---
lecture: 7
date: 2025-01-30
---
$(X, d)$ ball topology
Say $Y$ is a closed set in $X$ (in other words, $Y^{\complement}$ [[Open Sets|open]]).

Say we have a sequence $\{ X_{n} \}$ in $Y$ and that $x_{n} \to x \in X$.
Then $x \in Y$.

> [!info]- Proof
> Let's say $x \notin Y$.
> ![[Pasted image 20250130104559.png]]
> Which is a contradiction.
> Because $x$ cannot converge outside $Y$
> QED.

$(Y, d)$ complete if $(X, d)$ is complete.

# Proposition
Say $X$ is a [[Topological Space|topological space]], that is [[Hausdorff]].

If $B \subset X$ is [[Compact|compact]], then $B$ is closed.
## Proof
![[Pasted image 20250130110419.png]]
[[Hausdorff]] => $A_{y} \cap B_{y} = \emptyset$ , both are [[Open Sets|open]].
Then $\{ B_{y} \}_{y \in B}$ is an [[Open Cover|open cover]] of $B$

Compact => Pick out finite [[Subcover|subcover]] $\{ B_{y_{i}} \}_{i=1}^{n}$ of $B$.
Then $x \in \overbrace{\cap_{i=1}^{n} A_{y_{i}}}^{\text{Open}} \subset B^{\complement}$ since $\cap_{i=1}^{n} A_{y_{i}} \cap B_{y_{j}} = \emptyset$ (for any $j$)

So $B^{\complement}$ is [[Open Sets|open]], in other words, $B$ is closed.

QED.
# Heine-Borel Theorem
Say $A \subset \mathbb{R}^n$. Then $A$ is [[Compact|compact]] $\iff$ $A$ is closed and [[Bounded|bounded]].
## Proof
**For ->:** $A$ is closed since $\mathbb{R^n}$ is [[Hausdorff]] and $A$ is [[Compact|compact]], see [[#Proposition|the previous result]].
	It is bounded since we can cover it by finitely many balls.

**For <-:** Assume first that $A$ is an $n$-cube with boundary included.
	Say $A$ is not [[Compact|compact]]. If an [[Open Cover|open cover]] of $A$ has no finite [[Subcover|subcover]], then by halving sides of cubes we get a sequence of cubes contained in each other, each having no finite [[Subcover|subcover]]. The centres of these cubes form a [[Cauchy Sequence|Cauchy sequence]] with a limit $x \in A$.
	![[Pasted image 20250130120238.png]]
	$d(x_{n}, x_{m}) \lt \varepsilon, \; \forall n,\,m \gt N$. Show: $\mathbb{R}^n$ is complete.
	Any neighbourhood of $x$ from the cover will obviously contain a small enough cube, and will be a finite [[Subcover|subcover]].
	But this is a contradiction.

QED.

$A^{\complement}$ open. Say $\{ \cup_{n} \}$ is an [[Open Cover|cover]] of $A$. Then $\{ \cup_{n} \}, \, A^{\complement}$ is an open cover of the $n$-cube.
$\{ \cup_{n_{i}} \}_{i=1}^N, \, A^{\complement}$ cover of the $n$-cube.
Then $\{ \cup_{n_{i}} \}_{i=1}^N$ will cover $A$.
![[Pasted image 20250130121836.png]]
# Cor
$\mathbb{R}^n$ is locally compact [[Hausdorff]], and $\sigma$-compact
![[Pasted image 20250130122717.png]]

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# Complex Vector Space Exercise
Show that finite dimensional the complex vector space $V \implies V \simeq \mathbb{C}^n$ for some $n$.
$$x,y \in V$$
$$\implies x + y \in V$$
$$a \times x \in V, \; a \in \mathbb{C}$$
$\mathbb{C}^2$:
$$(x, y) + (z, w) \equiv (x+z, y+w)$$
$$a \times (x, y) \equiv (a x, a y)$$
Linear map:
## Proof
$n = \dim V$. Say $\{ v_{i} \}_{i=1}^n$ is a linear basis for $V$. Define [[Bijective|bijective]] [[Linear Map|linear map]] $L: \mathbb{C}^n \to V$ by $L((x_{1}, \dots, x_{n})) = \Sigma_{i=1}^n x_{i} \times v_{i}$. It is linear;
$L(a(x_{1}, \dots, x_{n}) + b(y_{1}, \dots, y_{n})) = L((ax_{1} + by_{1}, \dots, ax_{n} + by_{n})) = \Sigma_{i=1}^n(ax_{i} + by_{i})v_{i} =$
$\Sigma_{i=1}^n(ax_{i}v_{i} + by_{i}v_{i}) = a \times \Sigma_{i=1}^n x_{i}v_{i} + b \times \Sigma_{i=1}^ny_{i}v_{i} =$
$a \times L((x_{1},\dots,x_{n})) + b \times L((y_{1},\dots,y_{n}))$

**Surs.**
$v \in V$. Basis $\implies \exists \underbrace{x_{i}}_{\in \mathbb{C}}$ such that $v = \Sigma_{i=1}^n x_{i} v_{i} = L((x_{1}, \dots, x_{n}))$.
So $v \in \text{lm} L$ 

**Injective**
Say $L((x_{1}, \dots, x_{n})) = L((y_{1}, \dots, y_{n}))$ then
$\Sigma_{i=1}^n x_{i} v_{i} = \Sigma_{i=1}^n y_{i} v_{i} \xrightarrow{basis} x_{i} = y_{i}, \; \forall i$, so $x = y$
Injective for [[Linear Map|linear map]] $L \iff \ker L = \{ 0 \} \equiv \{ x \in \mathbb{C} | L(x) = 0 \}$.

$\ker L$ is a vector space
$\text{lm} \,L$ is a vector space

$L = 0 \to \ker L = \mathbb{C}^n$
$0 \in V$ (vectors spaces have to start from the origin, as that is where vectors themselves start from).
# Metric Space Exercise
Consider the unit circle $\mathbb{T} = S^1 = \{ (x,y) \in \mathbb{R}^2 | x^2 + y^2 = 1 \}$
$\mathbb{T} = \{ z \in \mathbb{C} \, | \, |\, z \,| = 1 \}$

Say there is a circle $z = (x, y)$, $x^2 + y^2 = 1$
This is a metric space for $d(z, z')$ = arclength between $z$ and $z'$.

Check:
1. $d(z', z) = d(z, z')$
2. $d(z, z') = 0 \iff z = z'$
3. $d(z, z'') \leq d(z, z') + d(z',z'')$
$d'(z, z')$ = length of straight line between $z$ and $z'$ $= \sqrt{ (x-x')^2 + (y - y')^2 } = \| \, z - z' \, \|$