Is path-connected space a continuous image of [0,1]?

Question

My intuition is that, from a point $a\in X$, draw a line from $a$ to $b\in X-\{a\}$, then another line from $b$ to $c\in X-\{a,b\}$, and so on. Now if $|X|=\mathfrak c$ then there are uncountable lines like these, can they be linked into a single path? Is there a limit on $X$'s cardinality?

Answer 1

The space of bounded, not necessarily continuous functions $\Bbb R\to \Bbb R$ has cardinality strictly greater than that of $\Bbb R$ (only a small tweak on Cantor's original diagonal argument works). With the topology generated by balls of the following form, for a function $f$ and real $r$ $$ B_f(r)=\{g\mid \forall x\in \Bbb R, |g(x)-f(x)|<r\} $$ you can still get a continuous path $h(t)(x)$ between any two functions $f$ and $g$: $$ h(t)(x)=(1-t)f(x)+tg(x) $$

Answer 2

A continuous image of $[0,1]$ is a Peano-continuum ,these are exactly all compact metric, connected and locally connected spaces $X$.

These all have size at most $\mathfrak{c}$, as is well-known.

On the other hand $\ell_\infty(\kappa)$ is a Banach space of size $\ge \kappa$, which is path-connected. Of course there are compact such spaces of any size as well, e.g. Tychonoff cubes $[0,1]^\kappa$ which are path-connected ( as a product of path-connected spaces) and of weight $\kappa$ and size $2^\kappa$.

The Warsaw circle is a compact metric and path-connected but not locally connected, so it's not a continuous image of $[0,1]$ by the first fact I mentioned.

Answer 3

Going somewhat along the same lines as Arthur's answer, it can be shown that there is no space $Y$ such that every path connected space is a continuous image of $Y$. The simple reason is that there is no bound on the cardinality of path-connected spaces. For every cardinal $\kappa$ consider the hedgehog space of spininess $\kappa$, let's denote it by $H (\kappa)$.

• To construct $H ( \kappa )$ consider the space $[0,1] \times \kappa$ where $[0,1]$ is given its usual topology, and $\kappa$ is discrete. That is, $[0,1] \times \kappa$ is a disjoint union of $\kappa$ copies of $[0,1]$. Then $H(\kappa) = ( [0,1] \times \kappa ) / \mathord{\sim}$, where $$ ( x,\xi ) \sim (y,\zeta) \Leftrightarrow x=y \mathrel{\&} \xi = \zeta, \text{ or } x = y = 0.$$ (That is, we identify all the $0$s from the different copies of $[0,1]$, which we will call spines of $H(\kappa)$.)

To see that $H(\kappa)$ is path connected, note that any two points on the same "spine" are obviously connected by a path, and two points on different spines are connected by a path that first goes through the $0$ point.

As $H(\kappa)$ has cardinality $\kappa \cdot 2^{\aleph_0} \geq \kappa$, given any space $Y$, $H(\kappa)$ cannot be a continuous image of $Y$ for any $\kappa > | Y |$.

Answer 4

For somewhat less obvious examples, consider the long line $X=\omega_1\times [0,1)$ and the space $Y$ obtained from the long line by "wrapping it around" (by making the nonexistent end touch the beginning).

$X$ and $Y$ both are path-connected and have cardinality of the continuum. $X$ is not compact, but is locally path-connected, while $Y$ is compact, but not locally path-connected (it's not path-connected near the origin).

The easiest way to see that they are not continuous images of $[0,1]$ nor $(0,1)$ is to note that they are inseparable (in fact, each contains an uncountable disjoint family of open sets), and it's easy to see that separability is preserved by continuous surjections.