Is any compact, path-connected subset of $\mathbb{R}^n$ the continuous image of $[0,1]$?

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

Solution 1:

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 $$ \{(x,y) : 0< x\leq 1, y = \sin(1/x) \} \bigcup \{(0,y): -1\leq y\leq 1\} \bigcup \{(x,0): -1\leq x\leq 0\} \bigcup \{(-1,y): -2\leq y\leq 0 \} \bigcup\{(x,-2): -1\leq x \leq 1\} \bigcup \{(1,y): -2\leq y\leq \sin(1) \}. $$

Solution 2:

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. :)