Connected set which is no-where path connected 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?
 A: 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.
A: 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.
A: 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.
A: 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.
