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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##.
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'##
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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