Do linear continua contain $\mathbb{R}$? Can a nontrivial connected space have only trivial path components?

Question

Consider a connected space $X$ under the order topology, such that $X$ contains at least two elements. Does there exist a subspace of $X$ homeomorphic to $\mathbb{R}$? If there isn't, then $X$ has only trivial path components.

Answer 1

Consider $X = [0,1]^{\omega_1}$ with the dictionary order. The property that $f, h \in X \Rightarrow \exists g : f < g < h$ follows readily so I'll skip that. The least upper bound property holds as well (Does $[0,1]^{\mathbb{\lambda}}$ have the least upper bound property under the dictionary order?) so we conclude $X$ is a linear continuum.

We prove for $f < g$ such that $f$ and $g$ are eventually $0$ that there is no path $\phi : [0,1] \to X$ such that $\phi(0) = f$ and $\phi(1) = g$.

Let $\alpha$ be an element of $\omega_1$ such that $f$ and $g$ are $0$ for all elements greater than $\alpha$. Consider $\beta \in \omega_1$ with $\beta > \alpha$ and $f_{\beta} \in X$ which agrees with $f$ everywhere except at $\beta$ and has $f_{\beta}(\beta) := f(\beta) + 1$.

Define $I_{\beta} := (f_{\beta+1}, f_\beta)$. Each $I_\beta$ is a disjoint open subset of $[f,g]$ which is nonempty by the linear continuum property. There are uncountably many such $\beta$ so $[f,g]$ cannot be homeomorphic to $[0,1]$ and we conclude there is no path connecting $f$ and $g$.

Now let $f$ and $g$ be arbitrary elements of $X$. We prove there exists $f_0, g_0$ which are eventually zero such that $f < f_0 < g_0 < g$. Let $\alpha$ be the first element such that $f(\alpha) \neq g(\alpha)$, so $f(\alpha) < g(\alpha)$. Pick $f_0$ and $g_0$ such that they agree with $f$ and $g$ before $\alpha$, such that $f(\alpha) < f_0(\alpha) < g_0(\alpha) < g(\alpha)$ - which is possible since $[0,1]$ is a linear continuum, and such that $f_0$ and $g_0$ are zero after $\alpha$. Then the relation holds by definition of the dictionary order.

Therefore the closed interval between any two points contains a closed interval which is not path connected and so cannot be path connected.

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I asked if $[0,1]^{\omega}$ under the dictionary order topology has only trivial path components and Niels Diepeveen answered this affirmatively (https://math.stackexchange.com/a/2318699/444923).

The nice thing about this linear continuum is that it has the same cardinality as the reals whereas $[0,1]^{\omega_1}$ has the cardinality of the power set of the reals. Therefore there's the uninsightful reason that $[0,1]^{\omega_1}$ has only trivial path components that its closed intervals have a different cardinality than that of a real interval.

Answer 2

Here's another way to get a counterexample. Let $L$ be any totally ordered set that does not embed in $\mathbb{R}$ (e.g. $L=\omega_1$, or any totally ordered set of cardinality greater than $\mathfrak{c}$). Define a sequence of totally ordered sets $Y_0\subset Y_1\subset Y_2\subset\dots$ as follows. Let $Y_0=\mathbb{Q}$ (or really any totally ordered set with more than one point). Given $Y_n$, define $Y_{n+1}$ to be $Y_n$ with a copy of $L$ inserted right after each element of $Y_n$. (If you like, $Y_{n+1}=Y_n\times (1+L)$ with the lexicographic order.)

Now let $Y=\bigcup_n Y_n$, and let $X$ be the Dedekind completion of $Y$. Note that $Y$ is a dense linear order, since if $x,y\in Y_n$ with $x<y$ then in $Y_{n+1}$ there is a whole copy of $L$ between $x$ and $y$. Thus $X$ is a complete dense linear order, and thus is connected in the order topology. But between any two points of $X$ there is a copy of $L$, and so no interval in $X$ is isomorphic to $\mathbb{R}$. It follows easily that no subspace of $X$ in the order topology is homeomorphic to $\mathbb{R}$.