- #1

- 1,462

- 44

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

- #2

member 587159

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

- #3

- 1,462

- 44

Because of the vertical asymptote at 0?Why is ##\log|x|## not continuous?

- #4

member 587159

Because of the vertical asymptote at 0?

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

- 334

- 61

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.I've learned that composition of continuous functions is continuous

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

WWGD

Science Advisor

Gold Member

- 5,518

- 4,198

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:

- #7

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,154

- 1,349

Share: