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I read the following from a website (https://www.askiitians.com/iit-jee-differential-calculus/limits-continuity-differentiability/preparation-tips.html):

"A function continuous on a closed interval [a, b] is necessarily bounded if both a and b are finite. This is not true in case of open interval."

My question is, why would the function become unbounded just by removing two finite numbers {a,b} from its domain? Is the statement even correct?

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    $\begingroup$ Because of the possibility of vertical asymptotes at these points $\endgroup$
    – MPW
    Commented Dec 20, 2018 at 14:56

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Yes, the statment is correct. Of course, if $f\colon[a,b]\longrightarrow\mathbb R$ is bounded, the its restriction to $(a,b)$ is still bounded. But, for instance, the function$$\begin{array}{rccc}f\colon&(a,b)&\longrightarrow&\mathbb R\\&x&\mapsto&\dfrac1{(x-a)(x-b)}\end{array}$$is unbounded (but it is continuous).

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  • $\begingroup$ According to the definition of bounded functions from Wikipedia, a function is bounded if there exists a real number M such that |f(x)| is less than or equal to M, where x lies in domain of f. For every point in the above function's domain (a,b) there will always exist a real number that is greater than the value of the function at that point. So shouldn't it be bounded on (a,b)? Since the only points where we will not be able to find the real number M are f(a) and f(b) and the domain does not include a and b $\endgroup$
    – Harsh
    Commented Dec 20, 2018 at 15:23
  • $\begingroup$ Take $M\in\mathbb{R}$. If, $M$ is large enough, then the equation $\bigl\lvert f(x)\bigr\rvert=M$ has two solutions in $(a,b)$ and the inequation $\bigl\lvert f(x)\bigr\rvert>M$ has infinitely many solutions there. $\endgroup$
    – José Carlos Santos
    Commented Dec 20, 2018 at 15:28
  • $\begingroup$ Isn't there an ambiguity in those definitions? For however close we bring x to a or b, f(x) will still be real, and we will still be able to define a real number M that is greater than f(x). In that case, f(x)=M and f(x)>M both will have no solution since we have defined M that way. $\endgroup$
    – Harsh
    Commented Dec 20, 2018 at 15:42
  • $\begingroup$ @Harsh for $f(x)$ to be bounded then you need $|f(x)| \le M$ for all $x$ in the domain of the function; you can vary $M$ for different $f$, but given $f$ and its domain you cannot vary $M$ with $x$ $\endgroup$
    – Henry
    Commented Dec 20, 2018 at 17:30
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Consider:

  • Is $\dfrac1{x(1-x)}$ continuous on $(0,1)$

  • Is $\dfrac1{x(1-x)}$ bounded on $(0,1)$?

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  • $\begingroup$ It is bounded on (0,1) and unbounded on [0,1], if I'm applying the definitions mentioned in my earlier comments on the above answer correctly. It's now clear that the statement in the question I asked is right, but I still don't understand where am I going wrong. $\endgroup$
    – Harsh
    Commented Dec 20, 2018 at 15:45
  • $\begingroup$ @Harsh: It is not bounded on $(0,1)$: for $x=0.01$ or $0.99$ you get about $101$; for $x=0.0001$ or $0.9999$ you get about $10001$ and it increases without limit close to the edges of $(0,1)$ $\endgroup$
    – Henry
    Commented Dec 20, 2018 at 17:17

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