MATH3401

Show: If \(\Omega_1,\Omega_2 \subseteq \C\) are open, then so is \(\Omega_1\cap \Omega_2\).

Proof: If \(\Omega_1\cap \Omega_2 = \emptyset\) done (\(\emptyset\) is open).

Otherwise, for any \(z\in \Omega_1\cap\Omega_2\) we have

Since \(\Omega_1,\Omega_2\) are open, set \(\varepsilon = \min \{\varepsilon_1, \varepsilon_2\}\) and note that \(\varepsilon>0\).

Thus

So \(B_{\varepsilon}(z)\subset \Omega_1\cap\Omega_2\). Since \(z\) was arbitrary in \(\Omega_1\cap\Omega_2\).

This implies \(\text{Int } \Omega_1\cap\Omega_2= \Omega_1\cap\Omega_2\). Hence \(\Omega_1\cap\Omega_2\) is open. \(\blacksquare\)

\(*\) A Point \(z\in \C\) is called an accumulation point of a set \(\Omega \subseteq \C\) if every deleted neighbouhood of \(z\) intersects \(\Omega\).

Example 1: \(\Omega=\left\{\dfrac{i}{2^n}\right\}_{n\in \N}\). The only accumulation point is 0.

Example 2: \(\Omega=B_1\). The set of accumulation points is \(\conj{B_1}\).

Let \(f\) be a \(\C\)-valued function defined on a deleted neighbourhood of \(z_0\in\C\).

\(\ds \lim_{z\ra z_0}f(z)=w_0\) says: Given \(\varepsilon > 0\) exists \(\delta > 0\) such that \[0<|z-z_0|<\delta \implies \left|f(z)-w_0\right|<\varepsilon.\]

Note: \(f\) does not have to be defined at \(z_0\).

Suppose \(f(z) = f(x+iy) = u(x,y) + i v(x,y) \), \[z_0= x_0+iy_0 \text{ and } w_0 =u_0+iv_0.\]

Theorem 1: \[ \lim_{z \ra z_0} f(z)=w_0 \iff \left\{ \begin{array}{rl} \ds\lim_{(x,y)\ra (x_0, y_0)} u(x,y) & = u_0,\\ \ds\lim_{(x,y)\ra (x_0, y_0)} v(x,y) & = v_0. \end{array} \right. \]

Theorem 2: Suppose \(\ds \lim_{z \ra z_0}f(z) = w_0\) and \(\ds \lim_{z \ra z_0}g(z) = \xi_0\) and \(\lambda\in \C\). Then

1. \(\ds \lim_{z \ra z_0}\left(f \pm g\right)(z) = w_0 \pm \xi_0\);
2. \(\ds \lim_{z \ra z_0}\left(\lambda \cdot f \right)(z) = \lambda \cdot w_0\);
3. \(\ds \lim_{z \ra z_0}\left(f \cdot g\right)(z) = w_0 \cdot \xi_0\);
4. \(\ds \lim_{z \ra z_0}\dfrac{f(z)}{g(z)} = \dfrac{w_0}{\xi_0}\), as long as \(\xi_0\neq 0\).