Representation Theorem For Compact set in R

Question

I am trying to prove following
A non empty bounded closed set S in R is either closed interval or can be obtained form closed interval by removing union of countable collection of open set whose end point belong to that set .

My Attempt: If closed interval then done .If Closed And bounded set but not interval Then Form bounded it must have some supremum and infimum and by closeness it must belongs to that set .
Now say M is sup S and m is inf S.$S\subset [m,M]=I$ {Proper set as we have done that case already}. $\exists x\in I $ such that $x\notin S$ So consider set T={x|x$\in I$, $x\notin S$} Which is non empty and bounded below by m.So it must has inf .That inf belong to S due to closeness .How to proceed further ?

Answer 1

I'm not happy with the sentence "I can visualize this that any compact set in ${\mathbb R}$ is either closed interval or finite point in ${\mathbb R}$". Whatever the exact intended meaning of this sentence: One of the most famous compact sets in ${\mathbb R}$, the Cantor set, has uncountably many components.

Nevertheless, the claim you want to prove is true and easy to prove. Take a compact set $K\subset{\mathbb R}$. The points $a:=\inf K$ and $b:=\sup K$ belong to $K$. The complement $\Omega:={\mathbb R}\setminus K$ is open. It consists of countably many disjoint open intervals, among them ${\mathbb R}_{<a}$ and ${\mathbb R}_{>b}$. Remove these from $\Omega$ and obtain $\Omega'$. Then $K=[a,b]\setminus\Omega'$, as desired.