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Although this seems intuitive, I don't quite see how to prove this.

A set $A$ is closed provided if $a_n \in A$ with $a_n \to p$, then $p \in A$. Since $A$ is bounded, then any nonempty $a_n \in A$ is also bounded so it has an LUB and GLB. But how do we know the LUB and GLB are within A?

Also, the empty set is both closed and bounded, but how can it contain its LUB and GLB when it is inherently empty? Perhaps the empty set must be excluded?

Edit: Title.

Mr Defect
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    Check the definition of LUB and GLB again. If I remember correctly, there is a nice sequence involved in their definitions. Try using that along with what you noted about closed sets. – Sinister Cutlass Sep 24 '15 at 00:10

2 Answers2

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The LUB and the GLB are only defined for non-empty sets, so you have to exculde the empty set.

Now, let $A$ be a non-empty closed and bounded set. Then $A$ has a LUB and a GLB. Put $l:=\sup A.$

Let's assume that $l\notin A.$

Now let $\varepsilon>0$ be arbitrary. Since $l-\varepsilon<l$ and since $l$ is the least upper bound of $A$ then $l-\varepsilon<a$ for some $a\in A.$ Since $l\neq a$ (because $l\notin A$) and $a<l$ (because $l$ is an upper bound of $A$) then $0<|l-a|<\varepsilon$ and hence $l$ is a limit point of $A.$ Since $A$ is closed then $l\in A,$ which contradicts our original assumption that $l\notin A.$ Thus $l\in A.$ The same proof applies for $\inf A.$

CIJ
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Proof without the method of contradiction.Let $x=\text {lub } A$. Case 1: For some $y<x$, the open interval $(y,x)$ is disjoint from $A$. Then $x \in A$ otherwise $y$ is an upper bound for $A$ which is less than $\text {lub } A$, which is absurd. Case 2 : For every $y<x$ we have $(y,x) \cap A \ne \phi$. Then for each $n \in N$ choose $z_n \in (x-2^{-n},x)\cap A$. Since $A$ is closed,and $\{z_n : n \in N\} \subset A $ and $\lim_{n \to \infty} z_n =x$, we have $x \in A$.

DanielWainfleet
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  • Nice proof, but you are using the method of contradiction when you consider the first case (it is a small detail though and the whole proof "doesn't use contradiction"). – CIJ Sep 24 '15 at 01:43
  • It is almost impossible to write a proof of anything without that method somewhere in it.What I mean is that I did not aim to derive a contradiction from assuming $x \not \in A$ at the outset. – DanielWainfleet Sep 24 '15 at 14:00