- #1
A problem on infimum and supremum.
- Thread starter Kinetica
- Start date
-
- Tags
- Supremum
Click For Summary
Homework Help Overview
The problem involves the concepts of infimum and supremum within the context of real numbers. The original poster presents a proof approach and seeks feedback on its correctness, while referencing a proof by contradiction as an alternative method.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants discuss different proof methods, including a straightforward approach and a proof by contradiction. There are questions about the definitions and properties related to lower bounds and the existence of an infimum.
Discussion Status
The discussion is active, with participants providing feedback on wording and definitions. Some express a preference for the clarity of the proof by contradiction, while others engage in clarifying the definitions and implications of the concepts involved.
Contextual Notes
There are mentions of deleted posts, indicating a dynamic conversation where participants reconsider their statements. The original poster expresses difficulty with wording, which is acknowledged by others in the thread.
- #2
ArcanaNoir
- 778
- 4
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.
- #3
stringy
- 90
- 1
Whoa, nevermind. Bah! I just typed a bunch of nonsense.
Rewrite!
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.
Rewrite!
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:
- #4
Kinetica
- 83
- 0
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?
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?
- #5
stringy
- 90
- 1
Yup, yup, yup. I realized what you meant right as I hit the post button. You have my apologies. See my edited post.
- #6
ArcanaNoir
- 778
- 4
Kinetica said:
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?
- #7
stringy
- 90
- 1
Wait, what?! There were two deleted posts? I thought that last bit was directed towards me.
- #8
ArcanaNoir
- 778
- 4
stringy said:
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.
- #9
Kinetica
- 83
- 0
Hahaha. Thank you guys for your help.
- #10
stringy
- 90
- 1
Well I make a fool out of myself on a daily basis in real life, so why not on the internet too? Lol. 
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.
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.
- #11
Kinetica
- 83
- 0
stringy said:
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!
- #12
ArcanaNoir
- 778
- 4
Kinetica said:
Avoid words. Use symbols. [itex]\Rightarrow[/itex] is your friend. :P
Similar threads
- · Replies 3 ·
- · Replies 3 ·
- · Replies 4 ·
- · Replies 1 ·
- · Replies 4 ·