Ultrafilter semigroups generated by direct sums (Q535216)

Let \(\lambda\) be an infinite cardinal. The direct sum of sets \(A_\alpha\) for \(\alpha<\lambda\) with distinguished elements \(0_\alpha\in A_\alpha\) is the subset~\(X\) of the Cartesian product \(\prod_{\alpha<\lambda}A_\alpha\) consisting of all~\(x\) with finite \(\text{supp}(x)=\{\alpha:x(\alpha)\neq0_\alpha\}\). Endow~\(X\) with the topology by taking as a~neighbourhood base at \(x\in X\) the subsets of~\(X\) of the form \(\{y\in X:y(\alpha)=x(\alpha)\) for all \(\alpha<\gamma\}\) where \(\gamma<\lambda\). Let \(\text{Ult}(X)\) denote the set of all nonprincipal ultrafilters on~\(X\) converging to \(0\in X\). For \(x,y\in X\) with \(\text{supp}(x)\cap\text{supp}{y}=\emptyset\) the partial operation \(x+y\in X\) is defined by \(\text{supp}(x+y)=\text{supp}(x)\cup\text{supp}(y)\) and \((x+y)(\alpha)=x(\alpha)\), if \(\alpha\in\text{supp}(x)\), \((x+y)(\alpha)=y(\alpha)\), if \(\alpha\in\text{supp}(y)\). This partial operation on~\(X\) induces a~semigroup operation on \(\text{Ult}(X)\). The authors prove that if the direct sums \(X\) and~\(Y\) are homeomorphic, then the semigroups \(\text{Ult}(X)\) and \(\text{Ult}(Y)\) are isomorphic.