Continuous surjection from $ [0, 1] $ to $\Bbb R^2$

Question

I need to find such path connected compact subset $X$ of $\Bbb R^2$ that the continuous surjection $[0, 1] \rightarrow X$ doesn't exists.

I know that circle $S^1$ isn't homeomorphic to $[0, 1]$ and that they have same cardinality, but I am not sure about surjection.

Thanks!

Think of the closure in the plane of the graph on $\sin(1/x)$ for $0 – 2017-02-15 — 2017-02-15

user id:8508 337k · Accepted Answer

Try the union of the line segments joining $[0,0]$ to $[1,1/n]$ for positive integers $n$ and to $[1,0]$.

user id:53993 5k · Accepted Answer

Hint: Every continuous image of $[0,1]$ is locally-connected.

user id:15500 128k · Accepted Answer

A variation on the topologist's sine curve gives an especially tantalizing example. Take the graph of $\sin(1/x)$ for $0

You cannot surject a circle onto that, but it still clearly divides the plane into an outside and an inside like a circle would.

3 Answers 3