# Path-conectedness of open balls implies path-connectedness

Let $$M$$ be a subset of a metric space $$(E,d)$$. I have just proved that $$M$$ path-connectedness implies $$M$$ connectedness. Now I need to show that if $$M$$ is connected and every open ball in $$E$$ is path-connected (in $$E$$), then $$M$$ is path-connected. I don't really know where to begin .Can you give me a hint?

EDIT: I'm really really confused. OK, given $$x\in S$$ I have to prove that the set $$S =\{y\in M:$$ there exists a path in $$S$$ between $$x$$ and $$y \}$$ is open and closed ( it's trivially non empty, because $$x\in S$$). I have been trying to prove it's open, I have to prove that for an arbitrary $$s\in S$$ there is a ball $$B$$ centered in $$s$$ such that $$B\bigcap M$$ is contained in $$S$$. However, although every $$B$$ is path-connected I cannot assure for any $$B$$ that $$B\bigcap M$$ is path-connected , which is what I need. Are you sure I can do it with these hypothesis? Please, I really need help with this.

EDIT2: Do we need $$M$$ to be an open set?

EDIT3: Tell me if I'm wrong, but I think the topologist's sine is a counterexample for the case when $$M$$ is not open. Your answers are OK assuming $$M$$ is open

We have to assume that $$M$$ is open or that every open ball in $$M$$ is path connected in $$M$$. Otherwise, as the OP himself pointed out, the topologist's sine curve provides a counterexample. Assuming that the hypothesis is modified here is a hint: let $$x \in M$$. Let $$S$$ be the set of all points $$y \in M$$ such that there is a path in $$M$$ from $$x$$ to $$y$$. It is fairly elementary to verify (using path connectedness of open balls) that $$S$$ is open and closed in $$M$$. Since $$M$$ is connected (and $$S$$ is not empty) it must be equal to $$M$$.

• So given $s\in S$ I need to find a ball centered in $s$ that is contained in S? It doesn't seem so easy – Seven Nov 24 '18 at 12:39
• The hypothesis talks about balls in E, it would be obvious if it talked about balls in M.I need to take balls of elements of M, but these are not for sure balls of E and, therefore, they are not for sure path-connected – Seven Nov 24 '18 at 13:05
• I'm really confused about these balls' things. I need more help. – Seven Nov 24 '18 at 13:08
• @Seven You are absolutely right. I had assumed that $M$ was open. I have revised my answer. Sorry for the oversight. – Kavi Rama Murthy Nov 24 '18 at 23:15

If $$M=\emptyset$$, then it is trivially connected. Otherwise, fix $$p\in M$$ and consider the set$$A=\left\{q\in X\,\middle|\,\text{ there is a path in M joining p and q}\right\}.$$Prove that:

1. $$A$$ is open;
2. $$A\neq\emptyset$$;
3. $$A^\complement$$ is open.

Since $$M$$ is connected, this will prove that $$A=M$$. In particular, $$M$$ is path-connected.

• Sir can $A^c$ be closed for any additional condition imposed? – Yadati Kiran Nov 24 '18 at 11:59
• @YadatiKiran The set $A^\complement$ will surely be closed, since $A$ is open. – José Carlos Santos Nov 24 '18 at 12:03