User:Dfeuer/Increasing Net Converges to Point iff Supremum

Theorem
Let $(D,\preceq)$ be a directed set.

Let $(L,\le)$ be a totally ordered set considered under the order topology.

Let $(x_i)_{i\in D}$ be a net in $G$.

Suppose that $(x_i)$ is increasing. That is, suppose that for all $i,j\in D$, $i\preceq j \implies x_i \le x_j$.

Let $x\in L$.

Then $(x_i)$ converges to $x$ iff $x$ is the supremum of $P = \{ x_i \mid i\in D \}$.

Forward implication
Suppose that $(x_i)$ converges to $x$.

Suppose for the sake of contradiction that for some $j \in D$, $x_j > x$.

Then since $(x_i)$ is increasing, $\forall k\in D: (k \succeq j\implies x_k\ge x_j > x)$.

But since $(x_i)$ converges to $x$, and $x_j > x$, $(x_i)$ is eventually less than $x_j$, a contradiction. Thus $x$ is an upper bound of $P$.

Let $m < x$.

Since $(x_i)$ converges to $x$, it is eventually greater than $m$.

Thus $m$ is not an upper bound of $P$.

We conclude that $x$ is the supremum of $P$.

Reverse implication
Suppose that $x$ is the supremum of $P$.

Let $b > x$.

Then $x_i < b$ for every $i\in D$, so $(x_i)$ is certainly eventually less than $b$.

Let $a a$.

But then since $(x_i)$ is increasing, $\forall k\in D: (k\succeq j\implies x_k \ge x_j > a)$, so $(x_i)$ is eventually greater than $a$.

Thus we conclude that $(x_i)$ converges to $x$.