- Thread starter
- #1

#### issacnewton

##### Member

- Jan 30, 2012

- 61

Here is the problem statement.

Let $X=Y = \{x\in \mathbb{R}\; :0<x<1\}$ . Define $ h\;:X\times Y\longrightarrow \mathbb{R}$

by $h(x,y)=2x+y$. For each $x\in X$, find $f(x) = \sup\{h(x,y)\; : y\in Y\}$.

Here is my attempt. I let $S=\{h(x,y)\; : y\in Y\}$. I claim that $\sup S = 2x+1$. To prove

this I first let x be arbitrary since we have to prove this $\forall x\in X$. And next thing

I have to prove is that $2x+1$ is the upper bound of $S$. I could do that. Next I let

$t$ be any other upper bound of $S$. I have to prove that $2x+1 \leqslant t$. Here

I start by assuming the negative. Assume $ t < 2x+1$. And now I am supposed to reach

a contradiction somewhere. Now since $t$ is the upper bound of $\{h(x,y)\; : y\in Y\}$, and since

$0<y<1$, I can claim that $2x+0.9 \leqslant t < 2x+1$, which means $0.9\leqslant t-2x <1$.

So we have $0<t-2x<1$. So I found a number which is between $0$ and $1$. I could exploit this

to come up with some contradiction. But I am stuck here. Any guidance will help.

Thanks