# How to prove that the complement of a closed set in $\mathbb{R^n}$ is open?

How does the argument go for: $\mathbb{R^n} \setminus A$ open, if $A$ closed?

I've tried to think of the closed set definitions, but I don't know how I'm supposed to apply them to an open set. Also was thinking of doing an antithesis ($\mathbb{R^n} \setminus A$ closed).

Here open set is a set that contains only interior points.
A closed set is a set that contains all its boundary points.

Interior and boundary points are displayed using the Ball definition.
(e.g. $x$ is an interior point of $A$ if $\exists r>0$ s.t. $B(x,r)\subset A$)

A closed set A has the equivalence theorem for closedness stating that one can construct a vector sequence $(x_k)_{k=1}^{\infty}$, $x_k \in A$ $\forall k$, for which $\lim_{k\rightarrow \infty}x_k=x$ and $x \in A$.

• This very much depends on your definitions of open and closed. – Did Nov 25 '15 at 10:24
• @Did I think I fixed that? – mavavilj Nov 25 '15 at 10:30
• Encourage you to state the definition you are using; otherwise people can hardly help you. – Benicio Nov 25 '15 at 10:31
• such a sequence can always be constructed. Just take $x_k=x\in A$ for each $k$. Do you mean to say that $A$ is (by your definition) closed if every element that serves as a limit of a sequence in $A$ is also an element of $A$? – drhab Nov 25 '15 at 10:46
• Missing "boundary points". (You know, we should not have to insist like that.) – Did Nov 25 '15 at 10:46

Assume $A$ is closed and let $x\in \mathbb{R}^n\backslash{A}$ arbitrary. We are looking for an $r>0$ such that $B(x,r)\subset \mathbb{R}^n\backslash{A}$, thus proving $\mathbb{R}^n\backslash{A}$ is open.
Since $A$ is closed, $x$ is not a limit point of $A$. So there is an $\epsilon>0$ such that $\mathrm{d}(x,y)>\epsilon$ for all $y\in A$. Take $r=\epsilon$ et voila.