Solution. (a) Closed, not open: it contains its boundary circle \(\left |z\right |=2\), and no disc about a boundary point stays inside. Boundary: \(\left |z\right |=2\).

(b) Open, not closed — a punctured disc. Every point has a small disc inside the set, but the centre \(i\) and the outer circle are missing. Boundary: \(\{i\}\cup \{\left |z-i\right |=1\}\).

(c) Closed, not open: the imaginary axis belongs to it but has no disc inside it. Boundary: the imaginary axis.

(d) Neither. It is not open, since no disc about \(1/n\) lies in a set of isolated points; and it is not closed, since \(0\) is a limit point not in the set. Boundary: the set together with \(0\).

Solution. Suppose \(S\) is closed, so \(\mathbb {C}\setminus S\) is open, and let \(w\) be a limit point of \(S\). If \(w\notin S\) then some disc about \(w\) lies in the complement, hence meets \(S\) nowhere — contradicting \(w\) being a limit point. So \(w\in S\).

Conversely, suppose \(S\) contains all its limit points and let \(w\in \mathbb {C}\setminus S\). Then \(w\) is not a limit point of \(S\), so some punctured disc about \(w\) misses \(S\); and \(w\notin S\) too, so the whole disc misses \(S\). Hence the complement is open and \(S\) is closed.

Solution. Let \(K_1,K_2\) be compact and let \(\mathcal {U}\) be an open cover of \(K_1\cup K_2\). It covers each \(K_j\) separately, so there are finite subcovers \(\mathcal {F}_1\) and \(\mathcal {F}_2\). Their union is finite and covers \(K_1\cup K_2\), so the union is compact.

For infinitely many, take \(K_n=\{n\}\) for \(n\in \mathbb {N}\). Each is a single point, hence compact, but \[\bigcup _{n\in \mathbb {N}}K_n=\mathbb {N}\] is unbounded, so by Heine–Borel it is not compact. The finiteness in the first part is essential: it is what allows the subcovers to be combined.

Solution. Fix \(a\in D\) and let \(A\) be the set of points of \(D\) reachable from \(a\) by such a chain. We show \(A\) is both open and closed in \(D\).

\(A\) is open: if \(z\in A\), take a disc \(B(z;r)\subseteq D\), possible since \(D\) is open. Any \(w\) in that disc is reached by appending the segment from \(z\) to \(w\), which lies in the disc and so in \(D\). Hence \(B(z;r)\subseteq A\).

The complement \(D\setminus A\) is open by the same argument: if \(z\) were reachable from a point of a disc about a non-reachable point, then appending one segment would make that point reachable too.

\(A\) is non-empty, containing \(a\). A non-empty subset of the connected set \(D\) that is both open and closed in \(D\) must be all of \(D\), so every point of \(D\) is reachable.

This is the fact used whenever ”\(f'=0\) on a domain implies \(f\) constant” is invoked: the constancy is propagated along such a chain.