1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Connected subset

  1. Oct 19, 2010 #1
    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. jcsd
  3. Oct 19, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What have you tried so far? Do you know what connected subsets of R look like?
  4. Oct 19, 2010 #3
    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.
  5. Oct 20, 2010 #4
    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!
  6. Oct 20, 2010 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    More specifically than just this, the only connected subsets of R are intervals
  7. Oct 20, 2010 #6
    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.
    Last edited: Oct 20, 2010
  8. Oct 21, 2010 #7
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Connected subset
  1. Subspace and subset (Replies: 10)

  2. Subset and subspace (Replies: 4)

  3. Subset and subspace (Replies: 1)

  4. Subsets question (Replies: 2)