I claim that if a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is continuous at a point ##a##, then there exists a ##\delta>0## and ##|h|<\frac{\delta}{2}## such that ##f## is also continuous in the ##h##-neighbourhood of ##a##.(adsbygoogle = window.adsbygoogle || []).push({});

Please advice if my proof as follows is correct.

Continuity at ##a## means this: for every ##\epsilon>0##, there exists a ##\delta>0## such that ##|x-a|<\delta \implies|f(x)-f(a)|<\epsilon##. --- (*)

We want to show this: for every ##\epsilon'>0##, there exists a ##\delta'>0## such that ##|x-(a+h)|<\delta' \implies|f(x)-f(a+h)|<\epsilon'##.

If we let ##\delta'## = ##\delta-|h|## and ##\epsilon=\frac{\epsilon'}{2}##, we will have ##|x-(a+h)|<\delta' \implies|x-a|<\delta\implies|f(x)-f(a)|<\epsilon\implies|f(x)-f(a+h)|<\epsilon'##

The details:

##|x-(a+h)|<\delta'##

##|x-(a+h)|<\delta-|h|##

##-\delta+|h|<x-a-h<\delta-|h|##

##-\delta\leq-\delta+|h|+h<x-a<\delta-|h|+h\leq\delta##

##|x-a|<\delta##

By (*),

##|f(x)-f(a)|<\epsilon##

##-\epsilon<f(x)-f(a)<\epsilon##

##-\epsilon<f(x)-f(a+h)+f(a+h)-f(a)<\epsilon##

##-\epsilon-f(a+h)+f(a)<f(x)-f(a+h)<\epsilon-f(a+h)+f(a)## --- (**)

By (*), when ##x=a+h##,

##\mid a+h-a\mid\,=\,\mid h\mid\,<\delta\implies\,\mid f(a+h)-f(a)\mid\,=\,\mid-f(a+h)+f(a)\mid\,<\epsilon##

Substituting this into (**), we get

##-2\epsilon<f(x)-f(a+h)<2\epsilon##

##\epsilon'<f(x)-f(a+h)<\epsilon'##

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# Continuity at a point implies continuity in the neighborhood

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