Let $V$ be a locally convex topological vector space (or even a Banach space), equipped with a closed partial order, or equivalently, a closed positive cone. It is well known that any directed net $\{v_\alpha\}$ tending to $v$ topologically also tends to $v$ order-theoretically, i.e. $\sup_\alpha v_\alpha = v$.
Under which condition is the converse true? That is, suppose that a directed net $\{v_\alpha\}$ has order-supremum $v$. Then under which conditions (on $\{v_\alpha\}$ and/or on the topology and order of $V$) does $\{v_\alpha\}$ tend to $v$ topologically? Are there known results (and/or counterexamples)?
Counterexample
In $\ell^1$, let the positive cone be $$P=\left\{x: \sum_{k=1}^n \frac{x_k}{k!}\ge 0, \ n=1,2,\dots\right\}$$ This is a proper cone since the only way for all these partial sums to be zero is when $x_k=0$ for all $k$. It is also closed, being an intersection of closed half-spaces.
Consider a sequence $z_j = -j^{-1}e_{j} + e_{j+1}$. It satisfies $z_j<0$ because $$ -\frac{1}{j\, j!} + \frac1{(j+1)!} < 0 $$ It is also increasing, as $$ z_{j+1}-z_{j} = \frac{1}{j}e_j - \frac{j+2}{j+1}e_{j+1}+e_{j+2} $$ and $$ \frac{1}{j\,j!} - \frac{j+2}{(j+1)(j+1)!} = \frac{(j+1)^2 - j(j+2)}{j(j+1)(j+1)!} = \frac{1}{j(j+1)(j+1)!} > 0 $$ If $z\ge z_j$ for all $j$, then necessarily $z\ge 0$. Thus $z_j$ converge to $0$ in the sense of order. But $\|z_j\|_1>1$, so the convergence is not topological.