Connected Subset of Real Numbers: Bounded Below, Unbounded Above

[tarheelborn]

If S is a connected subset of \mathbb{R} and S is bounded below, but not above, then either S=[a, \infty) or S=(a, \infty) for some a \in \math{R}.

 

[Office_Shredder]

What have you tried so far? Do you know what connected subsets of R look like?

 

[tarheelborn]

Yes I am familiar with the types of connected subsets of \mathbb{R}. I know that if a<b \in S then [a,b] \subseteq S. I think I need to find some point in [a,b] and show that a is the least greatest lower bound of that interval whether or not a is included in the interval. But I am not sure how to start.

 

[tarheelborn]

I see that if a is the glb for S, then a is either in S or s is approaching a for s in S. And I see that there is no upper bound, so the right end of the interval is infinity. But I need help on writing that up formally. Please. Thank you!

 

[Office_Shredder]

Yes I am familiar with the types of connected subsets of \mathbb{R}. I know that if a<b \in S then [a,b] \subseteq S.

More specifically than just this, the only connected subsets of R are intervals

 

[tarheelborn]

I know. I must not be stating my problem clearly. I need to prove that a connected subset of R that is bounded below but not above is equal to either [a, \infty) or (a, \infty) specifically. I am supposed to use a lemma that says if two points are in a connected subset of the real numbers, then all points in between these two points are also in the subset.

 

[tarheelborn]

I have gotten this far with the proof:


Since S is bounded below, S has a greatest lower bound, say a. Since S is not bounded above, I claim that S=(a, \infty)[/math] or S=[a, \infty). <br /> Case 1: Suppose a,x \in S such that a \neq x. Since a=g.l.b.(S), a&amp;lt;x. Now since S is unbounded, there is some s \in S such that s&amp;gt;x. but since a,s,x \in S, by previously proved lemma, [a,x] \in S and [x,s] \in S. Henc e S=[a, \infty).<br /> Case 2: Suppose a \notin S and suppose x \in S, x&amp;gt;a.<br /> <br /> Now I am not sure how to move on to say that everything approaching a is in S but a is not in S.