2 The results

We first present a very general result that connects NWC families of maps and resolvability.

Theorem 2.1 Let $\mathcal {F}$ be a NWC family of maps of the topological space X to the set Y. Moreover, $\mathcal {B}$ is a $\pi $ -base of X and $\mathcal {A}$ is a disjoint family of subsets of Y such that, putting $\mathcal {B}_f = \{B \in \mathcal {B} : |f[B]|> 1\}$ and $U_f = \bigcup (\mathcal {B} \setminus \mathcal {B}_f)$ for any $f \in \mathcal {F}$ , we have

$$ \begin{align*}\forall\,f \in \mathcal{F}\,\forall\,B \in \mathcal{B}_f\,\forall\,A \in \mathcal{A} \,(A \cap f[B \setminus U_f] \ne \emptyset).\end{align*} $$

Then X is $|\mathcal {A}|$ -resolvable.

Proof We are going to use transfinite recursion to produce a disjoint collection $\{D(A) : A \in \mathcal {A}\}$ of dense subsets of X.

To do that, we first introduce the following piece of notation. For $f \in \mathcal {F},\,\,B \in \mathcal {B}_f$ , and $A \in \mathcal {A}$ , we let

$$ \begin{align*}S(f, B, A) = f^{-1}[A] \cap (B \setminus U_f). \end{align*} $$

It follows from our assumptions that $S(f, B, A) \ne \emptyset $ .

We start our recursive construction by choosing $f_0 \in \mathcal {F}$ such that $\mathcal {B}_{f_0} \ne \emptyset $ and then define

$$ \begin{align*}D_0(A) = \bigcup \{S(f_0,B,A) : B \in \mathcal{B}_{f_0}\}\end{align*} $$

for all ${A \in \mathcal {A}}$ . Then $\{D_0(A) : A \in \mathcal {A}\}$ is disjoint because $D_0(A) \subset {f_0}^{-1}[A]$ for each $A \in \mathcal {A}$ , moreover $B \cap D_0(A) \ne \emptyset $ whenever $B \in \mathcal {B}_{f_0}$ and $A \in \mathcal {A}$ .

Now, assume that $\alpha> 0$ and we have already defined $f_\beta \in \mathcal {F}$ and the family $\{D_\beta (A) : A \in \mathcal {A}\}$ for all $\beta < \alpha $ , and consider

$$ \begin{align*}\mathcal{B}_\alpha = \bigcup \{\mathcal{B}_{f_\beta} : \beta < \alpha\}.\end{align*} $$

If $\mathcal {B}_\alpha = \mathcal {B},$ then our construction stops. Otherwise, we may pick $f_\alpha \in \mathcal {F}$ such that $\mathcal {B}_{f\alpha } \setminus \mathcal {B}_\alpha \ne \emptyset $ . In this case, we define

$$ \begin{align*}D_\alpha(A) = \bigcup \{D_\beta(A) : \beta < \alpha\} \cup \bigcup\{S(f_\alpha, B, A) : B \in \mathcal{B}_{f\alpha} \setminus \mathcal{B}_\alpha\}\end{align*} $$

for each $A \in \mathcal {A}$ .

Note that we clearly have $\bigcup (\mathcal {B}_{f\alpha } \setminus \mathcal {B}_\alpha ) \subset \bigcap \{U_{f_\beta } : \beta < \alpha \}$ , hence it may be verified by straightforward transfinite induction that $\{D_\alpha (A) : A \in \mathcal {A}\}$ is disjoint, moreover, we have $B \cap D_\alpha (A) \ne \emptyset $ whenever $A \in \mathcal {A}$ and $B \in \mathcal {B}_{f_\beta }$ with $\beta \le \alpha $ .

Of course, this construction must stop at some ordinal $\alpha $ , in which case putting $D(A) = \bigcup \{D_\beta (A) : \beta < \alpha \}$ the disjoint family $\{D(A) : A \in \mathcal {A}\}$ consists of dense subsets of X because $B \cap D(A) \ne \emptyset $ for any $B \in \mathcal {B}$ and $A \in \mathcal {A}$ and $\mathcal {B}$ is a $\pi $ -base of X.

Now, we deduce from Theorem 2.1 our promised result concerning $\pi $ -connected spaces.

Proof To apply Theorem 2.1, we of course put $Y = \mathbb {R}$ and $\mathcal {F} = C(X)$ . For $\mathcal {B}$ , we choose a $\pi $ -base of X consisting of connected sets, and let $\mathcal {A}$ be any disjoint collection of dense subsets of $\mathbb {R}$ with $|\mathcal {A}| = \mathfrak {c}$ .

Note that for any $f \in C(X)$ and $B \in \mathcal {B}_f$ , we have $|f[B]|> 1$ , hence $f[B]$ is a nondegenerate interval in $\mathbb {R}$ , being a non-singleton connected subset of $\mathbb {R}$ , and thus $A \cap f[B] \ne \emptyset $ holds for all $A \in \mathcal {A}$ . Consequently, we shall be done if we prove the following claim.

Claim For any $f \in C(X)$ and $B \in \mathcal {B}_f$ , we have

$$ \begin{align*}f[B] = f[B \setminus U_f].\end{align*} $$

To see this, pick any $t \in f[B]$ and note that $B \cap f^{-1}(t)$ is a non-empty proper closed subset of the connected subspace B. Properness follows from $|f[B]|> 1$ . Consequently, the boundary $H_t$ of $B \cap f^{-1}(t)$ in B is non-empty. Also, we have $H_t \subset B \cap f^{-1}(t)$ , the latter set being closed in B.

