topological vector lattice

A topological vector lattice $V$ over $ℝ$ is

• a Hausdorff topological vector space over $ℝ$,

• a vector lattice, and

• locally solid. This means that there is a neighborhood base of $0$ consisting of solid sets.

Proposition 1.

A topological vector lattice $V$ is a topological lattice.

Before proving this, we show the following equivalence on the continuity of various operations on a vector lattice $V$ that is also a topological vector space.

Lemma 1.

Let $V$ be a vector lattice and a topological vector space. The following are equivalent:

1. $\vee :V^2\to V$ is continuous (simultaneously in both arguments)

2. $\wedge :V^2\to V$ is continuous (simultaneously in both arguments)

3. ${}^{+}:V\to V$ given by $x^{+}:=x\vee 0$ is continuous

4. ${}^{-}:V\to V$ given by $x^{-}:=-x\vee 0$ is continuous

5. $|\cdot |:V\to V$ given by $|x|:=-x\vee x$ is continuous

Proof.

$(1\iff 2)$. If $\vee$ is continuous, then $x\wedge y=x+y-x\vee y$ is continuous too, as $+$ and $-$ are both continuous under a topological vector space. This proof works in reverse too. $(1\Rightarrow 3)$, $(1\Rightarrow 4)$, and $(3\iff 4)$ are obvious. To see $(4\Rightarrow 5)$, we see that $|x|=x^{+}+x^{-}$, since ${}^{-}$ is continuous, ${}^{+}$ is continuous also, so that $|\cdot |$ is continuous. To see $(5\Rightarrow 4)$, we use the identity $x=x^{+}-x^{-}$, so that $|x|=(x+x^{-})+x^{-}$, which implies $x^{-}=\frac{1}{2}(|x|-x)$ is continuous. Finally, $(3\Rightarrow 1)$ is given by $x\vee y=(x-y+y)\vee (0+y)=(x-y)\vee 0+y={(x-y)}^{+}+y$, which is continuous.∎

In addition, we show an important inequality that is true on any vector lattice:

Lemma 2.

Let $V$ be a vector lattice. Then $\mathrm{|}{a}^{\mathrm{+}}\mathrm{-}{b}^{\mathrm{+}}\mathrm{|}\mathrm{\le }\mathrm{|}a\mathrm{-}b\mathrm{|}$ for any $a\mathrm{,}b\mathrm{\in }V$.

Proof.

$|a^{+}-b^{+}|=(b^{+}-a^{+})\vee (a^{+}-b^{+})=(b\vee 0-a\vee 0)\vee (a\vee 0-b\vee 0)$. Next, $a\vee 0-b\vee 0=(b+(-a\wedge 0)\vee (-a\wedge 0)=((b-a)\wedge b)\vee (-a\wedge 0)$ so that $|a^{+}-b^{+}|=((b-a)\wedge b)\vee (-a\wedge 0)\vee ((a-b)\wedge a)\vee (-b\wedge 0)\le (b-a)\vee (-a\wedge 0)\vee (a-b)\vee (-b\wedge 0)$. Since $(b-a)\vee (a-b)=|a-b|$ and $a\vee 0$ are both in the positive cone of $V$, so is their sum, so that $0\le (b-a)\vee (a-b)+(a\vee 0)=(b-a)\vee (a-b)-(-a\wedge 0)$, which means that $(-a\wedge 0)\le (b-a)\vee (a-b)$. Similarly, $(-b\wedge 0)\le (b-a)\vee (a-b)$. Combining these two inequalities, we see that $|a^{+}-b^{+}|\le (b-a)\vee (-a\wedge 0)\vee (a-b)\vee (-b\wedge 0)\le (b-a)\vee (a-b)=|a-b|$.∎

We are now ready to prove the main assertion.

Proof.

To show that $V$ is a topological lattice, we need to show that the lattice operations meet $\wedge$ and join $\vee$ are continuous, which, by Lemma 1, is equivalent in showing, say, that ${}^{+}$ is continuous. Suppose $N$ is a neighborhood base of 0 consisting of solid sets. We prove that ${}^{+}$ is continuous. This amounts to showing that if $x$ is close to $x_0$, then $x^{+}$ is close to $x_0^{+}$, which is the same as saying that if $x-x_0$ is in a solid neighborhood $U$ of $0$ ($U\in N$), then so is $x^{+}-x_0^{+}$ in $U$. Since $x-x_0\in U$, $|x-x_0|\in U$. But $|x^{+}-x_0^{+}|\le |x-x_0|$ by Lemma 2, and $U$ is solid, $x^{+}-x_0^{+}\in U$ as well, and therefore ${}^{+}$ is continuous.∎

As a corollary, we have

Proposition 2.

A topological vector lattice is an ordered topological vector space.

Proof.

All we need to show is that the positive cone is a closed set. But the positive cone is defined as $\{x\mid 0\le x\}=\{x\mid x^{-}=0\}$, which is closed since ${}^{-}$ is continuous, and the positive cone is the inverse image of a singleton, a closed set in $ℝ$.∎