Statement: Let $I=[c,b]$ be a closed bounded interval and $f : I\to \mathbb R$ be continuous on $I$.Then $f$ is bounded on $I$.
Proof: Proving by contradiction, Let's assume that $f$ is not bounded in $I$
If $f$ is not bounded then by definition, there exist $a$ $\in I$ such that $|f(a)|>M$ for some $M>0$. Since $a$ is a cluster point in $I$ and therefore there exist a sequence $(x_{n})\subseteq I$ whose limit is $a$. By Sequential Criterion of limits and function being continuous concept sequence $(f(x_n))$ converges to $f(a)$ and this is true $\forall a\in I$.
As every convergent sequence is bounded,therefore this stands as an contradiction to the assumption we made.
Therefore $f$ is bounded in $I$

If my proof is right then it can be shown that with $I=(c,d)$ also function is bounded, which is wrong according to the many books and websites that I came across.So where have I gone wrong in my proof?
Thanks in advance.

  • $\begingroup$ What do you mean by "If $f$ is not bounded then by definition, there exist $a$ $\in I$ such that $|f(a)|>M$."? $\endgroup$
    – Math_user
    Apr 26, 2020 at 11:31
  • $\begingroup$ The theorem is a consequence of the continuous functions in closed and bounded interval. Throw any one of the conditions of the domain interval out, the result won't be same. $\endgroup$
    – Itachi
    Apr 26, 2020 at 11:46

3 Answers 3


First, you start by talking about a number $M$ without saying which number is that. And then you say that you reach a contradiction because the sequence $\bigl(f(x_n)\bigr)_{n\in\Bbb N}$ is bounded, without saying where is the contradiction.

And you don't need sites or textbooks to see that the map$$\begin{array}{ccc}(0,1)&\longrightarrow\Bbb R\\x&\mapsto&\frac1x\end{array}$$is unbounded.

  • $\begingroup$ regarding 'M' it is for some positive real number and secondly the proof never mentions (f(x_n)) is unbounded.So I don't still get where am I wrong? $\endgroup$ Apr 26, 2020 at 11:41
  • $\begingroup$ Now you have edited your question about $M$. And, yes, you did not claim that that sequence is unbounded. I've edited my answer. Why do you say that there is a contradiction? I don't see it. And you are still wrong by the example at the end of my answer. $\endgroup$ Apr 26, 2020 at 11:46
  • $\begingroup$ Thankyou, I get it now there is no contradiction.But if I directly start off my taking 'a' as a cluster point and continue my proof will that suffice to prove the theorem because at the end I proved $f$ to be bounded?(I am sorry if this sounds silly to you) $\endgroup$ Apr 26, 2020 at 12:08
  • $\begingroup$ You cannot possibly prove it because, as you wrote it yourself, then the proof would also apply to functions whose domain is an interval $(c,d)$. And then the statement is false. $\endgroup$ Apr 26, 2020 at 12:11
  • $\begingroup$ Thanks a ton( I get it why I am wrong).Is a closed interval only a necessary condition?If you consider $f(x)=1/x$ with domain (2,3) then $f$ is bounded. $\endgroup$ Apr 26, 2020 at 12:19

I think what You are trying to do is to assume $f$ is unbounded and conclude there must exist a sequence $\{x_n\}_{n=1}^{\infty}\subset I$ such that $|f(x_n)|>n$ for all $n\in\mathbb{N}$. This sequence has a convergent subsequence by the theorem of Bolzano-Weierstraß and since $I$ is closed, it's limit must be in $I$. Can You take it from here?

  • $\begingroup$ Well, this is almost identical to what is given in the book(and was able to understand it too ),but I was trying to prove it by myself. So am I right? If yes, can you answer the question at the end? $\endgroup$ Apr 26, 2020 at 11:48
  • $\begingroup$ Yes, I can, and @José Carlos Santo already did that: There is no contradiction in Your "proof" $\endgroup$ Apr 26, 2020 at 11:50

You need to understand the meaning of "$f$ is bounded on $I$". This means that "there is a constant $K>0$ such that $|f(x) |<K$ for all values of $x\in I$".

According to your approach the meaning of "$f$ is bounded in $I$" is "for every $x\in I$ the value $f(x) $ is bounded". The fact is that if $f$ is defined at some point $x$ then $f(x) $ is some specific real number and in some intuitive sense is bounded (the right term is $f(x) $ is finite). Based on your understanding every function is bounded on its domain.

A function does not become unbounded by taking values like $\pm\infty $ at some point. That's precisely because $\pm\infty $ are not exactly values which can be taken by some function. The famous example of unbounded function is $f(x) =1/x$ on $(0,1)$. For each $x\in(0,1)$ the value $f(x) =1/x$ is just a real number. The fact which makes it unbounded is that given any number $K>0$ we can find an $x$ with $|f(x) |>K$. Let's try with $K=10^{100}$ (pretty big number: one followed by hundred zeroes) and we see that if $x=10^{-1000}$ then $f(x) =10^{1000}>K $. You can see that this works for any chosen $K>0$ (choose $x$ with $0<x<1/K$).

You show that for each $a\in I$ we have a sequence $x_n\to a$ and $f(x_n) \to f(a) $ and convergent sequence being bounded lead to conclusion that $f(a) $ is bounded. As I mentioned above this is trivially the case as any specific real number is bounded / finite.

There are many proofs of the result you seek based on various forms of completeness of real numbers. Here is the one I like most based on nested interval principle.

Assume that $f$ is unbounded on $I_0=I$. Divide the interval $I_0$ into two closed intervals via midpoint and then $f$ is unbounded on at least one of these intervals (why?) say $I_1$. Repeat the procedure to get a sequence of closed intervals $I_n$ such that $f$ is unbounded on each $I_n$, $I_{n+1}\subseteq I_n$ and length of $I_n$ tends to $0$. Under these conditions the nested interval principle guarantees a unique point $c$ which lies in all intervals $I_n$. Further for every neighborhood $J$ of $c$ there is some interval $I_n$ such that $I_n\subseteq J$. By continuity of $f$ at $c$ there is a neighborhood $J$ of $c$ in which $f$ is bounded. And thus it is bounded in some $I_n\subseteq J$. This contradiction completes the proof.


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