-2
$\begingroup$

I have been working through Howie's book on semigroups, and while doing so I've come across this question:

"For every subset $K$ of an inverse semigroup $S$, and for every $s \in S$, show that $(Ks)ω = ((Kω)s)ω$"

$ω$ is defined by:

Let $S$ be an inverse semigroup, and let $H$ be a subset of $S$. The upper saturation or closure $Hω$ of $H$ in $S$ is defined by:

$$Hω = \{s \in S : (\exists h \in H) h \le s\}$$

Note these facts about $ω$ are given in the book:

$\bullet H$ is a subset of $Hω$

$\bullet$ If $H$ is a subset of $K$ then $Hω$ is a subset of $Kω$

$\bullet(Hω)ω=Hω$

I'm having a lot of trouble with starting this off, any help would be greatly appreciated. Thanks in advance.

  • 1
    Why you destroyed your question ?It is not acceptable here. – 2017-01-06
  • 0
    I really don't know either why this question was closed and I will ask for reopen it. – 2017-01-07

1 Answers 1

3

The property actually holds for any semigroup equipped with a preorder relation $\leqslant$ compatible with the multiplication. The fact that it is an inverse semigroup is not used.

Easy inclusion. Since $K \subseteq Kω$, one gets $Ks \subseteq (Kω)s$, whence $(Ks)ω \subseteq ((Kω)s)ω$.

Opposite inclusion. If $t \in ((Kω)s)ω$, there exists $u \in (Kω)s$ such that $u \leqslant t$. Now since $u \in (Kω)s$, there exists $r \in Kω$ such that $u = rs$. Finally, since $r \in Kω$, there exists $k \in K$ such that $k \leqslant r$. It follows that $ks \leqslant rs = u \leqslant t$ and thus $t \in (Ks)ω$. Thus $((Kω)s)ω \subseteq (Ks)ω$.