Proof. By the second note above, \(x\notin \overline {A}\) exactly when some \(\varepsilon >0\) gives \((x-\varepsilon ,x+\varepsilon )\subseteq A^{C}\). That is precisely the statement that \(x\) is an interior point of \(A^{C}\), i.e. \(x\in (A^{C})^{0}\). Taking complements in \((\overline {A})^C = (A^C)^0\) gives the second identity. □

Proof.

(i)

Suppose \(A\) is open, and let \(x_n\in A^C\) with \(x_n\longrightarrow x\). If \(x\) were in \(A\), there would be an \(\varepsilon >0\) with \((x-\varepsilon ,x+\varepsilon )\subseteq A\); since \(x_n\longrightarrow x\), some \(x_n\) would lie in this interval and hence in \(A\), which is impossible. So \(x\in A^C\), and \(A^C\) is closed.

Conversely, suppose \(A^C\) is closed, and let \(x\in A\). If no interval \((x-\varepsilon ,x+\varepsilon )\) were contained in \(A\), then for each \(n\) we could choose \(x_n\in A^C\) with \(|x_n - x|<\frac {1}{n}\). Then \(x_n\longrightarrow x\), and since \(A^C\) is closed, \(x\in A^C\), a contradiction. Hence \(x\) is an interior point of \(A\), and \(A\) is open.

(ii)

Apply (i) to the set \(A^C\), using \((A^C)^C = A\).

Proof. First, \(A^0\) is open. If \(x\in A^0\), there is \(\varepsilon >0\) with \((x-\varepsilon ,x+\varepsilon )\subseteq A\). Each point \(y\) of this interval has a smaller interval around it inside \((x-\varepsilon ,x+\varepsilon )\), hence inside \(A\), so \(y\in A^0\). Thus \((x-\varepsilon ,x+\varepsilon )\subseteq A^0\).

Next, let \(G\) be any open set with \(G\subseteq A\), and let \(x\in G\). Since \(G\) is open it is a neighbourhood of \(x\), and so is the larger set \(A\). Thus \(x\in A^0\), and \(G\subseteq A^0\). □

Proof. Let \(\{G_i: i\in I\}\) be a family of open sets, let \(G = \bigcup _{i\in I}G_i\), and let \(x \in G\). Then \(x \in G_i\) for some \(i\in I\). Since \(G_i\) is open, there exists \(\varepsilon > 0\) such that \((x - \varepsilon , x + \varepsilon ) \subseteq G_i \subseteq G\). Therefore \(G\) is open. □

Proof. Let \(G = \bigcap ^{n}_{i = 1} G_i\), where each \(G_i\) is open. If \(G= \emptyset \), it is open. Otherwise let \(x \in G\). Then \(x \in G_i\) for every \(i = 1,2,\dots ,n\), and since each \(G_i\) is open there exist \(\varepsilon _i > 0\) such that \(( x - \varepsilon _i, x + \varepsilon _i) \subseteq G_i\). Let \(\varepsilon = \min \{ \varepsilon _1, \varepsilon _2,\dots , \varepsilon _n\}>0\). Then \((x - \varepsilon , x + \varepsilon ) \subseteq G_i\) for every \(i\), so \((x - \varepsilon , x + \varepsilon ) \subseteq G\). Therefore \(G\) is open.

The theorem fails for infinite families. For example, if \(x\in \mathbb {R}\) and \(G_n = \left (x - \frac {1}{n}, x + \frac {1}{n}\right )\) for \(n\in \mathbb {N}\), then each \(G_n\) is open but \(\bigcap _{n\in \mathbb {N}} G_n = \{x\}\), which is not open. □

Solution. Let \(x\in (a,b)\) and put \(\varepsilon = \min \{x-a,\ b-x\}>0\). If \(|y-x|<\varepsilon \), then \(y>x-\varepsilon \geq x-(x-a) = a\) and \(y<x+\varepsilon \leq x+(b-x) = b\), so \(y\in (a,b)\). Thus \((x-\varepsilon ,x+\varepsilon )\subseteq (a,b)\). For example, at \(x = 0.9\) in \((0,1)\) we may take \(\varepsilon = 0.1\).

Solution. Interior. Each \(x\in (0,1)\) is interior, by the previous example. The point \(0\) is not interior, since every neighbourhood of \(0\) contains negative numbers, and \(2\) is not interior, since every neighbourhood of \(2\) contains points not in \(A\). So \(A^0 = (0,1)\).

Closure. The limit points of \(A\) are the points of \([0,1]\); the isolated point \(2\) is not a limit point but belongs to \(A\). So \(\overline {A} = [0,1]\cup \{2\}\).

Exterior. \(\operatorname {ext}(A) = (A^C)^0 = \bigl (\overline {A}\bigr )^C = (-\infty ,0)\cup (1,2)\cup (2,\infty )\).

Boundary. \(b(A)\) is what remains: \(b(A) = \{0,1,2\}\). Note that \(1\in b(A)\setminus A\) and \(2\in b(A)\cap A\).

Solution. Every interval contains both rational and irrational numbers. So no interval lies inside \(\mathbb {Q}\), giving \(\mathbb {Q}^0 = \emptyset \), and no interval lies inside \(\mathbb {Q}^C\), giving \(\operatorname {ext}(\mathbb {Q}) = (\mathbb {Q}^C)^0 = \emptyset \). Every real number is therefore a boundary point, and \(b(\mathbb {Q}) = \mathbb {R}\). The same argument gives \(b(\mathbb {Q}^C) = \mathbb {R}\).

Solution. We have \(G = (-\infty ,-\sqrt {2})\cup (\sqrt {2},\infty )\), a union of open intervals, which is open. (For instance, at \(x>\sqrt {2}\) the choice \(\varepsilon = x-\sqrt {2}\) works.) Since \(F = G^C\), Theorem 1.3.6 shows that \(F = [-\sqrt {2},\sqrt {2}]\) is closed.