Problem about finding sup of a set

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In summary: Hence we can assume that $0<t'<1$. I hope this clears things up. Let me know if you have any other doubts.
  • #1
issacnewton
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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
 
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  • #2
Re: problem about finding sup of a set

IssacNewton said:
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 $\{h(x,y):y\in Y\}$.
 
  • #3
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 ?
 
  • #4
Re: problem about finding sup of a set

IssacNewton said:
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?
 
  • #5
Re: problem about finding sup of a set

caffeinemachine said:
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
 
  • #6
Re: problem about finding sup of a set

IssacNewton said:
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 $\{h(x,y):y\in Y\}$.
 

FAQ: Problem about finding sup of a set

What is the definition of the supremum of a set?

The supremum of a set is the smallest number that is greater than or equal to all the numbers in the set. In other words, it is the least upper bound of the set.

How do you find the supremum of a set?

To find the supremum of a set, you must first determine if the set has an upper bound. If it does, then the supremum is simply the smallest number that is greater than or equal to all the numbers in the set. If the set does not have an upper bound, the supremum is undefined.

What is the difference between a supremum and a maximum?

A supremum and a maximum are two different concepts. A supremum is the smallest upper bound of a set, while a maximum is the largest element in a set. The supremum may or may not be an element of the set, while the maximum must be an element of the set.

Can a set have multiple supremums?

No, a set can only have one supremum. This is because the supremum is defined as the smallest upper bound of a set, and there can only be one smallest number.

Why is the concept of supremum important in mathematics?

The concept of supremum is important in mathematics because it is a fundamental concept in the theory of real numbers and analysis. It allows us to define limits, continuity, and other important concepts. It also has practical applications in fields such as economics, physics, and engineering.

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