Let $X$ be a real or complex vector space and consider the partially ordered sets $lc(X) \subseteq v(X) \subseteq t(X)$ of respectively locally convex topologies, vector topologies and all topologies on $X$ with partial order given by inclusion. Now, $t(X)$ is a complete lattice with infimum $\inf \mathcal{T}$ given by the intersection $\bigcap \mathcal{T}$ for $\mathcal{T} \subseteq t(X)$ and supremum $\sup \mathcal{T}$ given by the topology generated by $\bigcup \mathcal{T}$.
Question: Do $lc(X)$ and $v(X)$ also have more structure than just being partially ordered?
Here are a few facts and partial answers:
- if $\mathcal{T} \subseteq v(X)$ then $\sup \mathcal{T} \in v(X)$ (with supremum taken in $t(X)$)
- similarly, if $\mathcal{T} \subseteq lc(X)$ then $\sup \mathcal{T} \in lc(X)$ (with supremum taken in $t(X)$)
Basically, the reason is that $\sup \mathcal{T}$ is simply the initial topology for the identity maps $(X, \sup \mathcal{T}) \to (X, \tau)$, $\tau \in \mathcal{T}$ (which are linear). Thus, we should define the supremum operation in $lc(X)$ and $v(X)$ as that induced from $t(X)$. In particular, $v(X)$ and $lc(X)$ have a maximum. (One can also show that if the dimension of $X$ is uncountable then the maximum of $v(X)$ is not locally convex, i.e. $\sup v(X) \not\in lc(X)$.)
- if $\mathcal{T} \subseteq lc(X)$ then one could define $\inf \mathcal{T} \in lc(X)$ as follows: for each $\tau \in \mathcal{T}$, let $P_\tau$ denote the collection of all $\tau$-continuous seminorms. Then define $\inf \mathcal{T}$ as the locally convex topology generated by the intersection $\bigcap_\tau P_\tau$. Does this coincide with the infimum of $\mathcal{T}$ taken in $t(X)$?
- if $\mathcal{T} \subseteq v(X)$ then one could perform the above construction by considering $F$-seminorms instead of norms.
I am really unsure, whether the intersection of even two locally convex topology or vector topologies (i.e. the infimum in $t(X)$) coincides with the above defined infimum operations in $lc(X)$ resp. $v(X)$.
Since we don't require the topologies to be Hausdorff, $lc(X)$ and $v(X)$ are complete lattices.
Given a family $\mathcal{T}\subset lc(X)$, let
$$\mathcal{C} := \{ \sigma \in lc(X) : (\forall \tau \in \mathcal{T})(\sigma \subset \tau)\}.$$
The family $\mathcal{C}$ contains the indiscrete topology, hence it is not empty. Now let
$$\iota := \sup \mathcal{C}.$$
Since $\iota$ is an initial topology,
$$\operatorname{id} \colon (X,\tau) \to (X,\iota)$$
is continuous for every $\tau \in \mathcal{T}$, because
$$(X,\tau) \xrightarrow{\operatorname{id}} (X,\iota) \xrightarrow{\operatorname{id}} (X,\sigma)$$
is continuous for all $\sigma \in \mathcal{C}$. Hence $\iota \in \mathcal{C}$. So $\iota$ is a locally convex topology that is coarser than all $\tau \in \mathcal{T}$ and it is the finest such topology. Hence $\iota = \inf \mathcal{T}$. (And clearly, $\iota$ is the topology induced by the family of seminorms that are continuous for all $\tau \in \mathcal{T}$.)
The exact same argument works for $v(X)$ in place of $lc(X)$.
The question remains whether we have
$$\inf\nolimits_{v(X)} \mathcal{T} = \inf\nolimits_{t(X)} \mathcal{T}$$
for $\mathcal{T}\subset v(X)$, and
$$\inf\nolimits_{lc(X)} \mathcal{T} = \inf\nolimits_{v(X)} \mathcal{T} = \inf\nolimits_{t(X)} \mathcal{T}$$
for $\mathcal{T}\subset lc(X)$. Clearly we always have the inclusions $\inf_{lc(X)} \mathcal{T} \subset \inf_{v(X)} \mathcal{T}$ and $\inf_{v(X)} \mathcal{T} \subset \inf_{t(X)} \mathcal{T}$ for all $\mathcal{T} \subset lc(X)$ resp. $\mathcal{T} \subset v(X)$.
In general, we have $\inf_{v(X)} \mathcal{T} \subsetneqq \inf_{t(X)} \mathcal{T}$, since the intersection of two $T_1$-topologies is again a $T_1$-topology, but, as you have shown in the comments, if $\tau_1,\tau_2$ are two distinct Fréchet space (or just completely metrisable) topologies on $X$, then $\inf_{v(X)} \{\tau_1,\tau_2\}$ is not a $T_1$-topology (since a $T_0$ topology in $v(X)$ is Hausdorff).
I strongly expect that in general we also have $\inf_{lc(X)} \mathcal{T} \subsetneqq \inf_{v(X)} \mathcal{T}$, but so far I haven't found an example.