Lecture 12 - Induced Topologies

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 net in $\{x_i\}$ in $X$ converges to $x⟺f(x_i)\to f(x)\forall f\in F$.

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 $x\in \cap f_n^{-1}(A_n)\subset A$. Then $A_n$ is a neighbourhood of $f_n(x),\forall n$.

By assumption $f_n(x_i)\to f_n(x)$, so $f_n(x_i)\in A_n\forall i\ge i(n)$ for some $i(n)$. Pick $j$ such that $j\ge i(n)$ for all the finitely many $n$s.

Then $x_i\in \cap f_n^{-1}(A_n)\subset A$ for all $i\ge j$, so $x_i\to x$.

QED.

Corollary

$X$ 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 net $\{z_i\}$ that $z_i\to z$ in $Z$.

Then $\underset{=f(g(z_i))}{\underbrace{(f\circ g)(z_i)}}\to \underset{f(g(z))}{\underbrace{(f\circ g)(z)}}\forall f\in F$.

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.

$(x_i,y_i)\to (x,y)$

By the previous proposition a net $\{x_i\}$ in $\Pi X_{\lambda }$ converges to $x$ with respect to the product topology $⟺{\pi }_{\lambda }\to {\pi }_{\lambda }(x),\forall \lambda$.

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 family of functions $f:X\to Y_f$ is Hausdorff when all $Y_f$ are Hausdorff.

Proof

Say $x\ne y$ in $X$. Then $\exists f$ such that $f(x)\ne f(y)$ in $Y_f$.

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

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

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.

Corollary

A product of Hausdorff spaces is Hausdorff in the product topology.

$\Pi X_{\lambda }$
${\Pi }_{\lambda }$

$f:X\to Y_f$
$f:X_f\to Y$

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