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We know that the points of discontinuity of a monotone function on an interval $[a,b]$ are countable. Using this can we prov that:

  • Any bijection $f: \mathbb{R} \to [0,\infty)$ has infinitely many points of discontinuity.

If yes, how or otherwise how to prove the above result?

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    $\begingroup$ You don't even have a hunch? $\endgroup$ – Mariano Suárez-Álvarez Oct 28 '10 at 5:42
  • $\begingroup$ @Mariano: What did you mean. $\endgroup$ – anonymous Oct 28 '10 at 5:46
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Suppose to the contrary that $f$ has a finite number $n$ of discontinuities, at $x_1, x_2, \ldots x_n$. Then $\mathbb{R} - ( x_1, \ldots x_n )$ is a union of open intervals $I_1, \ldots I_{n + 1}$. $f$ restricted to each $I_m$ is continuous and injective, and therefore monotone, so each image $f|_{I_m}(I_m)$ is an open (in $\mathbb{R}$) interval $J_m$. The $J_m$ are non-empty, pairwise disjoint and contained in $(0, \infty)$; suppose that $J_{m_1} < J_{m_2} < \ldots < J_{m_{n + 1}}$. Then between each $J_{m_p}$ and $J_{m_{p + 1}}$ there is at least one point, which must be the image of some $x_q$. But this exhausts all the $x$'s, so that $0$ is not in the image of $f$.

BTW, can anybody tell me how to get the forum to display braces in math mode?

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    $\begingroup$ Prepend them with a backslash. \{a,b\} gives $\{a,b\}$ $\endgroup$ – J. M. is a poor mathematician Oct 28 '10 at 5:57
  • $\begingroup$ Huh. I tried that, and it didn't work in the preview. Let's try again: $\{ \}$ $\endgroup$ – Rotwang Oct 28 '10 at 6:10
  • $\begingroup$ @Rotwang: Try this \\{a\\} $\endgroup$ – anonymous Oct 28 '10 at 6:15
  • $\begingroup$ @Rotwang: Your test worked on my end. Try (hard) refreshing. $\endgroup$ – J. M. is a poor mathematician Oct 28 '10 at 8:02
  • $\begingroup$ It seems sometimes I need two backslashes and sometimes only one. I haven't figured out when for each. $\endgroup$ – Ross Millikan Oct 28 '10 at 13:02

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