Use the Bolzano-Weierstrass Theorem which is Every bounded sequence has a convergent subsequence to prove the following:

A continuous function defined on a closed, bounded interval must be bounded. That is, let $f$ be a continuous function defined on $[a,b]$. Then there exists a positive real number $B$ such that $|f(x)|≤ B$ for all $x ∈ [a,b]$.

could you please help me how can I prove it?

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    $\begingroup$ This is such a poor question that was simply copy-pasted (look at the in the question instead of fi), with no context and no effort shown. Did you read the How-to-ask page that is shown when you signed up, including the linked How to ask a good question page? $\endgroup$
    – user21820
    Apr 3, 2017 at 17:55

2 Answers 2


The prototypical proof is a proof by contradiction. We assume that $f$ is continuous and unbounded on $[a,b]$.

Then, for any $n$ there exists a number $x_n\in [a,b]$ such that $f(x_n)>n$.

Since $x_n$ is bounded, there exists a subsequence, say $x_{n_k}$ that converges to a number $x_0\in [a,b]$.

Inasmuch as $f$ is continuous on $[a,b]$, it is continuous at $x_0$ and hence $f(x_{n_k})\to f(x_0)$.

This means that $f(x_{n_k})$ is bounded which contradicts the statement that $f(x_{n_k})>n_k$.

Therefore, $f$ is bounded on $[a,b]$.

  • $\begingroup$ however even unbounded sequences can have a converging subsequence, so that makes this proof wrong $\endgroup$ Jun 10, 2020 at 0:11
  • $\begingroup$ @josephrock What on earth are you babbling about? The sequence $x_{n_k}\in [a,b]$ and is BOUNDED! So the proof is valid! $\endgroup$
    – Mark Viola
    Jun 10, 2020 at 2:27
  • $\begingroup$ i am saying a sequence that is unbounded can have a converging subsequence too , so this prove would work for the unbounded sequence too, however it is clear that an unbounded sequence cannot be bounded too $\endgroup$ Jun 10, 2020 at 2:29
  • $\begingroup$ there could be f(x) that are unbounded as the sequence x(nk) does not take into account all the x values $\endgroup$ Jun 10, 2020 at 2:36
  • $\begingroup$ @josephrock You said "so that makes this proof wrong." The proof is valid. The fact that an unbounded sequence can have a convergent subsequence is irrelevant. Not ALL unbounded sequences have a convergent subsequence. So, again, what are you talking about? $\endgroup$
    – Mark Viola
    Jun 10, 2020 at 2:37

Let $M=\sup f([a,b])$ ($M$ can eventually be $\infty$), we will show that this $M$ is reached for a certain $x\in[a,b]$ thus $M$ will be finite.

By definitions of $M$, $\exists (x_n)_{n\in\mathbb{N}}\in[a,b]$ such that $f(x_n)\rightarrow M$.

$x_n$ is bounded so we can apply Bolzano-Weierstrass theorem, let $x_{\phi(n)}\rightarrow d\in [a,b]$, where $(x_{\phi(n)})_{n\in\mathbb{N}}$ is a subsequence.

$f$ is continuous in $d$ so $f(x_{\phi(n)})\rightarrow f(d)$. By unicity of the limit $M=f(d)$ thus $M$ is finite. Can do the same with $m=\inf f([a,b])$. And finally take $B=\max(|m|,|M|)$ and $B<\infty$ as required.

  • $\begingroup$ the complete of this proof will be look like that ,Let m=inf f([a,b]) , we will show that this m is reached for a certain x∈[a,b] thus m is finite. By definitions of m, ∃(xn)∈[a,b] such that f(xn)→m. So , there exists a positive real number B=max(|m|,|M|) such that |f(x)|≤ B for all x ∈ [a,b]. is it correct? $\endgroup$
    – qwer tyui
    Apr 3, 2017 at 17:01
  • $\begingroup$ You still need to extract a convergent subsequence from $(x_n)$ with Bolzano Weierstrass. $\endgroup$
    – Bérénice
    Apr 3, 2017 at 17:03
  • $\begingroup$ like what you did above ? $\endgroup$
    – qwer tyui
    Apr 3, 2017 at 17:05
  • $\begingroup$ yes it is exactly the same reasonning $\endgroup$
    – Bérénice
    Apr 3, 2017 at 17:05

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