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$f:[0,1]\to [0,1]$ is defined in the following manner

$f(1)=1$ and if $a=.a_1a_2a_3a_4\dots$ which is decimal representation then $f(a)=.0a_10a_20a_3\dots$ we need to discuss continuity of $f$

consider a point $a=.315$

by definition $f(a)=.030105$

now the sequence $x_1=.3149,x_2=.31499,x_3=.314999,x_n=.31499\dots9$ ($n$times $9$),$\dots$ converges to $a$ but $f(x_n)$ clearly does not converges to $f(a)$, so can I say $f$ is not continuous on $[0,1]$?

Myshkin
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    With this argument, all you may say is that $f$ is not continuous at the point $0.315$. – user37238 Sep 17 '13 at 09:01
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    The map as described isn’t well-defined: $0.5=0.4\overline{9}$, but $f(0.5)=0.05$, while $f(0.4\overline{9})=0.04\overline{09}\ne0.05$. Was some provision made in the definition of $f$ to cover these cases by explicitly choosing one of the representations? – Brian M. Scott Sep 17 '13 at 09:01
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    $.099999\dots$ can be replaced by $.1$ – Myshkin Sep 17 '13 at 09:04
  • so $f$ is discontinuous only at terminating points? – Myshkin Sep 17 '13 at 09:06
  • @Senore: $f$ isn’t even defined to be a function on all of $[0,1]$, so there’s not much in point asking where it’s continuous. – Brian M. Scott Sep 17 '13 at 09:31

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