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If $f(x)$ takes a finite real value for all $x$ on the closed interval $[a,b]$, must there be a real number $M$ such that $M\geq f(x)$ for all $x$ on this interval? It seems that if not, there must be a point $c\in[a,b]$ such that $\lim_{x\to c}f(x)=+\infty$, and so $f(x)$ must be undefined at some point on this interval, but I don't know how to make this rigorous.

Edit: I see that $f(0)=0$, $f(x)=1/x$ on $(0,1]$ is a counterexample. I also see that I have been imprecise with terminology. Let me modify the question: Is there always a sub-interval $[a',b']$ with $a such that there is an upper bound for $f(x)$ on this sub-interval?

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    Your title is different from the question in the post: $M\ge f(x)$ does not mean that $M$ is maximum, it is an upper bound. In the other words, if such $M$ does not exist, $f$ is not bounded from above.2012-11-02
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    If $f$ is continuous, then it must be bounded, see [ProofWiki](http://www.proofwiki.org/wiki/Max_and_Min_of_Function_on_Closed_Real_Interval). It seems that in your post you use the assumption $f(c)=\lim\limits_{x\to c}f(x)$, which is an equivalent condition for continuity of a function $f \colon \mathbb R \to \mathbb R$.2012-11-02

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