I have been reading a lot about real analysis and infinite sets lately. Infinite sets I mean the most common ones ($\mathbb{N}$ and $\mathbb{R}$). I have come across many problems and there are two which look very similar in terms of solving technique but I could not seem to figure them out.
Problem 1: Is there a function ($f:$ $\mathbb{R} \rightarrow \mathbb{R}$) whose limit at every point is $ \infty $ ?
Problem 2: Can a function ($f:$ $\mathbb{R} \rightarrow \mathbb{R}$) have continuum many local min/max values ?
(We are looking for local extrema values not points, so the constant function only has 1 in this regard)
So both problems ask questions about some properties maximum cardinality. Obviously countably finite is achieveable for both questions ($\tan{x}$ and $x\sin{(1/x)}$). I think the answer is negative for both questions.
For Problem 1 I think all values have to be 'large' for this to happen, namely $\infty$ but could not exactly prove this. I also tried to make a bijection with rational numbers on how many points can have a limit at infinity, but also could not finish.
For problem 2 I tried bijection with rationals but I could not make it so that for every rational I pair it with one Local min/max.
I feel like the thought process should be roughly similar for both but I could not find it. I may be completely on the wrong foot either by guessing that the answer must be negative.
Can someone tell me if my ideas can be finished, or if not then what way would work for proving these, and also if my guess about the answer is right.
Best Answer
Here is a proof for Problem 1:
Assume that the limit of $f$ at every point is $\infty$. Then, for every $M\in \mathbb{Z}$, the set $$\{x\mid f(x) \leq M\}$$ cannot have any accumulation point (*), which implies that it is countable (**). But on the other hand $$\bigcup_{M\in\mathbb{Z}} \{x\mid f(x) \leq M\} = \mathbb{R}$$ which is a contradiction since the LHS is countable.
(*) If the set $\{x\mid f(x) \leq M\}$ had an accumulation point $x^*$, then there would exist a convergent sequence $x_n \to x^*$ with $f(x_n) \leq M$, but this contradicts $\lim_{x\to x^*} f(x) = \infty$.
(**) If a set $A\subseteq \mathbb{R}$ is uncountable but has no accumulation point, then for some $K\in \mathbb{N}$ the set $A\cap[-K,K]$ would be uncountable and have no accumulation point, contradicting Bolzano-Weierstrass.
For problem 2, you are looking at strict local extrema, right? (Else a constant function would have extrema everywhere)
For pairing the local extrema: it might work to consider the a local extremum $x$ as global extremum of the function restricted to some interval $(x-\varepsilon, x+\varepsilon)$. Then (by making it smaller if necessary) you can have the interval bounds be rational. So $x$ is global extremum of $f$ on some interval $(a,b)$, $a,b \in \mathbb{Q}$. Since strict global min/max are unique, we can have at most one such $x$ for any interval with rational bounds. But there are only countably many such intervals, hence the number of strict local extrema of $f$ must be countable.
So the idea would be to pair local min/max to a pair of rational numbers, instead of a single rational number.