the *NTIP* iff every ${{\mathcal A}}{\, \mbox{\rlap{$\sqsubseteq$}     \raise 2.3pt \hbox{\scriptsize $\mathrm{c}$}}\; }C(X)$ contains a non-trivial idempotent.

So, the NTIP is trivially false of connected spaces. If $X$ is not connected, then the CSWP implies the NTIP. The following is Lemma 3.5 of [@KU]; it is also easy to prove from the Bishop Antisymmetric Decomposition (see [@BIS], or Theorem 13.1 in Chapter II of [@GAM]).

\[lemma-restrict-to-perfect\] Assume that $X$ is compact and every perfect subset of $X$ has the NTIP. Then $X$ has the CSWP.

Among the totally disconnected spaces, the NTIP is strictly weaker than the CSWP (see [@KU]). However, the lemma implies the following corollary, which is used to reduce proofs of the CSWP to proofs of the weaker NTIP:

\[cor-idem\] If ${{\mathfrak K}}$ is a closed-hereditary class of compact spaces and every perfect space in ${{\mathfrak K}}$ has the NTIP, then every space in ${{\mathfrak K}}$ has the CSWP.

In particular, if ${{\mathfrak K}}$ is the class of compact scattered spaces, then this corollary applies vacuously, so all spaces in ${{\mathfrak K}}$ have the CSWP. If ${{\mathfrak K}}$ contains some non-scattered spaces, then, as in [@KU; @HK1], we produce idempotents using:

\[lemma-get-clopen\] Suppose that ${{\mathcal A}}{\, \mbox{\rlap{$\sqsubseteq$}     \raise 