arrow900
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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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ghostwalker
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(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.
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ghostwalker
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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.
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atsruser
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(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, \mathbb{N} has no largest element, neither does \mathbb{R}

Here are two examples.

1) \mathbb{Q} does not have the least upper bound property. For example, consider:

E =\{ x \in \mathbb{Q} : x^2 < 2 \}

Then E is bounded above (by 10, for example), is non-empty, but it does not have a supremum in \mathbb{Q}. (The only possible candidate is \sqrt{2} but that is not rational).

2) It is taken as an axiom that \mathbb{R} 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 \mathbb{R} then you can prove that Cauchy sequences converge to something in \mathbb{R}. (Or alternatively, if you assume that \mathbb{R} is Cauchy-complete i.e. that all Cauchy sequence converge to something in \mathbb{R}, then you can prove that it has the l.u.b property)

Now consider:

E =\{ x \in \mathbb{R} : x^2 < 2 \}

Then by the l.u.b. property, \sup E \in \mathbb{R} and it can be shown that \sup E = \sqrt{2}

Note that the supremum of a set may or may not be an element of that set. In the case above, \sup E \notin E
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arrow900
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(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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