Continuous extension of a function

Question

Can anybody help me with this problem?

Justify whether the following statement is true or false:

Every continuous function on $\Bbb Q\cap [0,1]$ can be extended to a continuous function on $[0,1]$.

Any help will be appreciated.

score 13 · Accepted Answer · answered Jul 23 '13 at 06:58

13

See what happens if you take the function

$$f:\Bbb Q\cap[0,1]:\to[0,1]:x\mapsto\begin{cases} 1,&\text{if }x>\frac{\sqrt2}2\\\\ 0,&\text{if }x<\frac{\sqrt2}2\;. \end{cases}$$

answered Jul 23 '13 at 06:58

Brian M. Scott

616,228

2

I have got it.f(x)=1/(x^2-0.5) will also work.Thanks very much for your help. – Ester Jul 23 '13 at 07:09
@Ester: You’re very welcome. – Brian M. Scott Jul 23 '13 at 07:11
just to add a bit, you can for sure tell that a continuous extension exists if the subset $A\subset[0,1]$ you're extending from is closed (though it's not a necessary condition) – W_D Jul 23 '13 at 07:13
@Alex: Yes, it does; that’s a consequence of the Tietze extension theorem. – Brian M. Scott Jul 23 '13 at 07:14

score 3 · Answer 2 · answered Jul 23 '13 at 07:39

To add to Brian's answer, if $(X,d)$ is a metric space and $D$ is dense in $X$ and $f : D \rightarrow Y$ is a uniformly continuous function where $Y$ is a complete, metric space, then $f$ has a unique extension to $X$ and the extended function is also uniformly continuous.

In particular, in your case, every uniformly continuous function can be extended to a uniformly continuous function on the reals.

Continuous extension of a function

2 Answers2