# Is every path connected $F_\sigma$ subset of a plane an image of $[0,1)$?

The title says it all. Let $$A$$ be a path connected $$F_\sigma$$ subset of a plane (or more generally $$\mathbb{R}^n$$). Recall that a subset is called $$F_\sigma$$ if it is a union of a sequence of closed sets.

Is it true that there is a continuous surjection from $$[0,1)$$ onto $$A$$? Equivalently, can $$A$$ be represented as a union of an increasing sequence of Peano continuums?

Note that we cannot drop "path", since $$\{y=\sin(\frac{1}{x}),x>0\}\cup \{(0,0)\}$$ is a connected $$F_\sigma$$ subset that cannot be represented as a union of an increasing sequence of continuums.

No, this fails even for compact subsets of $$\mathbb R^2$$. Namely, let $$X=C\times[0,1]\cup[0,1]\times\{0\}$$, where $$C$$ is the Cantor set. It is clearly path connected. $$X$$ cannot be an image of $$[0,1)$$, because the image of any interval $$[0,a],a<1$$ by this map can contain only finitely many points of $$C\times\{1\}$$ (because of compactness), and hence the image of $$[0,1)$$ can only contain countably many of them.
It might be of your interest that there is a complete topological classification of spaces which are images of $$[0,1)$$, namely they are the path-connected spaces which are countable unions of Hahn-Mazurkiewicz spaces (which means they are compact, Hausdorff, connected, locally connected, metrizable spaces), as shown here.
• I can understand that the image of $[0,a]$ can contain finite number of points of $C\times \{1\}$ due to local connectedness. What do you mean by "because of compactness"? Perhaps I am missing somethig
• @erz There are probably multiple ways to see it. The argument I had in mind is that for every point $(c,1)$ which the curve crosses, there is an open subset of $[0,a]$ on which the curve is contained in $\{c\}\times(0,1]$. Together with the preimage of the complement of $C\times\{1\}$ these form a cover of $[0,a]$ and we can conclude through compactness from there. Jun 16 at 10:06