# Verification regarding Boundedness theorem of continuous functions

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?

• What do you mean by "If $f$ is not bounded then by definition, there exist $a$ $\in I$ such that $|f(a)|>M$."? Apr 26, 2020 at 11:31
• 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. Apr 26, 2020 at 11:46

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.

• 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? Apr 26, 2020 at 11:41
• 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. Apr 26, 2020 at 11:46
• 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) Apr 26, 2020 at 12:08
• 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. Apr 26, 2020 at 12:11
• 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. 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?

• 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? Apr 26, 2020 at 11:48
• Yes, I can, and @José Carlos Santo already did that: There is no contradiction in Your "proof" 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 $$|f(x) | for all values of $$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).

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.