How to define a topology with a given convergence of nets. Suppose $X$ is a non-empty set and $\{\tau_{i}\}_{i}$ be a family of topologies on X. I want to define a topology $\tau$ on $X$ so that:

A net $\{x_\alpha\}_\alpha$ converges to $x$ w.r.to $\tau$ if and only if there exists some $i$ such that $\{x_\alpha\}_\alpha$ converges to $x$ w.r.to $\tau_i$.


*

*If I define $\mathcal{S} = \cup_i \tau_i$ then it is not a topology on $X$ but then we can take the weakest topology on X containing $\mathcal{S}$. The resulting topology is stronger than each $\tau_i$ and is the weakest possible one with this property. So convergencey w.r.to this topology immediately gives that w.r.to each $\tau_i$. But I think this is not the desired one above.

Is this a standard construction ? Any suggestion is appreciated. Thanks.
 A: Looking at it from the perspective of convergence spaces, (e.g. as defined here, a self-contained overview containing all relevant basic definitions in section 3, including the three net convergence axioms N1,N2,N3; it also mentions the conditions when a net convergence can be induced by a topology), then this idea doesn't seem to work either:
Say we take the set $X=\Bbb R$ and the two Sorgenfrey topologies $\tau_L$ with base $ \{(a,b]\mid a < b\}$ and $\tau_R$ with base $ \{[a,b)\mid a < b\}$, and their induced net-convergences $\to_L$ and $\to_R$. If we then use the rule that the net $(x_a)_{A \in A}$ converges to $x$ iff it converges for $\to_L$ or for $\to_R$, we see that the first two net convergence axioms (a constant net with value $x$ converges to $x$ and a quasi-subnet of a convergent net has the same limit) are easy to verify (for any two convergences on a set), but the axiom N3: if $(x_a)_{a \in A} \to x$ and $(y_a)_{a \in A} \to x$ then for any net $(z_a)_{a \in A}$ such that $z_a \in \{x_a,y_a\}$ for all $a \in A$, $z_a \to x$ as well, doesn't work as well: consider the sequences $x_n = \frac{1}{n}$, convergent to $0$ in $\tau_R$ and so in this combo-rule as well, and $y_n = -\frac1n$ also convergent to $0$ under $\tau_L$ and so in the combo-rule, while the mixed sequence $1,-\frac12, \frac13, -\frac14, \frac15, -\frac16, \ldots$ is not convergent to $0$ in this combo-rule, as it does not converge in $\tau_L$ nor in $\tau_R$.
So even in the more general setting of convergence spaces your proposed rule does not give a convergence space, even for two topologies, and so a fortiori there cannot be a topology that follows this rule either.
A: Taking the sup of a family of topologies in this way is indeed a standard construction which is employed in so-called final topologies (induced by maps).
It's a standard fact that the lattice of all topologies on a fixed set $X$ is complete.
It does not have your desired property: e.g. consider the left Sorgenfrey and right Sorgenfrey topologies on $\Bbb R$, with bases $\{(a,b]: a < b\}$ and $\{[a,b); a < b\}$ resp. The common upper bound is the discrete topology and a net only converges in that topology if it is eventually constant, while convergence in the Sorgenfrey topologies is much "wider" (upper and lower convergence).
Moreover, $\sup_i \tau_i$ has the property that if a net converges in it, it also converges in all $\tau_i$, which is already way stronger than you want. So $\inf_i \tau_i$, which is the intersection of the topologies seems a better candidate, which also doesn’t work in general ( look at that same example) but is a better approximation than the union idea. Maybe this rule does not always define a valid topology at all…
