Talk:Subset of Real Numbers is Interval iff Connected

This is the theorem that I was searching for ... I understand every thing except the last line in the proof ... But A ′ ∪B ′ =[a..b]

why A' U B' = [a,b] ??? anyone can explain this for me please ?? and I think there is a mistake in this line : Since a∈A ′ and A∪B=∅, it follows that b ′ >a. it should be A ∩ B = ∅ right ?--Noor 17:51, 13 June 2011 (CDT)


 * Good comment. It looks strange to me too. I will get back to it when I have time to think about it.
 * I have fixed your second point, it was a mistake. Should have been $\cap$.--prime mover 00:41, 9 June 2011 (CDT)

I added some lines that hopefully clarified the reason there was a contradiction at the end. Unrelated: we probably want to clarify that this is using the standard topology on the reals, and not just any topology. I'm not sure whether that should be made into a point here or if we should add to the definition page for the real numbers what the standard topology is. -- Qedetc 12:06, 9 June 2011 (CDT)
 * It's already there (indirectly) in the page you go to from Definition:Real Number Line. But it would do no harm to expand that point, as this is clearly a useful page which gets a fair few hits and therefore may well need clarification.
 * I have it in mind to draw a little explanatory picture, which I was in the process of doing this morning when I had a few minutes to look at it then. Having said that, it won't be tonight as I'm too tired for intricate work like that ...

Thank you very much about everything ... I like this proof .. but can I ask a question plz ..

Is it always when A and B are a separation of S then A and B are open and closed ?????? and why ??

And why do you say that (a ′′ ..b ′ ) is nonempty in the proof ? bcz it is an interval in real numbers ??--Noor 17:51, 13 June 2011 (CDT)


 * By definition, if $A$ and $B$ form a separation of $S$, then $A,B$ are disjoint open sets such that $S = A \cup B$. But, also by definition, closed sets are the same thing as complements of open sets. Since $A = S-B$ and $B = S-A$ (this uses disjointness), this shows that they're both closed as well.
 * (a ′′ ..b ′ ) is nonempty since $a'' < b'$, and there's always something between unequal reals, which is proven here Reals are Close Packed.
 * -- Qedetc 22:11, 12 June 2011 (CDT)

aha Yes Now I understand ur point, E is open iff complement of E is closed so, if S = A∪B and A,B are separation of S then they are disjoint open sets but complement of A is B and it is closed and complement of B is A and it is also closed so, both A & B apen and closed.

really, thanx for ur helping, answering and ur clarification.--Noor 17:51, 13 June 2011 (CDT)