniformly continuous.

As for an actual example, someone else has already suggested $\sin(1/x)$, so see their answer.

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