Is there finest topology which makes given vector space into a topological vector space?

Question

I we are given a vector space $(V,+,\cdot)$ over a field $\mathbb K$ (where $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$), is there the finest topology $\mathcal T$, such that $(V,\mathcal T)$ is a topological vector space?

I'll explain below what I have tried. This seems a very natural question to me, so I guess the answer (whether it is affirmative or negative) should be well known, at least to some local experts in topological vector spaces.

First guess would be the discrete topology, but discrete topology does not give a topological vector space. Indeed, if $\cdot \colon \mathbb K\times V\to V$ is continuous then for each fixed $v\in V$ the map $\alpha \mapsto \alpha\cdot v$ from $\mathbb K$ to $V$ should be continuous. Since the map is injective for $v\ne0$, this would imply that $\mathbb K$ has discrete topology too.

Perhaps we could get a topology such that the maps $\alpha \mapsto \alpha\cdot u$, $v \mapsto \beta\cdot v$, $v\mapsto v+u$ are continuous for each $u\in V$ and $\beta\in\mathbb K$ by iteratively taking the initial topology w.r.t. these maps, starting from the discrete topology on $V$ and the usual topology on $\mathbb K$ and iterating this proces. But even if this worked, we would get only separate continuity for $\cdot \colon \mathbb K\times V\to V$, and we want this map to be jointly continuous.

While looking for answer to my question, I found out about the existence of finest locally convex topology (Google Books), which is mentioned, for example in this question or this question. This is different from what I am asking here.

This is discussed at great length in chapter 26 of Schechter's Handbook of analysis and its foundations. Briefly, a vector space topology $X$ can be described by a collection of $F$-seminorms. The finest vector space topology on $X$ is then given by the collection of all $F$-seminorms on $X$. Similarly, by restricting attention to actual seminorms, one gets the finest locally convex topology described in the questions you link to. Alternative descriptions are in terms of suitable neighborhood bases as in the answer below. — Martin, Jul 18 '13 at 16:50

Etienne · Accepted Answer · 2013-07-18T09:06:43.910

5

I would say that there is a finest vector space topology. However, my argument may be completely wrong...

Let $(\tau_i)_{i\in I}$ be the family of all vector space topologies on $V$. Call a set $\mathcal E\subset V$ $elementary$ if it can be written as $\mathcal E=\mathcal U_{i_1}\cap \cdots \cap \mathcal U_{i_N}$, where $\mathcal U_{i_k}\in\tau_{i_k}$. Finally, let $\tau$ be the family of all subsets of $V$ which are unions of elementary sets.

Since the family of elementary sets contains $\emptyset$ and $V$ and is stable under finite intersections, it is rather clear that $\tau$ is a topology on $V$. Obviously, $\tau$ is finer than all $\tau_i$; so it remains to check that $\tau$ is a vector space topology.

Take $x,x'\in V$ and let $\mathcal W$ be a $\tau$-neighbourhood of $x+x'$. Choose an elementary set $\mathcal E=\mathcal U_{i_1}\cap \cdots \cap \mathcal U_{i_N}$ such that $x+x'\in \mathcal E\subset\mathcal W$. Since each $\tau_{i_k}$ is a vector space topology, one can choose $\mathcal V_k, \mathcal V'_k\in\tau_k$ such that $x\in\mathcal V_k$, $x'\in\mathcal V'_k$ and $\mathcal V_k+\mathcal V'_k\subset \mathcal U_{i_k}$. Then $\mathcal V=\bigcap_k \mathcal V_k$ and $\mathcal V'=\bigcap_k\mathcal V'_k$ are $\tau$-neighbourhoods of $x$ and $x'$ such that $\mathcal V+\mathcal V'\subset\mathcal E\subset \mathcal W$. This shows that addition is jointly continuous on $(V,\tau)\times (V,\tau)$.

Continuity of scalar multiplication can be checked in the same way.

