# Unions and Intersections of Closed Sets

Let $$S \subset \mathbf{R}^n$$. We call $$x \in \mathbf{R}^n$$ a boundary point of $$S$$ if for every $$\varepsilon > 0$$ the ball $$B(x, \varepsilon)$$ centered at $$x$$ of radius $$\varepsilon$$ contains at least one point $$y \in S$$ and at least one point $$z \notin S$$. The set of all boundary points of $$S$$ is called the boundary of $$S$$ denoted $$\partial S$$. We say that $$S$$ is closed if $$\partial S \subset S$$.

Now let $${S_\alpha}$$ be a collection of closed sets indexed by $$A$$ (that is $$\alpha$$ ranges over all possible values in $$A$$). We would like to show that the set
$S = \bigcap_{\alpha \in A} S_\alpha$
is closed. (Recall that $$x \in \bigcap_{\alpha \in A} S_\alpha$$ if and only if $$x \in S_\alpha$$ for all $$\alpha \in A$$.) To this end, we must show that every $$x \in \partial S$$ is also in $$S$$. To see this is the case, let $$x \in \partial S$$. Since $$x \in \partial S$$, for every $$\varepsilon > 0$$, there exist $$y \in S$$ and $$z \notin S$$ such that $$y, z \in B(x, \varepsilon)$$. Since $$y \in S$$, from the definition of $$S$$, we must have $$y \in S_\alpha$$ for all $$\alpha$$. Therefore, $$x \in S_\alpha \cup \partial S_\alpha$$ for each $$\alpha$$, hence $$x \in S_\alpha$$ for each $$\alpha$$ as $$\partial S_\alpha \subset S_\alpha$$. Thus we may conclude that
$x \in \bigcap_{\alpha \in A} S_\alpha = S$
which is what we needed to show.

Now we move on to consider unions of closed sets. First let $$S = \bigcup_{i = 1}^k S_i$$ be a finite union of closed sets. Suppose $$x \in \partial S$$. Again, for every $$\varepsilon > 0$$ there exists $$y \in S$$ and $$z \notin S$$ with $$y, z \in B(x, \varepsilon)$$. We would like to show that $$x \in S$$, hence $$S$$ is closed. Consider a sequence $$y_1, y_2, \ldots \in S$$ with \(|x – y_i|