Open and connected set in metric space

Question

In a normed space, we know that if a set is open and connected, it is path connected. Is it true for general metric space or general topological space?

Not true in general for topological spaces. Consider for example the "long line topology" which is built from gluing uncountably many $[0, 1)$s together + a positive infinity at one end. The simple fact is that even though it's connected, it's just too long for there to be a continuous map from $[0, 1]$ so that $0 \mapsto 0$ and $1 \mapsto \infty$. — daniel gratzer, Jan 18 '16 at 16:18

score 1 · Answer 1 · edited Jun 12 '20 at 10:38

1

For general topological spaces the answer is no. See here for a nice counterexample, the union of the graph of $\sin(1/x)$ and $\{0\}$:

http://topospaces.subwiki.org/wiki/Connected_not_implies_path-connected

If your space is locally path-connected, i.e. every point has a path-connected neighborhood, then connected implies path connected. This particularly holds for normed vector spaces, since any open ball around a point is path-connected.

So for metric spaces, connectedness and path-connectedness are equivalent, but for general topological spaces not.

edited Jun 12 '20 at 10:38

Community

1

answered Jan 18 '16 at 16:20

Stephan Mescher

421

how to prove any open ball around a point is path-connected? – 89085731 Jan 18 '16 at 16:58
Sorry, my original statement was wrong. The question of local path-connectedness is more sophisticated than I thought... – Stephan Mescher Jan 18 '16 at 17:17

score 0 · Answer 2 · edited Jan 18 '16 at 17:04

0

It is not true. The typical counterexample is the "topologist's sine curve". $$\{(t,\sin(1/t):t>0\}\cup \{(0,y):-1\le y\le 1\}.$$

edited Jan 18 '16 at 17:04

Martin Sleziak

53,687

answered Jan 18 '16 at 16:18

Tim Raczkowski

20,046

Open and connected set in metric space

2 Answers2