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I'm pretty stuck on the following question:

Let $f$ be an unbounded function from the closed interval $[a,b]$ to $R$. Prove that there exists a point $x$ in $[a,b]$ such that $f$ is unbounded in any neighborhood of $x$.

I understand that this comes naturally from the fact that $[a,b]$ is a closed interval - I'm sure that for any open/half open interval such a point $x$ does not have to exist. But I'm just not sure exactly why this is.

I tried assuming that such a point doesn't exist, which means that $f$ is bounded in any neighborhood of any point in $[a,b]$. The conclusion is that $f$ can't be unbounded (which is the contradiction I'm looking for), but I just can't take that extra step in order to conclude this (naturally I have to somehow derive this conclusion from the fact that the interval is closed).

Can somebody point me in a generally useful direction?

Thanks a lot in advance.

edit:

I think I came up with a general direction for a proof, but I'm not sure if it's the right way to go:

Let's look at the edges - $a$ and $b$. We need to ask - is there a neighborhood of $a$ and $b$ where $f$ is unbounded? If there is - we're done. If there isn't, this means that there is some $\epsilon_1$ neighborhood of $a$ and $b$ where $f$ is bounded, and so we'll define the closed interval $[a + \epsilon_1, b - \epsilon_1 ]$.

Now we'll ask the same question about $[a + \epsilon_1, b - \epsilon_1 ]$ - is there a neighborhood of the two edges where $f$ is unbounded? If there is - we're done. If not, there is some $\epsilon_2$ neighborhood of the two edges where $f$ is bounded, so we'll define the closed interval $[a + \epsilon_1 + \epsilon_2 , b - \epsilon_1 - \epsilon_2 ]$.

We'll continue in such a manner infinitely many times, until we can't do it any further. So (using Cantor's lemma) either we'll reach some single middle point x, or we'll reach some inner interval $[a', b'] \subseteq [a, b]$.

Now I claim that either way - we've found the point we're looking for - either x, a' or b' must be a point with no bounded neighborhoods. (but I'm not sure if there's still another step I need to take in order to show this)

What do you think?

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  • $\begingroup$ +1 for a very good question. Especially the fact the result in question does not deal with continuous functions is important as it emphasizes the key aspect of completeness. $\endgroup$ – Paramanand Singh Dec 20 '17 at 5:06
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For the sake of argument, suppose for every $x\in[a,b]$, $f$ is bounded in some neighborhood $U_x$ of $x$. Then $$ [a,b]=\bigcup_{x\in[a,b]} U_x $$ Now use compactness of $[a,b]$ to get a contradiction.

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  • $\begingroup$ I actually don't know what compactness is haha, I'm a complete beginner. could you please explain a bit further? $\endgroup$ – GSofer Dec 19 '17 at 16:35
  • $\begingroup$ Compactness tells you that you can choose finitely many $x_1,x_2,\cdots, x_n\in[a,b]$ such that $[a,b]=\cup_{k=1}^n U_{x_k}$. $\endgroup$ – Jack Dec 19 '17 at 17:04
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Well, the best approach here is to use the Nested Interval Principle. Since $f$ is unbounded on $[a, b] $, it is unbounded on at least one of the intervals $[a, (a+b) /2]$ and $[(a+b) /2,b]$. Call that interval $I_1$ and repeat the procedure to obtain a sequence of nested closed intervals $I_n$ such that $f$ is unbounded on each $I_n$. Since the length of $I_n$ tends to $0$ by Nested Interval Principle there is a unique point $c$ which lies in all intervals $I_n$. Now any neighborhood of $c$ contains some $I_n$ and hence $f$ is unbounded on any neighborhood of $I_n$.


The problem with your approach is that you never know when you can cover the entire interval by using steps of size $\epsilon$ which can be arbitrarily small. The nature of real line is not so simple to be handled by such naive arguments. You must resort to some form of completeness property of real numbers.

To convince yourself note that the result fails for rationals. The function $f:\mathbb{Q} \to\mathbb{Q} $ given by $f(x) =1/(x^{2}-2)$ is unbounded on the rational interval $I=\{x\mid x\in\mathbb{Q}, 1\leq x\leq 2\} $ but $f$ is bounded on every rational neighborhood (ie neighborhood containing only rational points) of any point in $I$. Your argument remains same even in this case also.

The argument of this answer is based on Nested Interval Principle which does not hold for the system of rational numbers. Thus in effect if we carry the process mentioned in the answer for rational intervals $I_n$ then there is no guarantee of a rational point $c$ which lies in all of $I_n$.

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  • $\begingroup$ That's a really cool idea. thanks! $\endgroup$ – GSofer Dec 20 '17 at 10:30
  • $\begingroup$ @GSofer: by the way you should also try to prove this using other forms of completeness like least upper bound property. $\endgroup$ – Paramanand Singh Dec 20 '17 at 10:39

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