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If $f:[0,1] \to \mathbb{R}^n$ is any continuous map, then the image $f([0,1])$ is a compact, path-connected set, which is easy to show using some elementary topology.

My question is the converse:

Namely, if $K \subset \mathbb{R}^n$ is compact and path-connected, then does there exist a continuous map $f:[0,1] \to \mathbb{R}^n$ such that $K = f([0,1])$?

My attempt at the problem:

I have a hunch that it might be true.

For any $k \in \mathbb{N}$, there exists a continuous surjection $f:[0,1] \to [0,1]^k$, which can be realized with a space filling curve. Therefore, any finite-dimensional cube can be realized as the continuous image of $[0,1]$.

Let $K \subset \mathbb{R}^n$ be compact and path-connected. My friend suggested that it would suffice to show that $K$ has the structure of a CW-complex with a finite number of cells, and then use the fact that any finite-dimensional cube is realized as the continuous image of $[0,1]$. However, I don't know if this is true.

Edit: It turns out that the answer is even more interesting than I anticipated, and is provided by the HM theorem: http://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn.E2.80.93Mazurkiewicz_theorem

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    $\begingroup$ This is false. Take the closure of the topologist's sine curve with the ends joined up by a curve. $\endgroup$
    – William Stagner
    Commented Apr 17, 2015 at 5:54
  • $\begingroup$ @Shalop can you share the source of the problem if you dont mind ;I am also looking for some problems in this subject $\endgroup$
    – Learnmore
    Commented Apr 17, 2015 at 6:06
  • $\begingroup$ @learnmore: A few years ago Wendelin Werner and associates proved that the frontier of the trace of a Brownian Motion has Hausdorff dimension $\frac{4}{3}$. Then they concluded that therefore the trace contains paths of dimension $\frac{4}{3}$. But a "path" is something which can be parametrized by a continuous map from $[0,1] \to \mathbb{R}$, and how do we know that the Brownian frontier is such a set? I thought of the above problem while thinking about how to prove this. $\endgroup$
    – shalop
    Commented Apr 17, 2015 at 6:13
  • $\begingroup$ @Shalop I mean that set but with an arc attaching to a point on the sine curve and a point on the y-axis, thus making it path-connected $\endgroup$
    – William Stagner
    Commented Apr 17, 2015 at 6:16
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    $\begingroup$ See also Hahn-Mazurkiewicz theorem and this post on MO. $\endgroup$
    – Martin Sleziak
    Commented Apr 17, 2015 at 7:37

2 Answers 2

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Here's a non-trivial example: the closure of the topologist's sine curve with the ends joined up.

If you want an explicit representation, take $$ \{(x,y) : 0< x\leq 1, y = \sin(1/x) \} \bigcup \{(0,y): -1\leq y\leq 1\} \bigcup \{(x,0): -1\leq x\leq 0\} \bigcup \{(-1,y): -2\leq y\leq 0 \} \bigcup\{(x,-2): -1\leq x \leq 1\} \bigcup \{(1,y): -2\leq y\leq \sin(1) \}. $$

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  • $\begingroup$ Maybe the post would look a bit better if you used picture from Wikipedia or one of many other pictures which can be found online. In fact, in this post there also a picture where the ends are joined. $\endgroup$
    – Martin Sleziak
    Commented Apr 17, 2015 at 7:30
  • $\begingroup$ As I have learned here, this space is also called Warsaw circle. $\endgroup$
    – Martin Sleziak
    Commented Apr 17, 2015 at 7:33
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Trivially, no. Let $K=\emptyset$. The empty set is closed as $\mathbb{R}^n$ is open, certainly bounded. And given any two points $x,y \in \emptyset$, there is a path connecting them. :)

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  • $\begingroup$ Great answer, I think maybe to make the question more interesting, it should add the non-empty hypothesis so this answer doesn't apply. Although of course, I don't want to diminish your correct contribution. $\endgroup$
    – Santropedro
    Commented Apr 12, 2017 at 0:37

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