A topological space \(X\) is called metacompact if for every open cover \(\mathcal C\) of \(X\), there is a subcover \(\mathcal C'\) satisfying the property that for every point \(p\in X\), there are only finitely many open sets in \(\mathcal C'\) containing \(p\).
Show that metacompactness is a homeomorphism invariant.
Let \(X\) be the integers with the topology \(\mathcal T:=\{U\subseteq X\mid 0\in U\}\cup\{\varnothing\}\). Show that this space is not metacompact.
Solution
1. (a) If \(f\colon X\to Y\) is a homeomorphism and \(X\) is metacompact, then \(Y\) is metacompact.
Let \(\mathcal C\) be an open cover of \(Y\). Then \(\{f^{-1}(U):U\in\mathcal C\}\) is an open cover of \(X\), so it has a point-finite subcover \(\{f^{-1}(U):U\in\mathcal C'\}\) for some \(\mathcal C'\subseteq\mathcal C\). Since \(f\) is onto, \(\mathcal C'\) covers \(Y\). For \(y\in Y\) and \(x=f^{-1}(y)\), \(y\in U\) if and only if \(x\in f^{-1}(U)\), so \(y\) lies in only finitely many members of \(\mathcal C'\). Applying this to \(f^{-1}\) gives the converse, so metacompactness is a homeomorphism invariant.
2. In \((\ZZ,\mathcal T)\), the open cover \(\mathcal C=\bigl\{\{0,n\}:n\in\ZZ\setminus\{0\}\bigr\}\) has no point-finite subcover.
Each \(\{0,n\}\) contains \(0\), so it is open, and each nonzero \(n\) lies in \(\{0,n\}\); so \(\mathcal C\) covers \(\ZZ\). The only member of \(\mathcal C\) containing \(n\neq0\) is \(\{0,n\}\), so every subcover is \(\mathcal C\) itself. The point \(0\) lies in all of the infinitely many members of \(\mathcal C\).
3. Q.E.D.
Step 1 proves (a), and step 2 shows that \((\ZZ,\mathcal T)\) is not metacompact, proving (b).