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Homework Help: A problem on infimum and supremum.

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem is attached. The official solution to this problem is a proof from the contrary.
    I decided to go the straight-forward way. Would you check if I am correct? Thank you in advance.

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Aug 2, 2011 #2
    It looks okay to me, although I'll confess I prefer the proof by contradiction. It seems somehow clearer. This method I had to think more about the definitions.
  4. Aug 2, 2011 #3
    Whoa, nevermind. Bah! I just typed a bunch of nonsense.


    First: to say that a nonempty set of reals T is bounded below means that there exists a [itex]v[/itex] such that [itex] v\leq t[/itex] for all [itex]t\in T[/itex]. You evidently know this, but you've just worded it strangely.

    Second: it is not the definition of infimum that implies the existence of a greatest lower bound, it is the least-upper-bound property of the reals. Or, rather, a corollary of it:

    Corollary: A nonempty set of reals bounded below has an infimum.

    This should have been proved in your book or perhaps is an exercise.

    I would also make it more clear that the elements in S are lower bounds for T.
    Last edited: Aug 2, 2011
  5. Aug 2, 2011 #4
    Hi ArcanaNoir!
    You said earlier that in my last sentence, it does not necessarily imply that one follows another. And you deleted the post. Did you change your mind?
  6. Aug 2, 2011 #5
    Yup, yup, yup. I realized what you meant right as I hit the post button. You have my apologies. See my edited post.
  7. Aug 2, 2011 #6
    Oh...you saw that?

    Yeah I changed my mind. I noticed the less/equal... and it follows from the definition of inf. That's what I meant about having to think harder to agree with the proof. The contradiction proof is more quickly accepted, you simply see the given contradiction, instead of having to be convinced something is always so. Know what I mean?
  8. Aug 2, 2011 #7
    Wait, what?! There were two deleted posts? I thought that last bit was directed towards me.
  9. Aug 2, 2011 #8
    Well that explains your second post. Lol

    It just goes to show that proof, although found to be viable, is not as clear as the contradiction, I think, since we both had to think twice.
  10. Aug 2, 2011 #9
    Hahaha. Thank you guys for your help.
  11. Aug 2, 2011 #10
    Well I make a fool out of myself on a daily basis in real life, so why not on the internet too? Lol. :smile:

    Here's another observation: The definition of infimum says that if v is any lower bound of T, then [itex] v \leq \inf T[/itex]. However, you've said that if v is ANY OTHER lower bound of T (I'm taking this to mean other than the inf itself!), then [itex] v < \inf T[/itex]. That's not wrong, but...

    I would change this. First state that every [itex] s \in S[/itex] is a lower bound of T. Then show that the inf must exist (by citing that corollary). Then, from the definition of infimum, we have[itex] s \leq \inf T[/itex]. This handles the case that [itex] \inf T \in S[/itex] a little bit better.
  12. Aug 2, 2011 #11
    You see, I am awfully bad with wording; and this is my biggest weakness. I don't know if I can ever improve that. Thanks for the note!
  13. Aug 2, 2011 #12
    Avoid words. Use symbols. [itex] \Rightarrow [/itex] is your friend. :P
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