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Hometopological vector lattice
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topological vector lattice
A topological vector lattice $V$ over $\mathbb{R}$ is

a Hausdorff topological vector space over $\mathbb{R}$,

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.
$(1\Leftrightarrow 2)$. If $\vee$ is continuous, then $x\wedge y=x+yx\vee y$ 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 $x^{}=\frac{1}{2}(xx)$ is continuous. Finally, $(3\Rightarrow 1)$ is given by $x\vee y=(xy+y)\vee(0+y)=(xy)\vee 0+y=(xy)^{+}+y$, 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 $a^{+}b^{+}\leqab$ for any $a,b\in V$.
Proof.
$a^{+}b^{+}=(b^{+}a^{+})\vee(a^{+}b^{+})=(b\vee 0a\vee 0)\vee(a\vee 0b% \vee 0)$. Next, $a\vee 0b\vee 0=(b+(a\wedge 0)\vee(a\wedge 0)=((ba)\wedge b)\vee(a\wedge 0)$ so that $a^{+}b^{+}=((ba)\wedge b)\vee(a\wedge 0)\vee((ab)\wedge a)\vee(b\wedge 0% )\leq(ba)\vee(a\wedge 0)\vee(ab)\vee(b\wedge 0)$. Since $(ba)\vee(ab)=ab$ and $a\vee 0$ are both in the positive cone of $V$, so is their sum, so that $0\leq(ba)\vee(ab)+(a\vee 0)=(ba)\vee(ab)(a\wedge 0)$, which means that $(a\wedge 0)\leq(ba)\vee(ab)$. Similarly, $(b\wedge 0)\leq(ba)\vee(ab)$. Combining these two inequalities, we see that $a^{+}b^{+}\leq(ba)\vee(a\wedge 0)\vee(ab)\vee(b\wedge 0)\leq(ba)\vee(a% b)=ab$. ∎
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 $xx_{0}$ is in a solid neighborhood $U$ of $0$ ($U\in N$), then so is $x^{+}x_{0}^{+}$ in $U$. Since $xx_{0}\in U$, $xx_{0}\in U$. But $x^{+}x_{0}^{+}\leqxx_{0}$ 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 $\{x\mid 0\leq x\}=\{x\mid x^{}=0\}$, which is closed since ${}^{}$ is continuous, and the positive cone is the inverse image of a singleton, a closed set in $\mathbb{R}$. ∎
Mathematics Subject Classification
06F20 no label found46A40 no label found Forums
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