Please let me know if this seems correct!

edited Jul 18 '13 at 09:06

answered Jul 18 '13 at 08:29

Etienne

13,636

A brief way to explain what you did is: you are simply taking the topology which has the union of all TVS topologies as a subbasis, that is the smallest topology which contains that union. – Mariano Suárez-Álvarez Jul 18 '13 at 08:49
In the other words, this is join of these topologies in the lattice of topologies. Which seems a reasonable choice to try. – Martin Sleziak Jul 18 '13 at 08:52
@Mariano. Yes, exactly. I should have said it in this way. – Etienne Jul 18 '13 at 08:52
You probably wanted to write $x+x'\in\mathcal E\subset \mathcal W$ instead of just $x+x'\in\mathcal E$. – Martin Sleziak Jul 18 '13 at 08:53
1

Also, one need only check that the preimage of each subbasic set of $\tau$ under the sum map $V\times V\to V$ is open, because of general properties of subbases, so you need not take an intersection. – Mariano Suárez-Álvarez Jul 18 '13 at 08:54
@Martin. Yes, thanks! – Etienne Jul 18 '13 at 09:07
@Mariano. You're right again, of course. – Etienne Jul 18 '13 at 09:07
To answer the last line of your post: Your proof seems correct to me. – Martin Sleziak Jul 18 '13 at 09:15
OK, thanks Martin. – Etienne Jul 18 '13 at 10:53

score 2 · Answer 2 · answered Feb 14 '23 at 05:14

In fact, your idea of iteratively taking the topology generated by the structure maps continuous does work--you just have to use the joint addition and scalar multiplication maps rather than the single-variable coordinatewise maps that you propose.

To be precise, given a topology $\tau$ on $V$, let $c(\tau)$ be the set of $U\in \tau$ such that $\sigma^{-1}(U)\subseteq V\times V$ and $\mu^{-1}(U)\subseteq \mathbb{K}\times V$ are open with respect to the product topologies given by $\tau$, where $\sigma:V\times V\to V$ is addition and $\mu:\mathbb{K}\times V\to V$ is scalar multiplication. Then any topological vector space topology on $V$ which is contained in $\tau$ must in fact be contained in $c(\tau)$. So now you can iterate this transfinitely: let $\tau_0$ be the discrete topology, $\tau_{\alpha+1}=c(\tau_\alpha)$, and $\tau_\alpha=\bigcap_{\beta<\alpha}\tau_\beta$ if $\alpha$ is a limit ordinal. By induction, any topological vector space topology on $V$ must be contained in $\tau_\alpha$ for all ordinals $\alpha$. But $(\tau_\alpha)$ is a descending sequence of topologies on $V$, so it must eventually stabilize (since there are more ordinals than topologies on $V$). That is, $\tau_\alpha=c(\tau_\alpha)$ for some $\alpha$, which means precisely that $\tau_\alpha$ makes $V$ a topological vector space. This $\tau_\alpha$ is then the finest topological vector space topology on $V$.

score 0 · Answer 3 · answered Feb 06 '24 at 11:19

Here is a more structural way to "construct" the finest vector topology.

Let $X$ be a vector space, let $(X_i)$ be topological vector spaces, and let $\mathcal{T}$ be the initial topology induced by linear maps $T_i \colon X \to X_i$. By noticing that $A_i \circ (T_i \times T_i) = T_i \circ A$, where $A_i$ is addition on $X_i$ and $A$ is addition on $X$, we see that $A$ is continuous. A similar argument shows that scalar multiplication on $X$ is continuous, so $\mathcal{T}$ is a vector topology.

Let $\mathfrak{T}$ be the set of vector topologies on $X$. The join in the lattice of topologies of a collection $(\mathcal{T}_i)$ of topologies from $\mathfrak{T}$ is the initial topology induced by the identity maps $X \to (X,\mathcal{T}_i)$, and this is a vector topology by the above, hence lies in $\mathfrak{T}$. In particular, $\bigvee \mathfrak{T}$ exists in $\mathfrak{T}$.

Is there finest topology which makes given vector space into a topological vector space?

3 Answers3

Linked