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Background: It's a fun exercise to try to construct a connected space $T$ such that no two points in $T$ can be connected with a path.

My solution to the puzzle was to use an order topology on a totally ordered set where the cardinality of any interval is $2^{2^{\aleph_0}}$, thus no two points can be connected with a path because the cardinality of $[0,1]$ is not big enough.

My question: Does there exist a connected space $T$ with cardinality equal to $2^{\aleph_0}$ which is no-where path connected? Equivalently, does there exist a connected space $T$ such that $|T| = 2^{\aleph_0}$ and any continuous function $f:[0,1]\rightarrow T$ is constant?

Dark Malthorp
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    I suspect something like the co-countable topology on the real line works, but it's too late in the evening to think about what a continuous path should look like in such a space, or whether they are all constant. – Arthur Jul 02 '22 at 21:04
  • For instance the pseudoarc, https://www.google.com/url?sa=t&source=web&rct=j&url=https://en.m.wikipedia.org/wiki/Pseudo-arc&ved=2ahUKEwiZk5L51tv4AhU_DkQIHWYvAhUQFnoECAQQAQ&usg=AOvVaw2_y7TMFSX4qpHfGTS-VAlk I am sure, this question was asked before. – Moishe Kohan Jul 03 '22 at 02:34
  • See here for an example of a linearly ordered space, which is connected but does not contain any non-trivial path. Obviously, its cardinality is $2^{\aleph_0}$. – Ulli Jul 03 '22 at 07:08

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Gustin's Sequence Space $X$ is connected, but totally pathwise disconnected. A definition can be found here and it is also example 125 in Counterexamples in Topology (found here) by Lynn Arthur Steen and J. Arthur Seebach. It is countable though, but can probably be used to construct a space with fitting cardinality. Maybe the compact-open-topology on $X^X$ will do it, but that will require a few lemmas on how connectedness and totally pathwise disconnectedness is inherited.

Steven Clontz
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Samuel Adrian Antz
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I tried to work out the idea of Arthur: Let $\mathbb{R}$ be endowed with the co-countable topology, that is $V \subseteq \mathbb{R}$ is open iff $V= \emptyset$ or $V^c:=\mathbb{R} \setminus V$ is at most countable. Clearly $\mathbb{R}$ is connected, since if $\mathbb{R} =V_1 \cup V_2$ with $V_1,V_2$ open and $V_1 \cap V_2 = \emptyset$ then $V_1 \subseteq V_2^c$. If $V_2 \not= \emptyset$ then $V_1$ is at most countable, forcing $V_1= \emptyset$.

Now let $a,b \in \mathbb{R}$, $a \not=b$ and assume $f:[0,1] \to \mathbb{R}$ is continuous with $f(0)=a,f(1)=b$. Let $(t_k)_{k \in \mathbb{N}}$ be any dense sequence in $[0,1]$ and set $M:=\{f(t_k):k\in \mathbb{N}\}$. Then $M^c$ is open (since $M$ is at most countable), hence $f^{-1}(M^c)$ is an open subset of $[0,1]$ with $f^{-1}(M^c) \subseteq [0,1]\setminus \{t_k:k \in \mathbb{N}\}$. Since $(t_k)_{k \in \mathbb{N}}$ is dense, this implies $f^{-1}(M^c) = \emptyset$. Thus $f^{-1}(M)=[0,1]$. Let $M_1:=M\setminus \{a\}, M_2 := \{a\}$. Then $M_1, M_2$ are closed (at most countable) and $M_1 \not= \emptyset$ ($a\not=b)$. Now $f^{-1}(M_1),f^{-1}(M_2)$ are closed, nonempty, $f^{-1}(M_1) \cap f^{-1}(M_2) = \emptyset$ and $$ f^{-1}(M_1) \cup f^{-1}(M_2)= f^{-1}(M)= [0,1]. $$ As $[0,1]$ is connected this is impossible.

Gerd
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  • How do you know, that $M^\complement$ is open? $M$ is the infinite ($k\in\mathbb{N}$) union of closed sets ${t_k}$, which does not have to be closed again. For example, if the subsequence $(\frac{1}{k})_{k\in\mathbb{N}}$ is included, but not $0$. – Samuel Adrian Antz Jul 03 '22 at 10:44
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    It is open with respect to the co-countable topology. $M=(M^c)^c$ is at most countable, hence $M^c$ is open. Note that $f:[0,1] \to \mathbb{R}$ is continuous with $\mathbb{R}$ endowed with the co-countable topology and $[0,1]$ endowed with the Euclidean topology. – Gerd Jul 03 '22 at 10:51
  • @SamuelAdrianAntz I edited the answer to stress this point. – Gerd Jul 03 '22 at 11:02
  • Ah yes, I was thinking about the cofinite topology. Could you also elaborate, how you concluded $f^{-1}(M^\complement)=\emptyset$ from $(t_k)_{k\in\mathbb{N}}$ being dense? I probably again oversee something trivial. – Samuel Adrian Antz Jul 03 '22 at 11:02
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    @SamuelAdrianAntz As $[0,1]$ is endowed with the Euclidean topology, each nonempty open subset contains an open interval $(\alpha,\beta)$. But this interval would contain some $t_k$, since $(t_k)$ is dense. But this is impossible. – Gerd Jul 03 '22 at 11:05
  • That's true, thanks a lot! – Samuel Adrian Antz Jul 03 '22 at 11:08
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An example which is a subset of the Euclidean plane is "Cantor's Leaky Tent", no 129 in Steen & Seebach. It has a dispersal point, i.e. a point such that $T \setminus \{p\}$ is totally disconnected. Any non-trivial connected subset must contain $p$ and so $\{p\}$ must be dense in any non-trivial path, which is impossible. $T$ has the cardinality of the continuum.

David Hartley
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I give a different presentation of the cofinite idea, that uses a wonderful theorem of Sierpinski.

Let $X$ be any homeomorph of $\Bbb N$ with the cofinite topology - all we need is countability. $X$ is both connected and locally connected, as evidenced by the fact that any cofinite, hence (countably) infinite, subset is connected (if $A$ is open and cofinite, its complement is by definition finite so it can never be open as the parent space. Countability is not needed here.

Countability comes into play in the following observation. By the Sierpinski Theorem on Continua, if a nonempty continuum $Y$ is expressed as a countably infinite disjoint union of closed subsets $\{F_n\}_{n\in\Bbb N}$, then exactly one $n_0$ is such that $F_{n_0}=Y$ and all other $\{F_n\}_{n\in\Bbb N\setminus\{n_0\}}$ are empty.

Any path $\gamma:[0,1]\to X$ pulls countable disjoint families of closed sets back into countable disjoint families of closed sets. In particular, every singleton is closed in $X$ and $F_n:=\gamma^{-1}(\{x_n\}),n\in\Bbb N$ is a countably infinite, disjoint family of closed subsets in $[0,1]$, where $\{x_n\}_{n\in\Bbb N}$ is any enumeration of $X$. By the Sierpinski theorem, and the standard fact that $[0,1]$ is a topological continuum (in Euclidean subspace topology), all but one of the $F_n$ are empty and exactly one is the entirety of $[0,1]$ - say $F_{n_0}$. Then $\gamma$ is the constant map to $x_{n_0}$.

This concludes that the only paths in $X$ are the trivial constant paths, so $X$ is not path-connected despite its global and local connectivity.

FShrike
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