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

1. Aug 15, 2011

1. The problem statement, all variables and given/known data
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.

3. The attempt at a solution

-sqrt(2) < a < 1 +2sqrt(2). for any a in A.

2. Aug 15, 2011

### HallsofIvy

Staff Emeritus
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 $2x+ y\sqrt{2}$.)

3. Aug 15, 2011

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. Aug 15, 2011

### ArcanaNoir

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: Aug 15, 2011
5. Aug 15, 2011

### vela

Staff Emeritus
The sup isn't in A the way A is defined.

6. Aug 15, 2011