## Connected subset

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 $$a \in \math{R}$$.

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 Blog Entries: 1 Recognitions: Homework Help What have you tried so far? Do you know what connected subsets of R look like?
 Yes I am familiar with the types of connected subsets of $$\mathbb{R}$$. I know that if $$a ## Connected subset I see that if a is the glb for S, then a is either in S or s is approaching a for s in S. And I see that there is no upper bound, so the right end of the interval is infinity. But I need help on writing that up formally. Please. Thank you! Blog Entries: 1 Recognitions: Homework Help  Quote by tarheelborn Yes I am familiar with the types of connected subsets of [tex] \mathbb{R}$$. I know that if $$a More specifically than just this, the only connected subsets of R are intervals  I know. I must not be stating my problem clearly. I need to prove that a connected subset of R that is bounded below but not above is equal to either [tex][a, \infty)$$ 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.
 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 $$S=(a, \infty)[/math] or [tex]S=[a, \infty)$$. Case 1: Suppose $$a,x \in S$$ such that $$a \neq x$$. Since $$a=g.l.b.(S)$$, $$ax$$. but since $$a,s,x \in S$$, by previously proved lemma, $$[a,x] \in S$$ and $$[x,s] \in S$$. Henc e $$S=[a, \infty)$$. Case 2: Suppose $$a \notin S$$ and suppose $$x \in S$$, $$x>a$$. Now I am not sure how to move on to say that everything approaching a is in S but a is not in S.