# Open connected subsets of path connected spaces

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?

• math.stackexchange.com/questions/766422/… Nov 3, 2018 at 0:48
• An open and connected subset of a locally path-connected space is path-connected. This is the theorem I think you are confused with. Nov 3, 2018 at 6:43

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.

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.