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I am thinking about this since couple hour. Is a continuous and bounded function $f:\mathbb R\to\mathbb R$ uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let $\epsilon>0$ and $a\in\mathbb R$. By continuity, there is a $\delta_a>0$ such that $$d(f(x),f(y))<d(f(x),f(a))+d(f(y),f(a))<\epsilon$$ if $d(x,y)\leq d(x,a)+d(y,a)<\delta_a.$ Then, if $\inf_a\delta_a>0$ I set $\delta=\inf_a\delta_a$ and we are done since $d(f(x),f(y))<\epsilon$ if $d(x,y)<\delta,$ but I didn't success to prove that $\delta\neq 0$.

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Consider $\sin(x^2)$. It is continuous and bounded but not uniformly continuous. This is because we can find arbitrarily small $\gamma$ so that $f(a)=-1$ and $f(a+\gamma)=1$. In other words if $\epsilon=1$ we cannot find a suitable $\delta$.

This is because the function oscillates faster and faster as $|x|$ grows.

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On the other hand what is true is that a continuous function $f:[a,b]\rightarrow \mathbb R$ is uniformly continuous.

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