How to prove that something is the supremum of a set?

In summary, the problem asks for the supremum of the set A={2x+sqrt(2)y : 0<x<1 , -1<y<2}. It is easy to show that for any x and y given in the defined domain, we have: -sqrt(2) < 2x+sqrt(2)y< 1+2sqrt(2). From this inequality, it is seen that 1+2sqrt(2) is an upper bound for the set. However, the question is how can I show that this is the supremum of the set, or in other words how can I show that this is the least element of the set of the upper bounds
  • #1
AdrianZ
319
0

Homework Statement


well, the problem asks me to find the supremum(lub) of the set A={2x+sqrt(2)y : 0<x<1 , -1<y<2}. It's easy to show that for any x and y given in the defined domain, we have: -sqrt(2) < 2x+sqrt(2)y< 1+2sqrt(2). well, from this inequality, It's clearly seen that 1+2sqrt(2) is an upper bound for the set. but the question is how can I show that this is the supremum of the set or in other words how can I show that this is the least element of the set of the upper bounds of the set A? I know what a supremum means and stuff like that but I don't know what I should precisely do when a problem asks me to find the supremum of the set.

The Attempt at a Solution



-sqrt(2) < a < 1 +2sqrt(2). for any a in A.
 
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  • #2
Did you mistype this? It is easy to see that [itex]1+ 2\sqrt{2}[/itex] is NOT an upper bound for the set [itex]\{2x+ \sqrt{2}y | 0< x< 1, -1< y< 1\}[/itex] because that set contains [itex]1.99+ 1.99\sqrt{2}[/itex] which is larger that [itex]1+ 2\sqrt{2}[/itex].
I assume you meant [itex]2+ 2\sqrt{2}[/itex]. (Or else the set is [itex]x+ y\sqrt{2}[/itex], not [itex]2x+ y\sqrt{2}[/itex].)
 
  • #3
Yea, thanks for noting that. I mistakenly typed 1+sqrt(2)2 instead of 2+2sqrt(2).
Now How can I show that's the least upper bound of the set?
 
  • #4
You might note that the function has no critical points and thus is increasing or decreasing everywhere, then note that it is increasing, and then note that the upper endpoint is [itex] f(1,2) \scriptsize{\text{ which is }} \normalsize 2+2\sqrt2 [/itex]

but this is just a thought, I don't have experience proving something is a supremum of a set. I do know that a function is a set, and this is a function, so it seems reasonable.

Oh, extra thought:
Since the set contains a largest value, the largest value is the supremum. I would include that.
 
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  • #5
ArcanaNoir said:
Since the set contains a largest value, the largest value is the supremum. I would include that.
The sup isn't in A the way A is defined.
 
  • #6
ArcanaNoir said:
but this is just a thought, I don't have experience proving something is a supremum of a set. I do know that a function is a set, and this is a function, so it seems reasonable.

Yea, that's my problem as well. I know what the supremum of A is. that part is obvious. but I don't know how to write it in a formal way.

vela said:
The sup isn't in A the way A is defined.
True.

well, this is what I've thought of. we know that because A is a bounded subset of R the supremum exists. since for any a in A we have: -sqrt(2)<2x+sqrt(2)y<2+2sqrt(2) then if a>=2+2sqrt(2) or a<=-sqrt(2) then a is not a member of A. therefore the set U={a in R: a>=2+2sqrt(2)} is the set of the upper bounds. by definition of U, It's clear that 2+sqrt(2) is the minimum of the set and therefore is the least upper bound or the supremum that we want.

but I'm still not so satisfied with all these explanations!
 

FAQ: How to prove that something is the supremum of a set?

1. What is the definition of supremum?

The supremum of a set is the least upper bound of the set, which means it is the smallest number that is greater than or equal to all the numbers in the set.

2. How do you prove that a number is the supremum of a set?

To prove that a number is the supremum of a set, you must show that it is an upper bound of the set and that it is the smallest upper bound. This can be done by showing that the number is greater than or equal to all the numbers in the set and that it is not possible to find a smaller upper bound.

3. Can there be more than one supremum for a set?

No, there can only be one supremum for a set. This is because the supremum is defined as the least upper bound, so there cannot be another number that is smaller and still satisfies the definition.

4. Can the supremum of a set be infinite?

Yes, the supremum of a set can be infinite. This is possible when the set contains infinitely many numbers and there is no upper bound that is finite.

5. Is the supremum of a set always a member of the set itself?

Not necessarily. The supremum of a set does not have to be a member of the set. It is possible for the supremum to be a limit point or an accumulation point of the set, but not a member of the set itself.

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