It is well known that a metric space is a continuous image of the closed unit interval if and only if it is compact, connected and locally connected. Is there a similar list of topological properties that characterizes those metric spaces which are continuous images of the Euclidean straight line L? In particular, is every closed and connected subset of a finitedimensional Euclidean space a contnuous image of L? I suspect that the answer is "no".

1$\begingroup$ As pointed out by Pietro Majer, you cannot expect a positive answer if you drop the local connectedness. The question whether every closed, connected and locally connected subset of a Euclidean space is the continuous image of the line seems more interesting. $\endgroup$ – Benoît Kloeckner May 25 '11 at 7:23

$\begingroup$ It seems to me that such a set should be necessarily arcwise connected (I'm not completely sure: I wrote a proof below: does it seem ok to you?). $\endgroup$ – Pietro Majer May 25 '11 at 20:16

$\begingroup$ Here it is; I guess it's the same argument. jstor.org/pss/2371224 $\endgroup$ – Pietro Majer May 25 '11 at 20:52

1$\begingroup$ On the other hand, there do exist closed connected subsets of the plane which are not locally connected, but are a continuous image of the real line. E.g, the comb space (en.wikipedia.org/wiki/Comb_space) and the closed infinite broom (en.wikipedia.org/wiki/Infinite_broom). $\endgroup$ – George Lowther May 26 '11 at 0:15

1$\begingroup$ And there are (nonclosed) connected and locally path connected subsets of the plane which are not a continuous image of the line. E.g., for any non (Lebesgue) measurable subset $S$ of the reals, then $A=((S\cup\mathbb{Q})\times\mathbb{R})\cup(\mathbb{R}\times\mathbb{Q})$ is connected and locally path connected, but is not measurable, so is not a continuous image of the line. In fact, if $S$ is not $F_\sigma$ than neither is $A$, so is not an image of the line. $\endgroup$ – George Lowther May 26 '11 at 0:30
Indeed, there are closed and connected subsets of a finitedimensional Euclidean space which are not arcwise connected, as it is a continuous image of L.
edit (after Benoît Kloeckner's comment) It seems to me that a complete metric space $X$ which is connected and locally connected is necessarily arcwise connected. The idea is that given two points $x$ and $y$ in $X$, and $\epsilon>0$, there is a finite sequence $ x=x^\epsilon_0,x^\epsilon_1,\dots ,x^\epsilon_n=y$ such that for $0\le i < n$ the points $x^\epsilon_i$ and $x^\epsilon_{i+1}$ belong to some connected open set $U^\epsilon_i$ of diameter less than $\epsilon$ (the reason is that given $x$, the set of all $y$ which are reachable this way is an open and closed nonempty set). So we can start with $\epsilon = 1$, and iterate the construction within each $U^\epsilon_i$, which is still connected and locally connected, finding new points between $x^\epsilon_i$ and $x^\epsilon_{i+1}$, taking $\epsilon=1,1/2,1/4\dots$. By completeness these dotted lines converge to a suitably parametrized arc joining $x$ and $y$.
edit. The preceding is indeed exactly Whyburn theorem (1931). In particular a closed, connected, locally connected, subset $A$ of the Euclidean space is arcwise connected. You further ask if it is a continuous image of the real line. If $A$ is bounded then you can even obtain it as a continuous image of the closed unit interval, via a construction à la Peano (incidentally, you can, of course, also choose the endpoints of the arc). More generally, continuous images of the unit closed interval are caracterized by the HahnMazurkiewicz theorem. If $A$ is unbounded, you may write it as a countable union of compact sets, each one image of an arc with domain $[k,k+1]$. These arcs glue together in a continuous function on $\mathbb{R}$ provided you choose the endpoints so that they match.
Finally, there are connected and locally connected subsets of the Euclidean space, in dimension at least 2, which are not countable union of compact sets, hence they are not continuous images of the real line. An example is $A=(\mathbb{R}\times\mathbb{R})\setminus (\mathbb{Q}\times\mathbb{Q})$, thanks to the Baire category argument (see George Lowther's comment above).

$\begingroup$ Thanks, Pietro for your answer which makes me feel stupid for not having thought of it myself. If I modify my second question by asking whether every arcwise connected subset of a finitedimensional Euclidean space is a continuous image of L, I still suspect that the answer would be "no". $\endgroup$ – Garabed Gulbenkian May 25 '11 at 19:18

$\begingroup$ However if it is also closed I think that you can make a construction à la Peano, producing a single arc that connects all points. $\endgroup$ – Pietro Majer May 25 '11 at 20:30

$\begingroup$ Pietro: sorry, I misread your answer. I now deleted mine. For the record: 1) A specific closed connected subset of the plane that is not path connected is the topologist's sine curve. 2) Your argument shows that a locally compact space is a continuous proper image of the real line if and only if it is separable, metrizable, connected and locally connected. 3) A related characterization is, a space is a continuous open image of the Baire space (=the subspace of the real line consisting of irrational numbers) if and only if it is Polish (=separable and metrizable by complete metric). $\endgroup$ – Sergey Melikhov May 27 '11 at 10:03