Show that if a subset $S$ of $R$ has a maximum element $y$ (in other words, $y \in S$ and $x \leq y$ for all $x \in S$), then $y =\sup S$.

Question

I'm doing a real analysis course and I've tried to demonstrate above rigorously. If anyone could give a small look over, it would be much appreciated.

Definition of Maximum element: $y \in S$, $x\leq y$ for all $x \in S$

Definition of the least upper-bound:

$1$: $x \leq M$ for all $x \in S$, and

$2$: $x \leq K$ for all $x \in S$ $\implies M \leq K$

By definition of a maximum element condition $1$ is fulfilled and thus proves $y$ is an upper bound to $S$ . Next take all upper bounds $K$ such that.

$x \leq y \leq K$

$y$ is a upper bound by $1$, so $K \geq y$ is verified, and thus $2$ is verified.

Therefore $\sup S = y$

Thanks!

Answer 1

No, you don't take all upper bounds $K$ such that $x\leqslant y\leqslant K$. First of all, this doesn't make sense (what is $x$?). And, above all, you can't put any restriction on the upper bound that you take when your goal is to prove that $y$ is the least upper bound.

So, let $K$ be an upper bound of $S$. Since $y\in S$, $y\leqslant K$. And, since $y$ itself is an upper bound of $S$, this proves that $Y$ is the least upper bound.

Answer 2

There is a $y \in S$ s.t.

$x\le y$ for $x \in S$.

1) $y$ is an upper bound √

2) Need to show that it is the least upper bound of $S$.

Assume there is an $a <y,$ real, s.t.

$x \le a$ for $x \in S$.

Since $y \in S$ we get

$y \le a<y$, a contradiction.

Hence $y=\sup S $