# 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 NonuniformlyContinuousFunction 2013-03-22 19:00:07 2013-03-22 19:00:07 pahio (2872) pahio (2872) 9 pahio (2872) Example msc 26A15 PointPreventingUniformConvergence ReductioAdAbsurdum