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Let $f:\mathbb{R}^n\to\mathbb{R}$ be a continuous function.

Prove or Disprove:

$\forall\ x\in\mathbb{R}^n$ and $\forall\ \epsilon>0\ \exists\ \delta>0$ such that if $\|x-y\|<\epsilon$ then $|f(x)-f(y)|<\delta$

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    **Hint:** bounded sets are relatively compact. – 2017-02-08

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Are you sure the position of $\epsilon$ and $\delta$ is correct?

If it is, then your statement is correct. Since by given $x$ and $\epsilon$, for any $\|x-y\|<\epsilon$, we have $y$ in a closed ball with radius $\epsilon$ centered at $x$. Since $f$ is continuous, it is bounded by $M$ on this closed ball, hence we have the $\delta=2M$.

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    I am 100% sure this is not uniform continuity to check. – 2017-02-08
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    @TrueTopologist Yes. Then since $\delta$ could depend on $x$ and $\epsilon$, basically you are proving the locally boundedness of continuous function on $\mathbb{R^n}$. – 2017-02-08