# Absolute function continuous implies function piecewise continuous?

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!

• Hint: What happens if $f$ only takes the values $1$ and $-1$? – Tobias Kildetoft Jan 24 '13 at 19:50
• @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} – tim_a Jan 24 '13 at 19:58
• 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. – Tobias Kildetoft Jan 24 '13 at 20:02
• 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. – tim_a Jan 24 '13 at 20:07
• No, what I mentioned gives you a way to construct a counter example. – Tobias Kildetoft Jan 24 '13 at 20:08

## 1 Answer

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.

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