The Lawson number of a semitopological semilattice

Abstract

For a Hausdorff topologized semilattice $X$ its $Lawsonnumber$ $\bar{\Lambda }(X)$ is the smallest cardinal $\kappa$ such that for any distinct points $x,y\in X$ there exists a family $\mathcal{U}$ of closed neighborhoods of $x$ in $X$ such that $|\mathcal{U}|\le \kappa$ and $\bigcap \mathcal{U}$ is a subsemilattice of $X$ that does not contain $y$. It follows that $\bar{\Lambda }(X)\le \bar{\psi }(X)$, where $\bar{\psi }(X)$ is the smallest cardinal $\kappa$ such that for any point $x\in X$ there exists a family $\mathcal{U}$ of closed neighborhoods of $x$ in $X$ such that $|\mathcal{U}|\le \kappa$ and $\bigcap \mathcal{U}=\{x\}$. We prove that a compact Hausdorff semitopological semilattice $X$ is Lawson (i.e., has a base of the topology consisting of subsemilattices) if and only if $\bar{\Lambda }(X)=1$. Each Hausdorff topological semilattice $X$ has Lawson number $\bar{\Lambda }(X)\le \omega$. On the other hand, for any infinite cardinal $\lambda$ we construct a Hausdorff zero-dimensional semitopological semilattice $X$ such that $|X|=\lambda$ and $\bar{\Lambda }(X)=\bar{\psi }(X)=cf(\lambda )$. A topologized semilattice $X$ is called (i) $\omega$-$Lawson$ if $\bar{\Lambda }(X)\le \omega$; (ii) $complete$ if each non-empty chain $C\subset X$ has $\inf C\in \overline{C}$ and $\sup C\in \overline{C}$. We prove that for any complete subsemilattice $X$ of an $\omega$-Lawson semitopological semilattice $Y$, the partial order ${\le }_X=\{(x,y)\in X\times X:xy=x\}$ of $X$ is closed in $Y\times Y$ and hence $X$ is closed in $Y$. This implies that for any continuous homomorphism $h:X\to Y$ from a compete topologized semilattice $X$ to an $\omega$-Lawson semitopological semilattice $Y$ the image $h(X)$ is closed in $Y$.

arXiv comment10 pages. arXiv admin note: text overlap with arXiv:1806.02868

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Cited by in corpus (1)

• Characterizing categorically closed commutative semigroups