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Let $f:\Bbb R \to \Bbb R$ be a continuous function. Prove that for every $x \in \Bbb R$ and for every $A>0$ there is a $B>0$ such that for every $y \in \Bbb R$, $\lvert y-x\rvert \le A \Rightarrow \lvert f(x)-f(y)\rvert \le B $.

It looks a lot like just regular defenition of continuity, but A and B "switched places". Any help in writing a formal proof? Thanks.

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    I tried to use the regular definition, but I'm pretty stuck. Can't figure out how to swap them. – 2017-02-05
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    In other words, the assertion is that $f([x-A, x+A])$ is bounded. – 2017-02-05
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    Continuous on all $\mathbb R \implies$ continuous on $[x - A, x + A] \implies $ uniformly continuous on $[x - A, x + A] \implies $ bounded on $[x - A, x + A] \implies $ bounded on $(x - A, x + A) $ – 2017-02-05

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