S is an ordered set and Y a nonempty bounded subset. If X subset Y such that for every y in Y there exists x in X that satisfies y <=x, then supX = supY.
How would I go about proving this?
My thoughts: Because for every y in Y there exists an x in X that satisfies y <=x, it implies that x is an upperbound for Y. In addition, Y contains X, then x is also an upperbound for X.
Are my thoughts correct and how would I go on from here?
On
Obviously $\sup Y$ is an upper bound for $X$ because $X$ is a subset of $Y$ so is just need to be shown that it's the smallest one.
But for that it's enough to show that there exists a sequence in $X$ that has $\sup Y$ as a limit.
Because $Y$ is the supremum there exists a sequence $(y_n)_{n\in\Bbb N}$ with $y_n \to \sup Y$.
Take now $x_n$ with $y_n \le x_n$ for all $n\in \Bbb N$ which exists by assumption then $x_n \to \sup Y$ by sandwich theorem and we are done…
Your way of thinking is quite correct, but you wrote that in a confusing way
Try this
Do you know there is a theorem that said "If $X\subset Y$ and $\sup Y$ is exist, then $\sup X\leq\sup Y$"
Now, since $Y$ is bounded, then $\sup Y$ is exist, so is $\sup X$. Suppose $\sup X<\sup Y$, then by the definition we have there is $y\in Y$ such that $y>\sup X$. And by the premise, we have there is $x\in X$ such that $x\geq y>\sup X$, and this contradicts the definition of $\sup X$ (because $x>\sup X$). Hence, $\sup X<\sup Y$ is impossible, then we have $\sup X=\sup Y$