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202210051329 Exercise 3.1

(a) Prove that the following are open sets:
i. \{ z\in\mathbb{C}:|z-1|<|z+\mathrm{i}|\};
ii. \mathbb{C}\backslash [0,1].
(b) Prove that the following are not open:
i. \{ z\in\mathbb{C}:\mathrm{Re}\, z\geqslant 0\};
ii. \{ z\in\mathbb{C}:|z|\leqslant 2,\mathrm{Re}\, z>1\}.

Extracted from H. A. Priestley. (2003). Introduction to Complex Analysis.

Background.

Definition. (open set)
A set S\subseteq \mathbb{C} is open if, given z\in S, there exists r>0 (depending on z) such that \mathrm{D}(z;r)\subseteq S.

Definition. (open disc)
The open disc centre a\in\mathbb{C} and radius r>0 is defined to be

\mathrm{D}(a;r):=\{ z\in\mathbb{C}: |z-a|<r \}.

(a) i.

Warm-up.

\begin{aligned} |z-1| & = |(x+\mathrm{i}y)-1| \\ & = \sqrt{(x-1)^2+(y)^2} \\ |z+\mathrm{i}| & = |(x+\mathrm{i}y)+\mathrm{i}| \\ & = \sqrt{(x)^2+(y+1)^2} \\ \end{aligned}

No, just let r=|z+\mathrm{i}| is done.

(a) ii.

Set-up.

In set notation a complex interval number may be represented in the form

\mathcal{A}=[a,b]+[c,d]\mathrm{i}=\{ x+\mathrm{i}y\, |\, a\leqslant x\leqslant b,c\leqslant y \leqslant d\}.

Extracted from R. Boche. (1966). Complex Interval Arithmetic with Some Applications.

Thus \mathbb{C}\backslash [0,1] is equivalent to

\{z=a+\mathrm{i}b\textrm{ where }a,b\notin [0,1]\},

as shown in the figure below:

Abortive attempt.

Lemma.

Let S\subset X be a non-empty subset and U\subset S. Then U is open in S if and only if U=V\cap S for some V\subset X which is open in X.

Proof. Necessity. \forall\, z\in U,\,\exists\, r>0 s.t. \mathrm{D}_S(z;r)\subset U. Let \displaystyle{V=\bigcup_{z\in U}\mathrm{D}_X(z;r)}. Then U=V\cap S. Note that V is open in X. Sufficiency. \forall\, z\in U\subset V,\,\exists\, r>0 s.t. \mathrm{D}_X(z;r)\subset V. Thus \mathrm{D}_S(z;r)=\mathrm{D}_X(z;r)\cap S\subset V\cap S\subset U.   \blacksquare

Now that U=\mathbb{C}\backslash [0,1], S=\mathbb{C}, and X=\mathbb{C} … quit this circular reasoning of no use.

Make use of the following

Lemma.

The union of any collection of open sets is open.

Proof.

Let \displaystyle{S=\bigcup_{\lambda\in I}S_\lambda} where S_\lambda is open in \mathbb{C} and I any index set. Then \forall\, z\in S, z\in S_\lambda for some \lambda and so \exists\, r>0 s.t. \mathrm{D}(z;r)\subset S_\lambda\subset S.   \blacksquare

Thereby

\mathbb{C}\backslash [0,1]=\mathbb{C}(-\infty ,0)\cup\mathbb{C}(1,\infty)

is open.

QED

(b)

A set S is not open whenever S^0 (/\textrm{int}\, S), the interior of S, falls short of its whole, i.e., S\supset S^0\neq S

For i. and ii. the unbelonging boundaries \partial S‘s to either are \{z\in\mathbb{C}:\textrm{Re }z=0\} and \{z\in\mathbb{C}:|z|=2,\mathrm{Re}\, z>1\}.