Two questions on path connected spaces

All spaces below are Tychonoff.

The following definition is taken from Small subsets of first countable spaces by van Douwen and Wage. It's apparently not the usual definition of property $D$, but equivalent to the usual definition when $X$ is first countable.

Definition. Say that $X$ has property $D$ if for any two disjoint closed subsets $A, B$ with $A$ countable and discrete, there is open disjoint $U, V$ with $A\subseteq U$ and $B\subseteq V$.

Lemma. If $A\subseteq X$ is countable closed and discrete, and $X$ has property $D$, then $A$ is $C$-embedded.

Proof: See exercise $3L.4$ in Rings of continuous functions by Gillman and Jerison. $\square$

Proposition 1. If $X$ is countably compact or normal, then $X$ has property $D$.

Proof: If $X$ is normal this is obvious, if $X$ is countably compact see exercise $3L.5$ of Rings of continuous functions by Gillman and Jerison. $\square$

Theorem 1. If $X$ has property $D$ and $x\in \beta X\setminus X$, then there is no sequence $(x_n)\subseteq X$ which converges to $x$.

Proof: $A = \{x_n : n\in\mathbb{N}\}$ is a closed discrete countable subset of $X$, and so a $C$-embedded subset of $X$, and so $C^\ast$-embedded subset of $\beta X$. If $x_n$ are distinct, let $f:A\to\mathbb{R}$ be such that $f(x_n) = (-1)^n$, then $f$ has no continuous extension to $\beta X$, contradiction. $\square$

Theorem 2. If $X$ is non-compact, locally compact and has property $D$ then $\beta X$ is not path-connected.

Proof: Let $p:[0, 1]\to \beta X$ be an arc such that $p(0)\in X$ and $p(1)\in\beta X\setminus X$. Since $p^{-1}(X)$ is not closed, there exists a sequence $t_n\in p^{-1}(X)$ with $t_n\to t$ and $p(t)\in \beta X\setminus X$. Contradiction. $\square$

Theorem 3. If $|\nu X|^\omega < |\beta X|$, then $\beta X$ is not path-connected.

Proof: Suppose $\beta X$ is path-connected. For each $x\in \beta X\setminus \nu X$ take an arc $p:[0, 1]\to \beta X$ with $p(0)\in X$. Since $\beta X\setminus \nu X$ contains no closed non-discrete sets of cardinality $ < 2^\mathfrak{c}$, there doesn't exist $0 < r < 1$ with $p([r, 1])\subseteq \beta X\setminus \nu X$, and so we can take a sequence $t_n\in p^{-1}(\nu X)$ with $t_n\to 1$. So for each $x\in \beta X\setminus \nu X$ there exists a countable set $S_x\subseteq \nu X$ such that $\overline{S_x} = S_x\cup \{x\}$. The map $x\mapsto S_x$ is injective into a set of cardinality $|\nu X|^\omega$, contradiction. $\square$

Theorem 4. If $\nu X$ has property $D$ and $X$ is not pseudocompact, then $\beta X$ is not path-connected.

Proof: Let $p:[0, 1]\to \beta X$ be an arc such that $p(0)\in X$ and $p(1)\in\beta X\setminus \nu X$. If $p^{-1}(\nu X)$ were closed, then $\beta X\setminus \nu X$ contains a copy of $[0, 1]$, which is impossible since any non-discrete closed subset of $\beta X$ in $\beta X\setminus \nu X$ is of cardinality $\geq 2^\mathfrak{c}$. So $p^{-1}(\nu X)$ is not closed, and we can take a sequence $t_n\in p^{-1}(\nu X)$ with $t_n\to t$ and $p(t)\in \beta X\setminus \nu X$. Since $\beta(\nu X) = \beta X$, this is impossible from above proposition. $\square$

Corollary 1. If $X$ is realcompact, normal and non-compact, then $\beta X$ is not path-connected.

Proof: $\nu X = X$ is has property $D$ and is not pseudocompact, so by theorem 4, $\beta X$ is not path-connected. $\square$

Proposition 2. A paracompact space $X$ is realcompact iff every closed discrete subset of $X$ is smaller than least measurable cardinal.

Proof: See theorem 3 in Measures in fully normal spaces by Katetov, and remarks after. $\square$

Corollary 2. If $X$ is non-compact paracompact space of cardinality smaller than least measurable cardinal, then $\beta X$ is not path-connected.

Proof of corollary: By proposition 2, $X$ is realcompact so $\nu X = X$ is normal. By corollary 1, $\beta X$ is not path-connected. $\square$

Here is a proposed counterexample to Question 1 which is second countable.

Suppose that $S$ is a once-punctured torus equipped with a complete hyperbolic metric of finite area. We take $\Lambda$ to be a compact geodesic lamination in $S$. That is, $\Lambda$ is a compact set which is also a disjoint union of simple geodesics. We arrange matters so that $\Lambda$ is non-empty and contains no loops. It follows that the path components of $\Lambda$ (its leaves) are all copies of $\mathbb{R}$ and there are uncountably many such. (In fact, an arc transverse to $\Lambda$ meets it in a Cantor set.) Thus $\Lambda$ is not path connected. There are two special leaves of $\Lambda$ called the boundary leaves; all other leaves are called interior leaves. Now, every leaf of $\Lambda$ is dense in $\Lambda$ (this is not obvious). Finally, the boundary leaves are approached from only one side, while interior leaves are approached from both sides. Finally, $S - \Lambda$ is homeomorphic to an "open disk minus a single point".

Let $f$ be any continuous map from $\Lambda$ to $\mathbb{C}$. We must show that $f(\Lambda)$ is path-connected. There are many cases to check:

• If $f$ is an embedding then it extends to give an embedding of $S$ into $\mathbb{C}$, a contradiction.
• If the images of two leaves of $\Lambda$ cross transversely (or the image of one leaf crosses itself) then we are done by the density of each leaf in $\Lambda$.
• If $f$ collapses some transverse Cantor set then we are done, for the same reason.
• If $f$ collapses a leaf (or even a ray in a leaf) then $f$ is constant and we are done.
• If $f$ does not collapse any rays, and does not identify points of distinct leaves, then (?) $f$ is homotopic to an embedding, and we reach a contradiction.
• If $f$ does identify points on distinct leaves, then it is either collapsing a transverse Cantor set or it is causing leaves to cross...

Hmm. This is not as tidy as I hoped - so I will leave it here in hopes it is useful to somebody. :)