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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.

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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.

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  • $\begingroup$ 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 $\endgroup$
    – erz
    Jun 16 at 4:55
  • $\begingroup$ @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. $\endgroup$
    – Wojowu
    Jun 16 at 10:06
  • $\begingroup$ Thank you! My point was that I couldn't accept your answer since I didn't fully understand it as stated. May I suggest you include this elaboration into the main text? $\endgroup$
    – erz
    Jun 16 at 11:18

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