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## Homework Statement

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.

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- Thread starter Kinetica
- Start date

- #1

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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.

- #2

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- #3

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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.

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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?

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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?

Oh...you saw that?

Yeah I changed my mind. I noticed the less/

- #7

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Wait, what?! There were two deleted posts? I thought that last bit was directed towards me.

- #8

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Wait, what?! There were two deleted posts? I thought that last bit was directed towards me.

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.

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Hahaha. Thank you guys for your help.

- #10

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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

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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.

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!

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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!

Avoid words. Use symbols. [itex] \Rightarrow [/itex] is your friend. :P

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