Actual Question is :

On Which of the following spaces is every continuous (real valued) function Bounded?

  1. $X_1=(0,1)$
  2. $X_2=[0,1]$
  3. $X_3=[0,1)$
  4. $X_4 =\{t\in [0,1] : t \text { is irrational}\}$.

I could see that $1,3$ are spaces in which not every continuous function is bounded.

I could see $2$ is a spacein which not evry continuous function is bounded.

For First option I have seen that $f(x)=\frac{1}{x}$ works as it is continuous but not bounded.

For Second option I see that continuous image of compact set is compact i.e., $f[0,1]$ is compact thus bounded.

For third option I could see that as $1$ is exculded in the domain, function is unbounded if it has a pole at $1$ So, I see that $f(x)=\frac{1}{x-1}$ would help.

For fourth option I had no clue (what i had does not help much i guess).

I would be thankful if someone can help me in this.

As this is not a serious problem I would request users to just give me hints and not to post as an answer.

Thank you.

  • 1
    $\begingroup$ This is hard to do without giving it away. You need a function that blows up at a single rational point. $\endgroup$ – copper.hat Nov 23 '13 at 5:39
  • $\begingroup$ Well, $\sqrt{2}$ is not rational... $\endgroup$ – copper.hat Nov 23 '13 at 5:43
  • $\begingroup$ Oh my bad... I actually meant to say $f(x)=\frac{1}{x-2}$... I do not really know at present that this is continuous at all irrationals... I would check that... $\endgroup$ – user87543 Nov 23 '13 at 5:44
  • $\begingroup$ Well, pick any irrational $\alpha$. Then you know that $|\alpha -\frac{1}{2}| >0$, so you can find a neighbourhood of $\alpha$ that is bounded away from $\frac{1}{2}$... $\endgroup$ – copper.hat Nov 23 '13 at 5:48
  • $\begingroup$ yes yes.. now i understood.. Thank you... $\endgroup$ – user87543 Nov 23 '13 at 5:49

You've got the basic idea - you can just generalize it.

Suppose $X\subset \mathbb{R}$ is such that every continuous function on $X$ is bounded. I claim that $X$ is both closed and bounded.

  1. Suppose $c \in \overline{X}\setminus X$, then the function $$ f(x)= \frac{1}{x-c} $$ will be unbounded. Hence, $X$ must be closed.

  2. If $X$ is not bounded, just take $f(x) = x$, then $f$ is unbounded.

Hence, $X$ must be compact.

Conversely, if $X$ is compact, every continuous function is bounded, so just check which of these sets are compact.

  • $\begingroup$ I really like the idea "I claim that X is both closed and bounded.".. This is more useful now... Thank you... :) $\endgroup$ – user87543 Nov 23 '13 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy