
#1
Oct1910, 07:29 PM

P: 123

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




#2
Oct1910, 08:18 PM

Mentor
P: 4,499

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




#3
Oct1910, 08:45 PM

P: 123

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




#4
Oct2010, 03:04 PM

P: 123

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!




#5
Oct2010, 06:11 PM

Mentor
P: 4,499





#6
Oct2010, 08:25 PM

P: 123

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)[/tex] or [tex](a, \infty)[/tex] 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.




#7
Oct2110, 12:24 PM

P: 123

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


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