ACIT4330-Page/content/Lectures/Lecture 12 - Induced Topologies.md

lecture, date

lecture	date
12	2025-02-24

Weakest Topology

Definition of Weakest Topology defined in the lecture.

Initial Topology

Definition of Initial Topology, defined in the lecture.

Proposition

Let X be a set with Initial Topology induced by a family, F, of functions.

Then a Nets in \{ x_{i} \} in X converges to x \iff f(x_{i}) \to f(x) \; \forall f \in F.

Proof

\Rightarrow:

Is obvious.

\Leftarrow:

Let A be a neighbourhood of x.

Hence there are finitely many Open Sets A_{n} \subset Y_{f_{n}} such that x \in \cap f^{-1}_{n}(A_{n}) \subset A. Then A_{n} is a neighbourhood of f_{n}(x), \; \forall n.

By assumption f_{n}(x_{i}) \to f_{n}(x), so f_{n}(x_{i}) \in A_{n} \; \forall i \geq i(n) for some i(n). Pick j such that j \geq i(n) for all the finitely many $n$s.

Then x_{i} \in \cap f^{-1}_{n}(A_{n}) \subset A for all i \geq j, so x_i \to x.

QED.

COR

X has Initial Topology induced by F. Say Z is a Topological Space, then:

g : Z \to X is Continuous f \circ g is Continuous \forall f \in F.

Proof

\Rightarrow:

Clear.

\Leftarrow:

Say we have a Nets \{ z_{i} \} that z_{i} \to z in Z.

Then \underbrace{(f \circ g)(z_{i})}_{= f(g(z_{i}))} \to \underbrace{(f \circ g)(z)}_{f(g(z))} \; \forall f \in F.

Hence g(z_{i}) \to g(z) by the proposition, so g is Continuous.

Product Topology

Definition of the Product Topology, defined in the lecture.

!Drawing 2025-02-24 11.58.37.excalidraw (x_{i}, y_{i}) \to (x,y)

By the previous proposition a net \{ x_{i} \} in \Pi X_{\lambda} converges to x with respect to the Product Topology \iff \pi_{\lambda} \to \pi_{\lambda}(x), \; \forall \lambda.

Note

\pi_{\lambda} are Continuous (obvious) and Open Sets \pi_{\lambda}(A) open in X_{\lambda} for A open in \Pi X_{\lambda}

Tychonoff

It is the Tychonoff Theorem (defined in lecture).

Separating Family

Defines Separating Points from the lecture.

Proposition

A set X with Initial Topology induced from a Separating Points of functions f : X \to Y_{f} is Hausdorff when all Y_{f} are Hausdorff.

Proof

Say x \neq y in X. Then \exists f such that f(x) \neq f(y) in Y_{f}.

Can separate f(x) and f(y) by neighbourhoods U and V such that U \cap V = \emptyset.

Then f^{-1}(U) and f^{-1}(V) will be disjoint neighbourhoods of x and y.

[!note] Reasoning for f^{-1}(U) and f^{-1}(V) being disjoint

z \in f^{-1}(U) \cap f^{-1}(V) f(z) \in U \cap f(z) \in V

Which is not possible

QED.

COR

A product of Hausdorff spaces is Hausdorff in the Product Topology.

\Pi X_{\lambda} \Pi_{\lambda}

f: X \to Y_{f} f : X_{f} \to Y

q : X \to X \setminus \sim q(x) = [x]