Prove the supremum exists

[silvermane]

Prove the supremum exists :)

Homework Statement

Let A = {x:x in Q, x^3 < 2}.
Prove that sup A exists. Guess the value of sup A.

The Attempt at a Solution

First we show that it is non-empty. We see that there is an element, 1 in the set, thus A is non-empty.
Now we show that A is bounded. We see that 2 is not in the set, thus there must be an upper bound on our set. Our set can be represented as (-oo, 2).
Since A is bounded above, then supA exists, and we are done.

(I just need help clarifying and making sure that I am following the correct logic here.)
=)

Also, would supA = 2?

Thanks for all your help in advance

 

[silvermane]

I don't need an answer or anything, I just need someone to confirm that what I've shown here is a good proof.

 

[╔(σ_σ)╝]

Hey,

The suprema is not 2. :-)

It is the cubed root of 2.

You have to show that any number that is greater than ,lets says, 2 is not in the set.

Saying that 2 is not in the set is merely saying that two is not in the set it doesn't prove 2 is an upperbound. A little bit more is required to show the set is bounded above.

:-)

 

[silvermane]

Hey,

The suprema is not 2. :-)

It is the cubed root of 2.

You have to show that any number that is greater than ,lets says, 2 is not in the set.

Saying that 2 is not in the set is merely saying that two is not in the set it doesn't prove 2 is an upperbound. A little bit more is required to show the set is bounded above.

:-)

Awe... :/
That's how our professor did these problems in class. I'm so confused now >.<

 

[╔(σ_σ)╝]

Well the suprema you gave is incorrect. I don't think your prof did that in class .

Ex

S={0,1,2,5}
4 is not in the set and it is not an upper bound.

My point is that if you say something is an upperbound you have to show that it is greater than every value in the set.

In your example I can see it is obvious what is and what is not an upper bound but this is analysis; we have to show everything.

I guess your prof wanted you guys to fill in the details or he simply doesn't require very detailed solutions. So I guess you can get away with what you did .:-)

 

[silvermane]

Well the suprema you gave is incorrect. I don't think your prof did that in class .

Ex

S={0,1,2,5}
4 is not in the set and it is not an upper bound.

My point is that if you say something is an upperbound you have to show that it is greater than every value in the set.

In your example I can see it is obvious what is and what is not an upper bound but this is analysis; we have to show everything.

I guess your prof wanted you guys to fill in the details or he simply doesn't require very detailed solutions. So I guess you can get away with what you did .:-)

lol I should probably be more precise when I stated my work :(
We're using the continuum property and the fact that there are infinite elements in our set, and once we find one not in our set, then there's no other greater ones in the set.
In this example, he doesn't want a proof proof, I think he just wants us to say it's "obvious" with observation, and etc.

But say we were to prove it, for enlightenment, how would I go about it that way?

and yes, I believe I meant to say that 2^3 was the supremum, I was just being noodle-headed :3

 

[╔(σ_σ)╝]

lol I should probably be more precise when I stated my work :(
We're using the continuum property and the fact that there are infinite elements in our set, and once we find one not in our set, then there's no other greater ones in the set.
In this example, he doesn't want a proof proof, I think he just wants us to say it's "obvious" with observation, and etc.

But say we were to prove it, for enlightenment, how would I go about it that way?

and yes, I believe I meant to say that 2^3 was the supremum, I was just being noodle-headed :3

You got the suprema wrong again .

It is 2^(1/3).

In the case you had to prove 2 is an upperbound you could start by using some axioms or simply state that for x<0, x^3 <0. So this automatically eliminates negative numbers as upper bounds.

Then you could show that x>2 , x^3 >8 which means it is not in the set.

Sorry, I am just been too picky. I guess we do not have to prove obvious things like these but I will be nice to at least state it without proof.

 

[╔(σ_σ)╝]

:-) The reason I am asking you to prove that there is an upper bound is that the question asked you to prove that the supremum exist. Which means you have to prove two things.
1) The set is non empty
2) It is bounded above.

If you were supposed to prove the suprema is x for example you can afford to wave your hand and say "clearly ... the set is bounded".
But when you are told to directly prove the suprema exist I think you need to prove those two thinks no matter how obvious it seems.