non-uniformly continuous function

We assert that the real function $x\mapsto\sin\frac{1}{x}$ is not uniformly continuous on the open interval $(0,\,1)$ .

For proving this, we make the antithesis that there exists a positive number $\delta$ such that

$|f(x_{1})\!-\!f(x_{2})|\;<\;1\quad\mbox{always when}\quad x_{1},\,x_{2}\in(0,% \,1)\;\;\mbox{and}\;\;|x_{1}\!-\!x_{2}|<\delta.$

Choose

$x_{1}\;=\;\frac{1}{\frac{\pi}{2}\!+\!2n\pi},\;\;x_{2}\;=\;\frac{1}{\frac{3\pi}% {2}\!+\!2n\pi}$

where the integer $n$ is so great that $x_{1}<\frac{\delta}{2}$ , $x_{2}<\frac{\delta}{2}$ . Then we have

$|x_{1}\!-\!x_{2}|\;\leqq\;|x_{1}|+|x_{2}|\;<\;\delta.$

However,

$f(x_{1})\!-\!f(x_{2})\;=\;1\!-\!(-1)\;=\;2.$

This contradictory result shows that the antithesis is wrong.

Title	non-uniformly continuous function
Canonical name	NonuniformlyContinuousFunction
Date of creation	2013-03-22 19:00:07
Last modified on	2013-03-22 19:00:07
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Example
Classification	msc 26A15
Related topic	PointPreventingUniformConvergence
Related topic	ReductioAdAbsurdum