Let $f(x)$ be a function such that it is unbounded on a closed interval $[a, b]$. Prove there exists a point from this interval in every neighborhood of which the function is unbounded.
Below is an idea based on the nested intervals. Let there be an interval $[a, b]$. Start with splitting this interval into two equal parts: $$ \left[a, \frac{a+b}{2}\right] \text{and}\left[\frac{a+b}{2}, b\right] $$
Since the function is unbounded there are 3 possible scenarios here:
- The function is unbounded on $\left[a, \frac{a+b}{2}\right]$
- The function is unbounded on $\left[\frac{a+b}{2}, b\right]$
- The function is unbounded on both parts of the interval.
In either case choose one of the intervals on which the function is unbounded. By repeating those steps we obtain a point which belongs to all the nested intervals. Let $c$ denote a common point.
But each interval was chosen in such a way that $f(x)$ in unbounded on that interval. We can now construct a sequence $\{x_k\}$ where $x_k \in [a,b]$ such that: $$ \lim_{n\to \infty}x_n = c \implies \lim_{n\to\infty} f(x_n) = \infty $$
Which would mean that $c$ is the point we are looking for and no matter what neighborhood of $c$ and $\epsilon > 0$ is chosen, $f(x)$ will always be unbounded for $x\in[c-\epsilon, c+\epsilon]$.
Please note this problem is given in the "Limit of the sequences" section and the limit of a function has not been introduced yet.
Apart from that I would like to ask for a verification or possibly other solutions just in case mine is not valid.
The missing item in your proof, and a point for which your professor might deduct some credit, would be "how do you know that $c$ is included in your interval (the set of points $[a,b]$)."
The point of the question is that you do know that $c$ is included in the interval because it is a limit point of a sequence of points in $[a,b]$ and a closed set includes all its limit points (by the definition of "closed set").