# Need help understanding Sets

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#1
I have been reading through the first chapter of Rudin's " Principles of mathematical analysis" and I got a bit confused with one of the definitions. I just need some clarification.

He states ,as a definition, that "An ordered set S is said to have the least upper bound property if the following is true:

**If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

I can't figure out why this must be true. I first thought that if we consider E to be the proper subset of S containing the largest element of S, then provided that ** implies that Sup E exists in S then it must be true that S has the least upper bound property.
But I am really uncertain about this, can some clarify why the existence of Sup S is defined in this way?

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6 years ago
#2
(Original post by arrow900)
I have been reading through the first chapter of Rudin's " Principles of mathematical analysis" and I got a bit confused with one of the definitions. I just need some clarification.

He states ,as a definition, that "An ordered set S is said to have the least upper bound property if the following is true:

**If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

I can't figure out why this must be true. I first thought that if we consider E to be the proper subset of S containing the largest element of S, then provided that ** implies that Sup E exists in S then it must be true that S has the least upper bound property.
But I am really uncertain about this, can some clarify why the existence of Sup S is defined in this way?
What you quoted - in red - is not defining the existance of Sup S.

That's covered by point 1.8 at the top of that page (in the 3rd edition)

You should also note that S may not have a largest element, .e.g the natural numbers, or the reals, etc.

What's quoted is saying:

"A set, S, has the least-upper-bound property" means "If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

If, and only if, the bit is green is true then, S has the least-upper-bound property.

I may have misunderstood where you're coming from, but hope that helps.
0
6 years ago
#3
Rephrasing the definition slightly may help.

A set, S, has the least upper bound property, iff every non empty proper subset of S that is bounded above, has a least upper bound that is in S.
0
6 years ago
#4
(Original post by arrow900)
I have been reading through the first chapter of Rudin's " Principles of mathematical analysis" and I got a bit confused with one of the definitions. I just need some clarification.

He states ,as a definition, that "An ordered set S is said to have the least upper bound property if the following is true:

**If E is a proper subset of S, E is not empty, and E is bounded above**, then Sup E exists in S"

I can't figure out why this must be true.
It's neither true nor false. It's a definition of the least upper bound property.

I first thought that if we consider E to be the proper subset of S containing the largest element of S, then provided that ** implies that Sup E exists in S then it must be true that S has the least upper bound property.
It's hard to see what you're saying here. E is not a *specific* subset of S. It's any non-empty proper subset. And also your statement "if we consider E to be the proper subset of S containing the largest element of S," indicates confusion. S may not have a "largest element". For example, has no largest element, neither does Here are two examples.

1) does not have the least upper bound property. For example, consider: Then is bounded above (by 10, for example), is non-empty, but it does not have a supremum in . (The only possible candidate is but that is not rational).

2) It is taken as an axiom that does have the least upper bound property since this then gives it the properties required for doing calculus. For example, if the l.u.b property holds for then you can prove that Cauchy sequences converge to something in . (Or alternatively, if you assume that is Cauchy-complete i.e. that all Cauchy sequence converge to something in , then you can prove that it has the l.u.b property)

Now consider: Then by the l.u.b. property, and it can be shown that Note that the supremum of a set may or may not be an element of that set. In the case above, 0
#5
(Original post by ghostwalker)
Rephrasing the definition slightly may help.

A set, S, has the least upper bound property, iff every non empty proper subset of S that is bounded above, has a least upper bound that is in S.
Yes that makes perfect sense. Its just that the definition in the textbook is phrased almost as if it is a theorem which caused the confusion. Thanks a lot.

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