What Steps Are Needed to Prove That a Number Is the Supremum?

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To prove that a number is the supremum, two conditions must be satisfied: it must be an upper bound for the set, and there must be points in the set that are arbitrarily close to it. In the example of proving that sup {1 - 1/n} = 1, it is established that 1 is an upper bound and that for every epsilon > 0, there exists an n such that 1 - 1/n is greater than 1 - epsilon. The discussion highlights the need for clear justification in demonstrating these conditions. Additionally, a clarification is made that 1 - e < x is not equivalent to 1/e < x. Understanding these steps is crucial for proving supremum effectively.
semidevil
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ok, so supermum is the least upper bound, but when you write down on paer, what am I trying to show...does that make sense? I mean, what am I trying to get to that will show something is the supermum.

so for example: prove sup {1 - 1/n} = 1 for all n in N.

so I start by saying that I know that 1 is an upper bound...

then, for all epsilon > 0, there exist n st n > epsilon, or 1/n < epsilon.

ok, now what...how else do I need to go to show that 1 is supermum...

how do I know that I have shown that 1 is the supermum??
 
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To show 1 is a supremum you need to show 2 things,
a) that 1 is an upper bound for your set
b) that you can find points in your set arbitrarily close to 1, That is for every \epsilon&gt;0 there is an x in your set where 1-\epsilon&lt;x.

That's all. It appears you've done both of these steps, but your organization seems unclear. Do you understand how what you've done relates to the above?


An alternate to b) above, you can instead prove
b*)if A is any other upper bound for your set, then 1<=A
 
Show that the number is an upperbound AND any number smaller is NOT an upperbound.
 
shmoe said:
To show 1 is a supremum you need to show 2 things,
a) that 1 is an upper bound for your set
b) that you can find points in your set arbitrarily close to 1, That is for every \epsilon&gt;0 there is an x in your set where 1-\epsilon&lt;x.

That's all. It appears you've done both of these steps, but your organization seems unclear. Do you understand how what you've done relates to the above?


An alternate to b) above, you can instead prove
b*)if A is any other upper bound for your set, then 1<=A

I dotn know what I've done to solve the problem...can you tell me what I did? lol...how haev I already solved the problem?

btw, is 1 - e < x the same as 1/e < x?
 
semidevil said:
so I start by saying that I know that 1 is an upper bound...

This is condition a), though you should should give some justification as to why 1 is an upper bound.

semidevil said:
then, for all epsilon > 0, there exist n st n > epsilon, or 1/n < epsilon.

This is b), but it needs some explanation. To prove b), let \epsilon &gt;0 and we want to find something in our set that's larger than 1-\epsilon. You've shown that there is an n\in\Bbb{N} where 1/n&lt;\epsilon. Now you know that 1-1/n is in your set and the last inequality tells us that 1-1/n&gt;1-\epsilon, and we've found an element in our set and condition b) is satisfied.


semidevil said:
btw, is 1 - e < x the same as 1/e < x?

No. Try e=1/2, x=1.
 
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