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I have a simple true/false question that I am not sure on how to prove it.

If $|f(x)|$ is continuous in $]a,b[$ then $f(x)$ is piecewise continuous in $]a,b[$

Anyone that can point me in the right direction or give a counterexample, even though I think it's true. Thanks in advance!

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  • $\begingroup$ Hint: What happens if $f$ only takes the values $1$ and $-1$? $\endgroup$ – Tobias Kildetoft Jan 24 '13 at 19:50
  • $\begingroup$ @Tobias So you think the statement is false? I don't see the problem with a function defined like this: \begin{cases} 1 & x <0\\-1 & x \geq 0\end{cases} $\endgroup$ – tim_a Jan 24 '13 at 19:58
  • $\begingroup$ What I mean is that if $f$ only takes those two values, then $|f|$ is automatically continuous. But there are nowhere continuous functions with just those two values. $\endgroup$ – Tobias Kildetoft Jan 24 '13 at 20:02
  • $\begingroup$ Ok, I think I understand what you mean. But the statement is in the other direction. If you already know that $|f|$ is continuous, does this imply that $f$ is piecewise continuous in any case. $\endgroup$ – tim_a Jan 24 '13 at 20:07
  • $\begingroup$ No, what I mentioned gives you a way to construct a counter example. $\endgroup$ – Tobias Kildetoft Jan 24 '13 at 20:08
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Here is a counter example to the statement:

Define $f(x)$ to be $1$ if $x$ is rational and $-1$ if $x$ is irrational. Now $f$ is not continuous anywhere, but $|f|$ is identically $1$ and thus continuous.

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  • $\begingroup$ Thank you for helping me! Do you know if there is a way of defining such a function with symbolic math software like Maple? $\endgroup$ – tim_a Jan 24 '13 at 20:21
  • $\begingroup$ You were starting light for the OP. $\endgroup$ – mrs Jan 24 '13 at 20:31

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