Proof.

Let \(\, z = x + i y \in \mathbb {C}\)
\(\lim \, B'(z,\varepsilon ) \cap S \neq \emptyset \)
There are rational numbers \(\, x_0,y_0\ni x < x_0 < x + \frac {\varepsilon }{\sqrt {2}}\,\) and \(\, y < y_0 < y + \frac {\varepsilon }{\sqrt {2}}\)
So \(\, x_0 + iy_0 \in S\,\) and \(\, x_0 + iy_0 \in B'(z,\varepsilon )\) \[x_0 +iy_0 \in S\cap B'(z,\varepsilon )\] Hence \(z\) is a limit point of \(S\) □

Proof. Let \(z\in \overline {S}^c\) so \(z\in S\Big (\therefore \overline {S}\supseteq S\Big )\). Also \(z\) is not a limit point of \(S\) which implies
\(\exists \, \varepsilon > 0 \ni B(z;\varepsilon ) \cap S = \emptyset \). (We want to show that \(B(z;\varepsilon )\cap \overline {S} = \emptyset \))
Suppose that \(\, B'(z; \varepsilon ) \cap \overline {S}\neq \emptyset \,\) then there is a limit point of \(S\) in \(B'(z;\varepsilon )\) say \(w\) \[\left |w - z\right | < \varepsilon \] Let \(0 < \delta < \varepsilon - \left |w - z\right |\). Then \(B'(w,\delta ) \cap S \neq \emptyset \). (\(\therefore w\) is a limit point)
\(\implies \, B'(z;\varepsilon ) \cap S\neq \emptyset \,\) which is a contradiction. So \[B(z;\varepsilon ) \cap \overline {S} = \emptyset \implies B(z;\varepsilon ) \subseteq \overline {S}^c\] \(\implies \, \overline {S}^c\,\) is open in \(\mathbb {C} \implies \overline {S}\) is a closed set. □

Proof.

(i) \(\, \implies \,\) (ii) \(\, \implies \,\) (iii) \(\,\implies \,\) (i)

(ii) \(\, \implies \,\) is by definition of closure \(\big (\, \overline {S} = S\cup \big \{\text {limit points of}\, S\big \}\, = S\big )\).

(iii) \(\, \implies \,\) (i) \(\,\overline {S}\) is a closed set and if \(S = \overline {S}\), \(S\) is a closed set.

(i) \(\, \implies \,\) (ii) Suppose \(S\) is closed. (will show: limit points of \(S\) cannot be in \(S^c\)). Suppose \(z \in S^c\) is a limit point of \(S\). Since \(S^c\) is open \(\, \exists \, \varepsilon > 0 \ni B(z; \varepsilon ) \subseteq S^c\). \(\implies \, B(z;\varepsilon ) \cap S = \emptyset \) which contradicts the assumption that \(z\) is a limit point.
\(\implies \, S\) contains all its limit points.  □

Solution. \(S\) is an open disc together with the integers marked on the real axis. The two pieces behave differently, and that is the point of the exercise: isolated points contribute to the boundary but never to the interior.

Interior points

Every \(z\) with \(\left |z\right |<1\) has a disc about it inside \(S\), by the argument of the previous example. An integer \(n\) with \(\left |n\right |\geq 1\) has none: any disc about it contains points off the real axis, and those belong to \(S\) only if they lie in the unit disc, which they do not. Hence \[\operatorname {int}S=\{z:\left |z\right |<1\}.\]

Limit points

Every point of the closed disc \(\left |z\right |\leq 1\) is a limit of points of the open disc, so all of them qualify. An integer \(n\) with \(\left |n\right |\geq 1\) does not: the disc of radius \(\tfrac 12\) about it meets \(S\) in \(n\) alone. So the set of limit points is \(\{z:\left |z\right |\leq 1\}\).

Boundary

A boundary point is one every disc about which meets both \(S\) and its complement. For the disc that is the unit circle; each integer \(\left |n\right |\geq 1\) also qualifies, since every disc about it contains \(n\) itself and plenty of points outside \(S\). Hence \[\partial S=\{z:\left |z\right |=1\}\cup \{n+0i:\left |n\right |\geq 1\}.\] So \(S\) is neither open — it contains no disc about \(z=1\) — nor closed, since it omits the circle \(\left |z\right |=1\).