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So here is my proof, which after looking up others seems to be too simple, or not rigourous enough, though I don't see why (hence I am asking!):

We take the contrapositive, and so prove that if $f$ is unbounded on $[a,b]$, then it is not continuous on this interval.

By hypothesis, $\exists c \in [a,b] : \displaystyle\lim_{x\to c} f(x) = \infty \implies f(c)$ is undefined. (or, $ \displaystyle\lim_{x\to c} f(x) \not= f(c)$ so we get discontinuity instantly)

Since $f(c)$ doesn't exist, the definition of continuity cannot be applied so $f$ must be discontinuous at at least $x=c$, as required.

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  • $\begingroup$ $f(c)$ is defined. It has a finite value. But you can use the fact that $f$ is unbounded about $c$ to show $f$ is not continuous at $c$. Also, you should explain why such a $c$ exists (use the compactness of $[a,b]$). $\endgroup$ – David Mitra Aug 12 '13 at 16:38
  • $\begingroup$ How can $f(c)$ be defined? If all numbers in the image of the interval are defined (and so finite), then one simply picks the largest and one gets the bound! $\endgroup$ – FireGarden Aug 12 '13 at 16:42
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    $\begingroup$ "By hypothesis, $\exists c \in [a,b] : \displaystyle\lim_{x\to c} f(x) = \infty$"... No, this is not the hypothesis. The hypothesis is that $f$ is unbounded. If you think this implies what you wrote, you should show that it does. $\endgroup$ – Did Aug 12 '13 at 16:46
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    $\begingroup$ $f$ unbounded on $S \subseteq \Bbb R$ doesn't imply that $\exists c \in S: \displaystyle\lim_{x\to c} f(x) = \infty$ (or $-\infty$). Ex1: $S= (0,1], f(x)= \frac 1x$. Ex2: $S= \Bbb R, f(x)=x$ $\endgroup$ – walcher Aug 12 '13 at 16:48
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    $\begingroup$ @FireGarden that's exactly the point, though. Somewhere you need to use that $[0,1]$ is closed rather than half-open/open. To be more specific, if your lemma $\exists c \in [a,b] : \lim_{x \to c} f(x) = \infty$ is true because $[a,b]$ is a closed interval, can you prove it? $\endgroup$ – Thomas Belulovich Aug 12 '13 at 17:12
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The contrapositive you wrote down is incorrect, but the basic idea there can be adapted to give a proof, as I show below.

To be precise, what you wrote down is strictly stronger than the genuine contrapositive. To know that such a $c$ even exists, you need to use the sequential compactness of $[0,1]$. See the wiki on the Heine-Borel Theorem.

The original statement is that for a continuous map $f : [0,1] \rightarrow \mathbb{R}$, there exists a constant $M > 0$ such that $|f(x)| \leq M$ for all $x \in [0,1]$.

The structure of this sentence might seem peculiar from a natural-language standpoint, but it's precise enough that we can form a contrapositive: if a map $f : [0,1] \rightarrow \mathbb{R}$ has the property that for any $M > 0$ there exists $x \in [0,1]$ such that $|f(x)| > M$, then $f$ is not continuous.

Let's give a proof now. Suppose $f$ has this property. For each $M \in \mathbb{N}$, let $x_M$ denote any choice of an element of $[0,1]$ for which $|f(x_M)| > M$. The compactness of $[0,1]$ ensures the existence of a convergent subsequence $x_{M_k} \rightarrow x^* \in [0,1]$, where $M_k \rightarrow \infty$.

Apply the continuity of $f$ at $x^*$: $$ \infty > |f(x^*)| = \lim_{k \rightarrow \infty} |f(x_{M_k})| \geq \lim_{k \rightarrow \infty} M_k = \infty $$ which is a contradiction.

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    $\begingroup$ I don't understand. How can the statement $\infty > \infty$ really make sense? I don't see why this is definitely a contradiction. $\endgroup$ – FireGarden Aug 12 '13 at 17:46
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    $\begingroup$ That sentence is a bit terse. What's happening is that there is a finite real number $|f(x^*)|$ with the property that it is greater than any natural number. $\endgroup$ – A Blumenthal Aug 12 '13 at 17:48
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This addresses the proof proposed by the OP, showing that there exists some positive functions $f$ unbounded on $[0,1]$ such that the assertion $\lim\limits_{x\to c}f(x)=+\infty$ does not hold.

Choose some irrational $z$ in $[0,1]$ and define $f$ on $[0,1]$ by $f(x)=0$ if $x$ is irrational and $f(x)=1/|x-z|$ if $x$ is rational.

Then, $f$ is unbounded while, for every $c$ in $[0,1]$ there exists some $x$ as close to $c$ as one desires such that $f(x)=0$. Hence $\lim\limits_{x\to c}f(x)=+\infty$ does not hold.

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Your argument is basically correct but the claim of the the existence of a $c$ such that $\displaystyle\lim_{x\to c} f(x) = \infty$ needs to be modified and justified. This does not follow immediately from the hypothesis that $f$ is unbounded. Namely, the unboundedness of $f$ implies the existence of a sequence $(x_i)$ such that $f(x_i)$ tends to infinity. However, the sequence may not be a convergent sequence. The fact that it contains a convergent subsequence is the key step in the proof. The existence of such a subsequence follows from the compactness of the interval. It follows that the limit $\displaystyle\lim_{x\to c} f(x)$ does not exist. Hence the function is not continuous at $c$.

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