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 \(|x – y_i| < 1/i\). Such a sequence exists by choosing \(\varepsilon = 1, 1/2, 1/3, \ldots\). Since each \(y_i \in S\), \(y_i \in S_j\) for some \(j\). Since there are infinitely many \(y_i\) and only finitely many \(S_j\), some \(S_j\) must contain infinitely many of the \(y_i\). I will leave it to my diligent readers to verify that for this \(S_j\) we have \(x \in \partial S_j \subset S_j\), and therefore \(x \in S\). This shows that \(S\) is closed. Notice that the previous argument relies crucially on the fact that \(S\) is a union of finitely many closed sets. What if \(S = \bigcup_{i = 1}^\infty S_i\) is an infinite union? Can we still conclude that \(S\) is closed? To see this is not the case, consider the sets \[ S_i = [1/i, 1 - 1/i] \] where \(i\) is \(2, 3, 4, \ldots\). Is the union of these sets closed?