Post Undeleted by Joel David Hamkins
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For any irrational number $x$, let $f(x)$ be the real number arising from the integer part of $x$, together with every other digit of the rest of the expansion of $x$.

This is surjective, since one may interleave the digits of any real $y$ with any nonrepeating pattern, and thereby find an irrational $x$ with $f(x)=y$. This is continuous, since in order to determine finitely many digits of $f(x)$, it suffices to specify finitely many digits of if $x$.

Oops, it's not actually continuousx_n\to x$, because of non-unique representationthen$f(x_n)\to f(x)$. 2 added 78 characters in body For any irrational number$x$, let$f(x)$be the real number arising from the integer part of$x$, together with every other digit of the rest of the expansion of$x$. This is surjective, since one may interleave the digits of any real$y$with any nonrepeating pattern, and thereby find an$x$with$f(x)=y$. This is continuous, since in order to determine finitely many digits of$f(x)$, it suffices to specify finitely many digits of$x\$.

Oops, it's not actually continuous, because of non-unique representation.

Post Deleted by Joel David Hamkins
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