Proposition 1   A topological vector lattice $V$ is a topological lattice.

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. $(1\Leftrightarrow 2)$ . 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 $(3\Leftrightarrow 4)$ 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.

Lemma 2   Let $V$ be a vector lattice. Then $|a^+-b^+|\le |a-b|$ for any $a,b\in V$ .

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 $|a^+-b^+|\le (b-a)\vee (-a\wedge 0)\vee (a-b)\vee (-b\wedge 0) \le (b-a)\vee (a-b)=|a-b|$ .

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.

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 $\lbrace x\mid 0\le x\rbrace = \lbrace x\mid x^-=0\rbrace$ , which is closed since $^-$ is continuous, and the positive cone is the inverse image of a singleton, a closed set in $\mathbb{R}$ .