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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. $v(X)$.

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  • $\begingroup$ Does your definition of vector topologies require that the topology is Hausdorff? $\endgroup$ – Daniel Fischer Mar 11 '16 at 11:08
  • $\begingroup$ No. The intersection of two Hausdorff (vector) topologies is not necessarily Hausdorff, so that the subcollection of Hausdorff topologies on $X$ forms only a semilattice. Note also that if $\tau_1$ and $\tau_2$ are two comparable completely metrizable vector topologies on a vector space $X$ (e.g. Fréchet space topologies) then $\tau_1 = \tau_2$ (by the open mapping theorem). Similarly and more generally, the intersection of two distinct Fréchet topologies cannot be Hausdorff. $\endgroup$ – yadaddy Mar 11 '16 at 13:28
  • $\begingroup$ Here the proof for completion: Let $\tau_1$ and $\tau_2$ be Fréchet space topologies on $X$ with Hausdorff intersection $\tau_0 := \tau_1 \cap \tau_2$. Then $id_{1,0} : (X, \tau_1) \to (X, \tau_0)$ is continuous. Since $\tau_0$ is Hausdorff it follows that $id_{1,0}$ has closed graph. By strengthening $\tau_0$ on the range to $\tau_2$ it follows that $id_{1,2} : (X, \tau_1) \to (X, \tau_2)$ has closed graph. Since $\tau_1$ and $\tau_2$ are Fréchet, we have the closed graph theorem which implies that $id_{1,2}$ is continuous. Similarly, $id_{2,1}$ is continuous and thus $\tau_1 = \tau_2$. $\endgroup$ – yadaddy Mar 11 '16 at 13:42
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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.

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  • $\begingroup$ I try to abstractify your construction. If $(L, \leq)$ is a complete lattice and $S \subseteq L$ then $(S, \leq)$ is a poset but in general not necessarily a lattice. Even if it is a lattice it is not necessarily a sublattice of $(L, \leq)$. In case $(S, \leq)$ is a complete join-semilattice then it is already a complete lattice when setting $\inf A := \sup \bigcap_{a \in A} \{ x \in L \mid x \leq a \}$ for $A \subseteq S$. Even if the suprema in $S$ and $L$ coincide this is not enough to deduce that $S$ is a sublattice of $L$ (since the corresponding infima do not need to coincide). $\endgroup$ – yadaddy Mar 30 '16 at 9:01
  • $\begingroup$ In the general abstract case, $\bigcap_{a\in A} \{ x\in L \mid x \leq a\}$ might not intersect $S$. In our setting, $S$ contains the smallest element of $L$, so the intersection always meets $S$. But indeed, it can be that $\sup \bigcap_{a\in A} \{ x\in S \mid x \leq a\} \neq \sup \bigcap_{a\in A} \{ x\in L \mid x \leq a\}$ even if the suprema in $S$ and $L$ coincide. $\endgroup$ – Daniel Fischer Mar 30 '16 at 9:22
  • $\begingroup$ Oh yes, sorry, I meant "... setting $\inf A := \sup \bigcap_{a \in A} \{ x \in S \mid x \leq a \}$ for $A \subseteq S$", since we want to use the supremum of the complete join-semilattice $S$. $\endgroup$ – yadaddy Mar 30 '16 at 9:25
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    $\begingroup$ Yes, then one just needs that $S$ contains a smallest element to make $S$ a complete lattice. But as you say, it need not be a sublattice even if $S$ is a sub-semilattice of $L$. $\endgroup$ – Daniel Fischer Mar 30 '16 at 9:29

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