TITLE: Basic Open set question
QUESTION [9 upvotes]: In $\mathbb{R}^2$, if I have an open set, call it $U$, and a point $u_0\in U$. Is the set $U^{'} = U - \{ u_0 \}$ is open as well? Or perhaps it is non-open & non-closed?
My intuition is that it still remains open, but I'm not sure on how to formalize this. Is it enough if I say that since for each $u\in U$ exists a neighbourhood $V_u$ such that $V_u\subseteq U$, so it is also true for each $u\in U- \{u_0 \}$, thus making $U^{'}$ open as well? Not sure if this is formal enough (or perhaps I missed something and it's wrong)
And assuming this is correct, can I remove more than one point? i.e. remove any $u_0,...,u_n$ from $U$ and still it will remain an open set?
Thanks!

REPLY [5 votes]: It is still open because a point is closed. So, in formal terms: The complement of a point is open and an open set without a point is the same as the interesction of your set and the complement of the point