# Exercise 3.6.6 Introduction to Real Analysis by Jiri Lebl

Suppose $$S \subset R$$ and $$f: S \to R$$ is an increasing function. Prove:

a) If $$c$$ is a cluster point of $$S \cap (c, \infty)$$, then $$\lim_{x \to c^+} f(x) < \infty$$.

b) If $$c$$ is a cluster point of $$S \cap (-\infty, c)$$ and $$\lim_{x \to c^-} f(x) = \infty$$, then $$S \subset (-\infty, c)$$.

For a), I know that $$\lim_{x \to c^+} f(x) = \inf \{f(x) : x\in S, x>c\}$$. I need to show that this is bounded in some way.

For b), I know that $$\lim_{x \to c^-} f(x) = \sup \{f(x) : x \in S, x .

This question is difficult to me. I appreciate if you give some help.

For part (a), Since $$f$$ is increasing, we can say $$f(x) \geq f(c)$$ whenever $$x > c$$. Can you take it from here?
Take $$s \in S$$. Suppose for contradiction that $$s \geq c$$. Then $$f(s) \geq f(c) \geq f(x)$$ for all $$x < c$$. Can you see why this is a problem?
• For a), $f(c)$ is a lower bound for $\{f(x) : x\in S, x>c\}$, but how does this help to show that $\{f(x) : x\in S, x>c\}$ is bounded from above? For b), how do we know that $f(s) < \infty$? – shk910 Dec 22 '19 at 4:03
• $f(c)$ might not even be defined, so we can't use it. – bjorn93 Dec 22 '19 at 4:42
• That's a good point. The proof will take a little more work, but the idea is to replace $c$ by a point $y \in S$ less than, but arbitrarily close to $c$, which must exists since $c$ is a cluster point. You'll probably have to use an $\epsilon$ at some point. – Charles Hudgins Dec 22 '19 at 5:05
I think for the b), if not overthinking it, it can be reasoned in this way: Say, $$x_{0}\in S$$ and $$x_{0}\geq c$$, then $$f(x)\leq f(x_{0})$$ for all $$x\in S\cap(-\infty,c)$$, so $$\lim_{x\rightarrow c^{-}}f(x)\leq f(x_{0})<\infty$$, a contradiction.