Show every bounded infinite set has a maximum limit point and a minimum limit point. Show every bounded infinite set has a maximum limit point and a minimum limit point.
Here is my thought even if it is not formal
Let $S$ be bounded and infinite set.
Bolzano–Weierstrass theorem: Every bounded and infinite set has a limit point.
Since it is bounded by completeness property(Can I apply?) the set has least upper bound(Sup(S)) and greatest lower bound(Inf(S)).
 Now my claim is that maximum limit point$=Sup(S)$ and minimum limit point$=Inf(S.)$
I need someone to tell me how to proceed.
 A: Can you use the fact that the set of limit points is closed? If so, you can say that there are sequences in it that converge to its sup and its inf. Since its closed, both inf and sup are in it.
A: Let $L$ be the set of limit points (which is bounded, since $S$ is bounded). Let's show that $\sup L\in L$.
Given $\varepsilon>0$, take $l\in L$ with $|\sup L-l|<\varepsilon/2$, and since $l$ is a limit point, we can take $s\in S$ with $|s-l|<\varepsilon/2$. Thus, $|\sup L-s|\leq|\sup L-l|+|l-s|<\varepsilon$. This means that $\sup L\in L$ (this is actually the argument that show that $L$ is closed). Similarly, $\inf L\in L$.
A: Let $S'$ denote the set of all limit points of $S$. Here are some facts.

*

*If $S$ is any subset of $\mathbb R$, then $S'$ is a closed set.


*If $S$ is bounded, then $S'$ is bounded.


*If $S$ is bounded and infinite, then $S'$ is nonempty.


*A closed and bounded subset of $\mathbb R$ is compact.


*A nonempty compact subset of $\mathbb R$ has a maximum element and a minimum element.
It follows directly from the listed facts that, if $S$ is a bounded infinite subset of $\mathbb R$, then $S'$ has a maximum element and a minimum element. Which of those facts do you already know, and which ones do you need proved?
