# I Continuity of composition of continuous functions

1. May 9, 2017

### Mr Davis 97

I've learned that composition of continuous functions is continuous. $\log x$ and $|x|$ are continuous functions, but it seems that $\log |x|$ is not continuous. Is this the case?

2. May 9, 2017

### Math_QED

Why is $\log|x|$ not continuous?

3. May 9, 2017

### Mr Davis 97

Because of the vertical asymptote at 0?

4. May 9, 2017

### Math_QED

But how can you apply the theorem you just stated if $\log(x)$ isn't even defined at $x = 0$? If you start with false assumptions, surely you will get wrong results.

5. May 10, 2017

### zwierz

In this theorem you must also trace carefully domains of the functions. The exact formulation is as follows. Let $X,Y,Z$ be topological spaces and $f:X\to Y,\quad g:Y\to Z$ be continuous functions. Then the function $g\circ f:X\to Z$ is a continuous function.
In your case $f(x)=|x|,\quad g(y)=\log y,\quad X=\mathbb{R}\backslash\{0\},\quad Y=\{y\in\mathbb{R}\mid y>0\},\quad Z=\mathbb{R}$

6. May 13, 2017

### WWGD

But remember that the composition may not even be defined at some points, let alone be continuous. Try specifying the theorem a bit more carefully, as Zwierz , I think, suggested. EDIT: You may want to consider the inverse image of open/closed set is open/closed definition as well.

Last edited: May 13, 2017
7. May 15, 2017

### mathwonk

as suggested above, it is crucial to have a complete and precise statement of a theorem, including hypotheses. in your case the theorem says roughly, if f is continuous at the point p, and if also g is continuous at (and defined on a neighborhood of) the point f(p), then gof is (defined on a neighborhood of and) continuous at the point p. try to see what the points p and f(p) are in your case that give a problem, if any.