Welcome to our community

Be a part of something great, join today!

[SOLVED] Problem about finding sup of a set

issacnewton

Member
Jan 30, 2012
61
Hello

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
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: problem about finding sup of a set

Hello

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
The thing marked in red is in my opinion incorrect. You can ask your doubts about that.

Here's how you can proceed. To show that $2x+1\leq t$. Say $t=2x+t'$ where WLOG, we assume that $0<t'<1$. Assume on the contrary, just as you did, that $t<2x+1$. This gives $t'<1$. Say $1-t'=\delta$. Note that $\delta>0$. Put $y_0=1-\delta/2$. So $y\in Y$. But now $2x+y_0>2x+t'=t$, contradicting the fact that $t$ was an upper bound of $\{h(x,y):y\in Y\}$.
 

issacnewton

Member
Jan 30, 2012
61
Re: problem about finding sup of a set

Since t is the upper bound of $\{h(x,y)\; : y\in Y\}=\{2x+y\;:y\in Y\}$, I let $y=0.9$, then I have $2x+0.9 \leqslant t$. Why is this wrong ?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: problem about finding sup of a set

Since t is the upper bound of $\{h(x,y)\; : y\in Y\}=\{2x+y\;:y\in Y\}$, I let $y=0.9$, then I have $2x+0.9 \leqslant t$. Why is this wrong ?
Oops! I am sorry. That was no mistake. But anyway.. did you see the solution I posted?
 

issacnewton

Member
Jan 30, 2012
61
Re: problem about finding sup of a set

Oops! I am sorry. That was no mistake. But anyway.. did you see the solution I posted?

Yes I saw your solution. I had trouble understanding it. How can you claim $t = 2x+t'$ ?
And further how does $0<t'<1$ ? Can you connect the logical gaps please ?

Thanks
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: problem about finding sup of a set

Yes I saw your solution. I had trouble understanding it. How can you claim $t = 2x+t'$ ?
And further how does $0<t'<1$ ? Can you connect the logical gaps please ?

Thanks
I am not claiming that $t=2x+t'$. I merely introduced an extra variable $t'$ to make things neater. You can do without it too.

We can assume, without loss of generality, that $0<t'<1$ since if $t\geq 1$ then clealy $2x+1\leq 2x+t'$. If $t'\leq 0$ then $2x+t'=t$ cannot be an upper bound to $\{h(x,y):y\in Y\}$.