If is bounded below but not bounded above, then is either or for some .
I know from a previously proved theorem that if is a connected subset of and , , then . And it seems intuitive that is the greatest lower bound for this set, but I am not sure how to begin to prove it.
Surely you left something out.
Did you mean that is connected?
I thought the same thing, but I am not sure. I have rewritten the problem as it was presented to me, but I feel sure that it must also mean S is a connected subset.
Surely. Otherwise it is false.
Originally Posted by tarheelborn
There is a well known theorem: The only nontrivial connected sets of real numbers are:
Only the last two fit the description of .
OK, then. If is a connected subset of and is bounded below, but not above, then is either or for some .
I am trying to prove the two parts of that theorem described in the last post. I believe I need to show that since is bounded below, it has a greatest lower bound and that said bound doesn't actually need to be included in the interval. But I am not sure how to get that started.
I have gotten this far with the proof:
Since is bounded below, has a greatest lower bound, say . Since is not bounded above, I claim that or .
Case 1: Suppose such that . Since , . Now since is unbounded, there is some such that . but since , by previously proved lemma, and . Henc e .
Case 2: Suppose and suppose , .
Now I am not sure how to move on to say that everything approaching a is in S but a is not in S.
I will help with this point.
If is connected then the closure, , is also connected.
So no limit point can be separated from the set.
Thus both are connected.