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!
$\endgroup$ 23 Answers
$\begingroup$Yes, any singleton set $\{u_a\}$ is closed in $\mathbb{R}^n$ (in the standard topology), and so its complement $\mathbb{R}^n - \{u_a\}$ is open.
So, in your setting, $$U - \{u_0\} = U \cap (\mathbb{R}^2 - \{u_0\})$$ is an intersection of two open sets and is thus open.
Like you suggest, the same works if we remove any finite number of points $u_a$ from $U$.
$\endgroup$ $\begingroup$For your argument to work you need to show that the neighborhoods $V_u$ do not contain $u_0$. If you know that $\mathbb R^2$ is Hausdorff, then you can modify how you chose your neighborhoods and then the proof should work. By induction, you could then remove a finite number of points from $U$ and still have an open set.
You don't need the full strength of the Hausdorff condition however: Being a $\mathrm T_1$ space would suffice.
$\endgroup$ $\begingroup$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
$\endgroup$
