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

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Homework Help Overview

The problem involves finding the supremum (least upper bound) of the set A = {2x + sqrt(2)y : 0 < x < 1, -1 < y < 2}. The original poster expresses understanding of the concept of supremum but is uncertain about how to formally prove that a specific value is the supremum of the set.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the identification of upper bounds for the set and question the correctness of the initially proposed upper bound. There are attempts to clarify the definition of the supremum and how to demonstrate that a proposed upper bound is indeed the least upper bound.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying the definition of the supremum. Some guidance has been offered regarding the properties of the function involved, but there is no consensus on the formal proof required to establish the supremum.

Contextual Notes

There is a recognition that the supremum may not be an element of the set A as defined, and participants note that the set is bounded, which implies the existence of a supremum. However, there is uncertainty about how to articulate the proof formally.

AdrianZ
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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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Did you mistype this? It is easy to see that 1+ 2\sqrt{2} is NOT an upper bound for the set \{2x+ \sqrt{2}y | 0&lt; x&lt; 1, -1&lt; y&lt; 1\} because that set contains 1.99+ 1.99\sqrt{2} which is larger that 1+ 2\sqrt{2}.
I assume you meant 2+ 2\sqrt{2}. (Or else the set is x+ y\sqrt{2}, not 2x+ y\sqrt{2}.)
 
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?
 
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 f(1,2) \scriptsize{\text{ which is }} \normalsize 2+2\sqrt2

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.
 
Last edited:
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.
 
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!
 

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