# Continuous map from two intervals to the closed topologist's sine curve?

The closed topologist's sine curve is the set

$\{(x,y)\in\mathbb{R}^2 : x=0, -1\le y\le 1\}\cup\{(x,y)\in\mathbb{R}^2 : y=\sin\frac{1}{x}, 0<x\le \pi\}.$

Clearly it is not possible to continuously map the closed unit interval onto this set, as it fails to be path connected. But what about the disjoint union of two intervals?

I would guess not, I would hope the topologist's sine curve is more badly behaved, but I can't see how to show it's not possible. Is the continuous image of a locally (path) connected space, necessarily locally (path) connected under some reasonable assumptions?

• Any continuous image of closed interval (more generally, of any compact space) is compact. The "crazy curve" is not closed in $\mathbb R^2$, hence it is not compact. – Martin Sleziak May 10 '12 at 8:37
• That argument shows you can't do it the obvious way (one interval onto the vertical part, one onto the crazy curve) ... but I don't see yet an argument that you can't do it at all. – Marcel Besixdouze May 10 '12 at 8:40
• Well, this was part of your question before last edit, so I addressed it in my last comment. Anyway, candidates for images of your two intervals are closed connected subsets of $X$. One of them has to contain the right endpoint. The only connected closed subsets containing this point are the whole space and $X\cap(\varepsilon,\infty)\times\mathbb R$ for some $\varepsilon>0$. – Martin Sleziak May 10 '12 at 8:44
• Yes that does it. Thank you. I was hoping to gain a more general insight, but this does solve the question I posed. – Marcel Besixdouze May 10 '12 at 9:45
• @MarcelTuring: You can do it using a disjoint union of a closed and a half-open interval, if that helps. (Which implies you can also do it using a disjoint union of $\aleph_0$ closed intervals.) – Dejan Govc May 10 '12 at 12:17

Maybe this is the more general insight you were looking for. If $X$ is a space with a finite number of path-components where each path-component is compact, and $f$ is a continuous map, then the path-components of $f(X)$ are also compact.
To see that this is true, note that the continuous image of a path-connected set is also path-connected. This means that any path-component of $f(X)$ must be a union of images of path-components of $X$. Since continuous images of compact sets are compact, and finite unions of compact sets are also compact, it follows that $f(X)$ can have only compact path-components.