The well-known Hahn-Mazurkiewicz theorem characterizes those nonempty Hausdorff spaces $X$ that admit a continuous surjection $\alpha: [0, 1] \to X$ from the closed unit interval: it is necessary and sufficient that $X$ be a compact, connected, locally connected metrizable space. 


My Googling skills did not enable me to locate a nice way of characterizing Hausdorff spaces $X$ that admit a continuous surjection $(0, 1) \to X$ from an *open* interval. Is there one? 



continuous images of open intervals
Call $A$ an HM-space if there is a continuous surjection $I\to A$ (where $I$ is the closed interval $[0,1]$). Note that if $A$ is an HM-space, then it is path-connected.


Theorem: If $X$ is path-connected, then the following are equivalent:


1. $X = \bigcup\_{n=1}^\infty A\_n$ where each $A\_n$ is an HM-space
2. there is a continuous surjection $\mathbb{R} \to X$
3. there is a continuous surjection $(0,\infty) \to X$
4. there is a continuous surjection $[0,\infty) \to X$.


Proof:
Clearly (2) and (3) are equivalent; let's prove (3) implies (1). 
If $f:(0,\infty) \to X$ is a continuous surjection, let $A\_n = f([n-1,n])$. Then 
$A\_n$ is evidently an HM-space and $X$ is the indicated countable union.


Now we show (1) implies (4). if $X = \bigcup\_{n=1}^\infty A\_n$ with each $A\_n$ an HM-space, then we can choose surjective paths $\alpha\_n: [2(n-1),2n-1]\to A\_n$; write $x\_n = \alpha\_n(0)$ and $y\_n= \alpha\_n(1)$. Since $X$ is path-connected, we can find paths $\beta\_n:[2n-1,2n]\to X$ from $y\_n$ to $x\_{n+1}$. Concatenating these paths gives the desired surjection $0,\infty) \to X$.


Finally, since there is a continuous surjection $\mathbb{R}\to [0,\infty )$, (4) implies (2).$\quad \square$


Just musing: there is a partial order on spaces defined by $X< Y$ if there is a continuous surjection $Y \to X$. Hahn-Mazurkowicz classifies $\{ X \mid X < I\}$. The equivalence of (2), (3) and (4) shows that half-open intervals are "equivalent" to open ones (i.e., both $(0,1)< [0,1)$ and $[0,1) < (0,1)$); and closed ones are different, being compact. Finally, note that a Peano space (i.e., HM-space) $X$ is "equivalent" in this sense to $I$ if and only if it is the domain of a nonconstant continuous function $f:X\to \mathbb{R}$.


