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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.

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  • $\begingroup$ No, that's not right. For instance, let $S = \{0,3\} \cup [1,2]$. Then $\sup S = 3$, but $3$ is not a limit point of $S$. Instead, you want the least upper bound (resp. greatest lower bound) of the limit points of $S$. $\endgroup$ – TonyK Nov 28 '14 at 20:39
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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.

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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, $\inf L\in L$.

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