Talk:Neighbourhood of Point Contains Point of Subset iff Distance is Zero

Is it true that this theorem only applies to Limit Points? Because if $x$ is an isolated point of $A$, then it can't be a limit point of $A$. But $d \left({x, A}\right) = 0$ because by definition of an isolated point $x \in A$. So $\ds \inf_{y \mathop \in A} \map d {x, y} = \map d {x, x} = 0$.

I think this theorem should apply to the closure of $A$. --HumblePi (talk) 15:07, 15 February 2017 (EST)


 * You could well be right. The exercise itself (will change accordingly) states: $\map d {x, A} = 0$ iff every neighbourhood of $x$ contains a point of $A$.


 * It's been a while since I went near topology so I can't guess whether the two statements (i.e. the one about limits and the one about neighborhoods) are equivalent or not, and my head's not in the right space at the moment. I'll leave it up to you to chec over what it says now and what it ought to be titled, and how it is to be proved. --prime mover (talk) 16:51, 15 February 2017 (EST)


 * That's great and all, but I actually don't know how to rename a page on this site, and I couldn't find anything on the Help page, so I'm afraid I won't be able to accomplish that. :( --HumblePi (talk) 17:11, 15 February 2017 (EST)


 * Under "More". I believe you should have authorisation. Let me know if not. --prime mover (talk) 17:46, 15 February 2017 (EST)


 * I think that "Point in Closure iff Distance is Zero" is a good title. --HumblePi (talk) 17:14, 15 February 2017 (EST)


 * ... except the wording of the theorem doesn't mention closure. --prime mover (talk) 17:46, 15 February 2017 (EST)