There exist a sequence x_n E S s.t. x_n -> sup S

[kingwinner]

Homework Statement

Consider sequence of real numbers.
Theorem: If a= sup S, then there exist a sequence x_n E S such that x_n ->a

Proof:
Take ε = 1/n and find x_n E S such that 0 ≤ a - x_n < 1/n.
Now show x_n -> a.
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I am very very confused about this proof.
1) Why are they taking ε = 1/n? What motivates this?

2) It seems to me that n is simply a "subscript" of the sequence x_n and it's a bit weird to talk about ε = 1/n. Is there any relationship between the "n" in ε = 1/n and the "n" in the sequence x_n? Are they the SAME "n"?

3) In the proof, they say "find x_n E S such that 0 ≤ a - x_n < 1/n", but how do we know that such a thing even EXISTS?

4) At the end of the proof, they say "show x_n -> a", but HOW??

Homework Equations

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The Attempt at a Solution

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Can someone please explain the proof in greater detail?
Any help is much appreciated! :)

 

[union68]

If there were NOT an element of S so that for all n, 0 \leq a - x_n &lt; 1/n, is a really the supremum of S?

Note that this inequality can also be written

a- \frac{1}{n} &lt; x_n \leq a

which may be more illustrative. The point of choosing \epsilon = 1/n is because this will shrink the neighborhood around a as n gets larger. This should give you some indication as to the construction of the sequence.

 

[kingwinner]

OK, and here is my attempt to rewrite and understand the proof in a more presentable manner.

Proof:
Take ε = 1/n.
Now a = sup S => for EACH n=1,2,... there exists an x E S such that a -1/n < x ≤ a.
For EACH n=1,2,... PICK one such point and CALL it x_n,
so that a -1/n < x_n ≤ a for all n=1,2,...
Now we need to prove that this sequence x_n->a.
By definition, x_n->a iff
for ALL ε>0, there exists an integer N such that n≥N => |x_n - a|< ε.
So for ANY given ε>0, I must be able to choose an N that works.
But I am stuck here. How can we choose such an N in our case?

Thank you! :)

 

[union68]

It looks like you're on the right track. Given \epsilon &gt;0, can you manipulate the quantity 1/n in some manner to help you? Remember, n gets really, really big as you go out in the sequence. What does this say about 1/n?

You have literally constructed the sequence already in the first part of your proof...

 

[kingwinner]

I've constructed the sequence, but I am not sure how to prove it "rigorously" that the sequence converges to a.

Definition: x_n->a iff
for ALL ε>0, there exists an integer N such that n≥N => |x_n - a|< ε.

Now,
a -1/n < x_n ≤ a for all n=1,2,...
=> -1/n < x_n - a ≤ 0 < 1/n for all n=1,2,...
=> |x_n - a| < 1/n for all n=1,2,...

But at the beginning of the proof we said that ε = 1/n? So I am a little confused...why is ε changing for different values of n? This is really werid...shouldn't epsilon be given to be some fixed positive number?
Also, given ε, how should I choose N such that n≥N => |x_n - a|< ε?

Can someone please help me out?
Thanks!

 

[union68]

Mmm'kay, I don't really like how they set \epsilon = 1/n. That seems strange.

Rather, for any given \epsilon&gt;0, there always exists an n large enough so that 1/n &lt; \epsilon. But, you've already constructed a sequence such that

a - \frac{1}{n} &lt; x_n \leq a

for all n. How is \left|x_n - a \right| &lt; \epsilon related to this inequality given that 1/n &lt; \epsilon? What should you choose for your N?

Try not to get too bogged down in the definitions; i.e., try to look at the big picture before diving into epsilons and stuff.

 

[kingwinner]

OK, so I think taking N>(1/ε) works.

At the beginning of the proof, they said that ε = 1/n and ε changing for different values of n? This is werid. I believe that in the definition of limit, ε should be given to be some FIXED arbitrary positive number?

When they say "ε = 1/n" at the beginning of the proof, is this the "same" ε as the one that appears at the end when we're trying to prove the limit using definition? or are they totally unrelated?

thanks!