3.1 Generating Open Sets
- 1.
- The empty set is open
- 2.
- The entire complex plane is an open set.
- 3.
- If \(\, S_1, \, S_2\,\) are open sets then \(S_1 \cap S_2\) is open.
- 4.
- If \(\big \{S_{\alpha }\big \}_{\alpha \in A}\,\) (\(A\) is some index set) is an arbitrary collection open sets in \(\mathbb {C}\) then \(\, \bigcup \limits _{\alpha \in A}S_{\alpha }\,\) is an open set.
- 1.
- \(\, \big \{z \in \mathbb {C}:\, \left |z - a\right | > r\big \}\,\) is an open set.
- 2.
- \(\, \big \{z \in \mathbb {C}:\, \left |z - a\right | > r_1\big \} \cap \big \{z \in \mathbb {C}:\, \left |z - a\right | < r_2\big \}\,\) where \(\, r_1 < r_2\).
\(= \big \{z \in \mathbb {C}:\,r_1 < \left |z - a\right | < r_2\big \}\,\) is an open set.
Example 3.7. Show that the upper half plane \(\pi = \big \{ z \in \mathbb {C}:\, \operatorname {Im}(z) > 0\big \}\) is open.
Solution. Let \(z_0\in \pi \) and write \(y_0=\operatorname {Im}(z_0)>0\). Take \(r=y_0\) and let \(D=\{z:\left |z-z_0\right |<r\}\).
If \(z\in D\) then, since the imaginary part of a complex number never exceeds its modulus, \[\left |\operatorname {Im}(z)-\operatorname {Im}(z_0)\right | =\left |\operatorname {Im}(z-z_0)\right |\leq \left |z-z_0\right |<y_0 ,\] so \(\operatorname {Im}(z)>y_0-y_0=0\) and hence \(z\in \pi \).
Thus \(D\subseteq \pi \): every point of \(\pi \) carries a disc lying wholly inside \(\pi \), which is the definition of an open set. Note the choice \(r=y_0\) is the largest that always works — the distance from \(z_0\) to the real axis.
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