# 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.

• 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 $S$. – TonyK Nov 28 '14 at 20:39

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, $\inf L\in L$.