# 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 boundsf

## 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.

Did you mistype this? It is easy to see that $1+ 2\sqrt{2}$ is NOT an upper bound for the set $\{2x+ \sqrt{2}y | 0< x< 1, -1< y< 1\}$ 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.

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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.

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.

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!