Does every open and connected subset of path connected topological space has to be path connected? Statement should be false as there is a similar theorem but for Euclidean spaces, however I can't think of a counterexample. What about the same statement, but for path connected metric spaces?
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$\begingroup$ math.stackexchange.com/questions/766422/… $\endgroup$– John DoumaNov 3, 2018 at 0:48
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$\begingroup$ An open and connected subset of a locally path-connected space is path-connected. This is the theorem I think you are confused with. $\endgroup$– Henno BrandsmaNov 3, 2018 at 6:43
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The classical example of a connected (metric) space that is not path-connected is the topologist's sine curve. I will give an example based on this.
Consider the graph of the $\sin(\frac{1}{x})$ function on $(0,1]$.
Let $X$ be the space which consists of this graph together with the vertical line segment connecting $(0,-1)$ and $(0,1)$, and the curve in red:
$X$ is a metric space that is path connected. You can also clearly see this as an open subset of $X$:
This is an open connected subspace of the $X$ that is not path connected.
This counter example is a metric space. It applies as well for the general case of topological spaces.
answered Nov 3, 2018 at 0:56
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Consider the space $$X=\left\{(x,y)\in\Bbb R^2\,:\, (x=0\land y\le 2)\lor \left(x\ne 0\land y=\sin\frac1{\lvert x\rvert}\right) \lor (y\ge 0\land x^2+y^2=4)\right\}$$
I.e. a topologist sine, plus an appropriate vertical half-line, plus a half circle "path-connecting" the curve to the tip of the half-line. Then, $X\setminus \{(0,2)\}$ is connected, but not path-connected.