Now, take any $B' \in \mathcal {B} \setminus \mathcal {B}_f$ , then $f[B'] = \{t'\}$ for some $t' \in \mathbb {R}$ . If $t' \ne t$ , then we have $B' \cap f^{-1}(t) = \emptyset $ , hence $B' \cap H_t = \emptyset $ as well. If, on the other hand, $t' = t$ , then $B'\cap B$ is in the interior of $B \cap f^{-1}(t)$ in B, hence we have $B' \cap H_t = \emptyset $ again. This, however, means that $H_t \cap U_f = \emptyset $ , consequently, $t \in f[H_t] \subset f[B \setminus U_f]$ , completing the proof of the claim.

A closer inspection of this proof reveals that we did not use the full force of continuity for the members f of the NWC family $\mathcal {F} = C(X)$ . What we used was that f preserves connectedness for $B \in \mathcal {B}$ , i.e., $f[B]$ is connected in $\mathbb {R}$ for $B \in \mathcal {B}$ , moreover, that $f^{-1}(t)$ is closed in X for all $t \in \mathbb {R}$ .

3 Discussion and questions

Corollary 2.2 is clearly sharp in the sense that the cardinal $\mathfrak {c}$ cannot be replaced with anything bigger because there are very nice crowded and locally connected spaces of cardinality $\mathfrak {c}$ . The natural question arises, however, if such a space X is maximally resolvable, i.e., $\Delta (X)$ -resolvable, where $\Delta (X)$ is the minimum cardinality of a non-empty open set in X.

In a recent arXiv preprint [Reference Lipin4], Lipin proved that if $2^{\mathfrak {c}} = 2^{(\mathfrak {c}^+)}$ , then there is a locally connected and pseudocompact Tykhonov space X such that $\Delta (X)> \mathfrak {c}$ but X is not $\mathfrak {c}^+$ -resolvable. So, at least consistently, Corollary 2.2 is sharp in that sense as well.

As we wrote in the Introduction, Costatini [Reference Costantini1] proved that crowded $\pi $ -connected regular spaces are $\omega $ -revolvable. As usual, in his treatment regular implies $T_1$ . But his proof actually works for all crowded $\pi $ -connected quasi-regular spaces as well, without the use of any additional separation axiom. Recall that a space is quasi-regular if for every non-empty open U, there is a non-empty open V such that $\overline {V} \subset U$ . The only slight modification we need is that in this case, the definition of crowded has to be replaced by the following assumption: there is no finite indiscrete open subspace. For $T_0$ -spaces, this assumption is clearly equivalent with the usual one, i.e., not having any isolated points.

Hewitt [Reference Hewitt3] gave an example of a regular connected space X of cardinality $\omega _1$ . Let $Y=\lambda _f(X)$ be the superextension of X consisting of all finitely generated maximal linked systems consisting of closed subsets of X. Then Y also has size $\omega _1$ , since the collection of all finite subsets of X has size $\omega _1$ , moreover, Y is Hausdorff, connected and locally connected (see [Reference Verbeek7, IV.3.4(v) and (viii)]).

We claim that Y is also regular and adopt the terminology of [Reference Verbeek7]. To prove this, let $m\in Y$ be arbitrary, and let A be a closed subset of Y not containing m. Let $F\subseteq X$ be a finite defining set for m, and let $n = m{\restriction }F$ . There is a finite collection of open subsets $\mathcal {U}$ of X such that $m\in \bigcap _{U\in \mathcal {U}} U^+ \subseteq Y\setminus A$ . For each $U\in \mathcal {U}$ , let $N_U\in n$ be such that $N_U\subseteq U$ . By regularity of X, we may pick an open neighborhood $V_U$ of $N_U$ whose closure is contained in U. Then $\bigcap _{U\in \mathcal {U}} V_U^+$ is a closed neighborhood of m that misses A.

Hence Y is a connected, locally connected regular space which by Costantini’s result is $\omega $ -resolvable. (In this special case, there is also a direct proof of this.) But Y has cardinality $\omega _1$ , hence it is not $\mathfrak {c}$ -resolvable if the Continuum Hypothesis fails. So, for crowded locally connected regular spaces, a positive answer to the following question is the best we can hope for.

We cannot resist mentioning here that it is still a widely open question whether non-singleton connected regular spaces are resolvable (see e.g., [Reference Pavlov and Pearl6] for details and references). This seems to be one of the most central open problems in the area.

We call a space X nowhere 0-dimensional if no non-empty open subspace of X is 0-dimensional. Clearly, any crowded locally connected, even $\pi $ -connected, space is nowhere 0-dimensional.

Assume that X is a nowhere 0-dimensional Tykhonov space. We may assume that X is a subspace of $\mathbb {R}^I$ for some set I. For every $i\in I$ , let $\pi _i\colon \mathbb {R}^I \to \mathbb {R}$ be the ith projection. Then, for every nonempty open subset U of X, there exists $i\in I$ such that $\pi _i(U)$ has nonempty interior. For otherwise U would be a subspace of a product of 0-dimensional spaces, and so would itself be 0-dimensional. Hence the collection of projections $\mathcal {P} = \{\pi _i{\restriction }X : i \in I\}$ is NWC in a (very) strong sense. This observation lead us to the following problem.