# MATH3401

Lecture 12

### Topology

$$*$$ $$\Omega \subseteq \C$$ is piecewise affinely path connected if any two points in $$\Omega$$ can be connected by a finite number of line segments in $$\Omega$$, joined end to end.

### Topology

For open sets in $$\C$$, the two definitions are equivalent

(not so in general: see for example the comb space, to be confirmed/cleaned-up).

### Topology

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

$$z\in \Omega_1 \Rightarrow \exists\varepsilon_1 \gt0 \text{ such that } B_{\varepsilon_1}(z)\subset \Omega_1$$ ... ♣️

$$z\in \Omega_1 \Rightarrow \exists\varepsilon_2 \gt 0 \text{ such that } B_{\varepsilon_2}(z)\subset \Omega_2$$ ... 🙂

### Topology

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

Thus

$$B_{\varepsilon}(z)\subset \Omega_1$$ by ♣️,

$$B_{\varepsilon}(z)\subset \Omega_2$$ by 🙂.

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. $$\blacksquare$$

### Topology

$$*$$ An open, connected subset of $$\C$$ is called domain.

$$*$$ An set whose interior is a domain is called a region.

### Topology

$$*$$ 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 $$\conj{B_1}$$.

### Limits

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 $$z_0$$.

### Limits

Examples:

$$f(z) = \left\{ \begin{array}{ll} 0 & z\neq 0,\\ 1337 & z=0. \end{array} \right.$$    $$\ds\lim_{z\ra 0}f(z)=0$$;

$\ds \lim_{z\ra 0}\dfrac{\sin z}{z}=1.$

Remark: If a limit exists, it is unique.

#### Limit Theorems

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

#### Limit Theorems

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 $$\xi_0\neq 0$$.