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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 $\lim_{x \to c^-} f(x) = \sup \{f(x) : x \in S, x <c\}$.

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

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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?

By the way, I'm not trying to be cute with these questions. If you want a little more direction, let me know.

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  • $\begingroup$ 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 $f(s) < \infty$? $\endgroup$ – shk910 Dec 22 '19 at 4:03
  • $\begingroup$ $f(c)$ might not even be defined, so we can't use it. $\endgroup$ – bjorn93 Dec 22 '19 at 4:42
  • $\begingroup$ 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. $\endgroup$ – Charles Hudgins Dec 22 '19 at 5:05
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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.

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