Show that if $f:X\to\textbf{R}$ is a continuous function, so is the function $|f|:X\to\textbf{R}$ defined by $|f|(x) = |f(x)|$.

Question

Show that if $f:X\to\textbf{R}$ is a continuous function, so is the function $|f|:X\to\textbf{R}$ defined by $|f|(x) = |f(x)|$.

MY ATTEMPT

According to the definition of continuity, for every $x_{0}\in X$ and every $\varepsilon > 0$, there is a $\delta > 0$ such that \begin{align*} x\in X,\,d(x,x_{0}) < \delta \Rightarrow ||f(x)| - |f(x_{0})|| \leq |f(x) - f(x_{0})| < \varepsilon \end{align*} whence we conclude that $|f|$ is also continuous.

Is the wording of my proof correct? I am curious if there is another way to solve it too.

Answer 1

The version of the proof my answer was based on:

MY ATTEMPT

According to the definition of continuity, for every $x_{0}\in X$ and every $\varepsilon > 0$, there is a $\delta > 0$ such that for every $x\in X$ it holds \begin{align*} d(x,x_{0}) < \delta \Rightarrow ||f(x)| - |f(x_{0})|| \leq |f(x) - f(x_{0})| < \varepsilon \end{align*} whence we conclude that $|f|$ is also continuous.

Its basically right since it has the key step (application of the reverse triangle inequality $||a|-|b||\le |a-b|$) but you compressed the entire proof into one line that doesn't make much logical sense. What you wrote is not the definition of continuity of $f$, and not of $|f|$, but a sentence that combines everything I mentioned thus far.

Properly written, maybe as a text book might have, it should first have the definition of continuity of $f$, then the above inequality, and then you point out that $|f|$ satisfies the definition of being continuous.