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.
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?