4.6 KiB
lecture, date
| lecture | date |
|---|---|
| 7 | 2025-01-30 |
(X, d) ball topology
Say Y is a closed set in X (in other words, Y^{\complement} Open Sets).
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. !Which is a contradiction. Because
xcannot converge outsideYQED.
(Y, d) complete if (X, d) is complete.
Proposition
Say X is a Topological Space, that is Hausdorff.
If B \subset X is Compact, then B is closed.
Proof
!
Hausdorff => A_{y} \cap B_{y} = \emptyset , both are Open Sets.
Then \{ B_{y} \}_{y \in B} is an Open Cover of B
Compact => Pick out finite 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, in other words, B is closed.
QED.
Heine-Borel Theorem
Say A \subset \mathbb{R}^n. Then A is Compact \iff A is closed and Bounded.
Proof
For ->: A is closed since \mathbb{R^n} is Hausdorff and A is Compact, see #Proposition.
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. If an Open Cover of A has no finite Subcover, then by halving sides of cubes we get a sequence of cubes contained in each other, each having no finite Subcover. The centres of these cubes form a Cauchy Sequence with a limit x \in A.
!
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.
But this is a contradiction.
QED.
A^{\complement} open. Say \{ \cup_{n} \} is an Open 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.
!
Cor
\mathbb{R}^n is locally compact Hausdorff, and $\sigma$-compact
!
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 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 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:
d(z', z) = d(z, z')d(z, z') = 0 \iff z = z'd(z, z'') \leq d(z, z') + d(z',z'')d'(z, z')= length of straight line betweenzandz'= \sqrt{ (x-x')^2 + (y - y')^2 } = \| \, z - z' \, \|