Intersection of weak topologies

Question

Let $X$ be a real vector space, let $X^\star$ be its algebraic dual, and fix two vector subspaces $\mathscr{A}, \mathscr{B}\subseteq X^\star$ such that $\mathscr{C}:=\mathscr{A}\cap \mathscr{B} \neq \{0\}$.

Question. Is it true that a convex subset $S\subseteq X$ is closed in the both the weak topologies generated by $\mathscr{A}$ and $\mathscr{B}$ (that is, $S$ is closed in $(X,\sigma(X,\mathscr{A})$ and $(X,\sigma(X,\mathscr{B}))$) if and only if $S$ is closed in $(X,\sigma(X,\mathscr{C}))$?

The "if" part should be trivial by the fact that $\mathscr{C}\subseteq \mathscr{A}$ and $\mathscr{C}\subseteq \mathscr{B}$, so that $\sigma(X,\mathscr{C})\subseteq \sigma(X,\mathscr{A})$ and $\sigma(X,\mathscr{C})\subseteq \sigma(X,\mathscr{B})$ (without the hypothesis that $S$ is convex). What about the converse?

Ps. Here, the weak topology $\sigma(X,\mathscr{A})$ is the coarser topology on $X$ such that each function in $\mathscr{A}$ is continuous.

Answer 1

Building upon the example in the comments of two distinct Fréchet topologies $\sigma$ and $\tau$ on a vector space $X$, let us take $\mathscr A=(X,\sigma)'$ (the continuous dual, $\mathscr B=(X,\tau)'$, and $\mathscr C=\mathscr A \cap \mathscr B$. Then $S=\{0\}$ is closed in $\sigma(X,\mathscr A)$ and in $\sigma(X,\mathscr B)$ (because weak topologies of Hausdorff LCS are Hausdorff) but it is not closed in $\sigma(X,\mathscr C)$ because otherwise that latter locally convex topology would be Hausdorff and coarser than $\sigma$ and $\tau$ and this would imply that the identical map $(X,\sigma)\to(X,\tau)$ and its inverse had closed graph so that $\sigma=\tau$ by the closed graph theorem.

It might be possible that $\mathscr C=\{0\}$ in this situation. But this could be remedied by considering a product $\mathbb R\times X$. An additional assumption which might change the game would be that $\mathscr C$ separates points, i.e., $\sigma(X,\mathscr C)$ is Hausdorff. I do not have a clue how to prove something in this case (and I am in fact rather sceptical that the claim then becomes true) -- but it excludes the example.

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Two additional remarks:

The example shows that final topologies (i.e., very simple categorical colimits) depend very much on the category. For the final locally convex topology, I believe that your claim is true.

The existence of distinct Fréchet (or even Banach) topology follows from the fact that separable Fréchet spaces have Hamel dimension $c$, the cardinality of the reals. If you believe in the continuum hypothesis, this is very easy because they cannot be countably dimensional by Baire's theorem and the dimension cannot be bigger than $c$ because one can eaily construct a linear injection into $\mathbb R^{\mathbb N}$. However, the continuum hypothesis is not needed as can be seen, e.g., in Corollary 2.2.5 in the book Barrelled Locally Convex Spaces of Bonet and Pérez Carreras.

Answer 2

This is a long comment.

Consider the case $0\in S.$ And topology $X_{\mathcal{A}+\mathcal{B}} = \sigma(X, \mathcal{A}+\mathcal{B}).$ Let $S^\circ = \{y\in \mathcal{A}+\mathcal{B}, \langle S, y\rangle \ge -1\}$ polar of $S.$ As $S$ is closed in $\sigma(X, \mathcal{A}),$ it should be $(S^\circ\cap \mathcal{A})^\circ = S.$ $((S^\circ\cap \mathcal{A})\cap \mathcal{B})^\circ$ is convex closure of union of $S$ and $\mathcal{B}^\circ$ in $X$ if $\mathcal{A}$ and $\mathcal{B}$ are closed in $\sigma(\mathcal{A}+\mathcal{B}, X).$ So if $\mathcal{B}^\circ = \{0\}$ (separating) then $((S^\circ\cap \mathcal{A})\cap \mathcal{B})^\circ = S$ and $S$ is closed for $\sigma(X, \mathcal{A}\cap\mathcal{B}).$ Similarly for $\mathcal{A}$ separating. Careful check shows that separateness is not important here ($S$ should contain $B^\circ$ and $A^\circ$ if $S$ is closed in $X_{\mathcal{A}},X_{\mathcal{B}}$), only "closedness" in sum.

Remark. I have in mind the following result: $\sigma(X/M^\circ, M)$ is quotient topology of $\sigma(X, \mathcal{A})$ if and only if $M$ is weak*-closed in $\mathcal{A}$ for $\sigma(\mathcal{A}, X).$ So maybe it is possible to derive necessary conditions for the equivalence to hold for all convex $S.$ Sufficient condition is "closedness".