# Abstract Nonsense

## Munkres Chapter 2 Section 19 (Part II)

9.

Problem: Show that the axiom of choice (AOC) is equivalent to the statement that for any indexed family $\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ of non-empty sets, with $\mathcal{A}\ne\varnothing$, the product $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$

Proof: This is pretty immediate when one writes down the actual definition of the product, namely:

$\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=\left\{\bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_{\alpha}:\bold{x}(\alpha)\in U_\alpha,\text{ for every }\alpha\in\mathcal{A}\right\}$

So, if one assumes the AOC then one must assume the existence of a choice function

$\displaystyle c:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }c\left(U_\alpha\right)\in U_\alpha\text{ for all }\alpha\in\mathcal{A}$

So, then if we consider $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}\Omega$ as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function\$

$i:\mathcal{A}\to\Omega$

where clearly since we have already indexed out set we have that $i:\alpha\mapsto U_\alpha$. So, consider

$c\circ i:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}}U_\alpha$

This is clearly a well-defined mapping and $\left(c\circ i\right)(\alpha)=c\left(U_\alpha\right)\in U_\alpha$ and thus

$\displaystyle c\circ i\in\prod_{\alpha\in\mathcal{A}}U_\alpha$

from where it follows that $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$

Conversely, let $\Omega$ be a class of sets and let $i:\mathcal{A}\to\Omega$ be an indexing function. We may then index $\Omega$ by $\Omega=\left\{U_{\alpha}\right\}_{\alpha\in\mathcal{A}}$. Then, by assumption

$\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha\ne\varnothing$

Thus there exists some

$\displaystyle \bold{x}\in\prod_{\alpha\in\mathcal{A}}U_\alpha$

Such that

$\displaystyle \bold{x}:\mathcal{A}\to\bigcup_{\alpha\in\mathcal{A}},\text{ such that }\bold{x}(\alpha)\in U_\alpha$

Thus, we have that

$\displaystyle \bold{x}\circ i^{-1}:\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}\to\bigcup_{\alpha\in\mathcal{A}}$

is a well-defined mapping with

$\displaystyle \bold{x}\left(U_\alpha\right)\in U_\alpha$

For each $\alpha\in\mathcal{A}$. It follows that we have produced a choice function for $\Omega$ and the conclusion follows. $\blacksquare$

Remark: We have assumed the existence of a bijective indexing function $i:\mathcal{A}\to\Omega$, but this is either A) a matter for descriptive set theory or B) obvious since $\text{id}:\Omega\to\Omega$ satisfies the conditions. This depends on your level of rigor.

10.

Problem: Let $A$ be a set; let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of spaces; and let $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an indexed family of functions $f_\alpha:A\to X_\alpha$

a) Prove there is a unique coarsest topology $\mathfrak{J}$ on $A$ relative to whish each of the functions $f_\alpha$ is continuous.

b) Let

$\mathcal{S}_\beta=\left\{f_\beta^{-1}\left(U_\beta\right):U_\beta\text{ is ope}\text{n in}X_\beta\right\}$

and let $\displaystyle \mathcal{S}=\bigcup_{\alpha\in\mathcal{A}}$. Prove that $\mathcal{S}$ is a subbasis for $\mathfrak{J}$.

c) Show that the map $g:Y\to A$ is continuous relative to $\mathfrak{J}$ if and only if each map $f_\alpha\circ g:Y\to X_\alpha$ is continuous.

d) Let $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ be defined by the equation

$f(x)=\left(f_\alpha(x)\right)_{\alpha\in\mathcal{A}}$

Let $Z$ denote the subspace of $f\left(A\right)$ of the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$. Prove taht the image under $f$ of each element of $\mathfrak{J}$ is an open set in $Z$.

Proof:

a) We first prove a lemma

Lemma: Let $\mathfrak{J}$ be a topology on $A$, then all the mappings $f_\alpha:A\to X_\alpha$ are continuous if and only if $\mathcal{S}\subseteq\mathfrak{J}$ where $\mathcal{S}$ is defined in part b).

Proof:Suppose that all the mappings $f_\alpha:A\to X_\alpha$ are continuous. Then, given any open set $U_\alpha\in X_\alpha$ we have that $f_\alpha$ is continuous and so $f_\alpha^{-1}\left(U_\alpha\right)$ is open and thus $f_{\alpha}^{-1}\left(U_\alpha\right)\in\mathfrak{J}$ from where it follows that $\mathcal{S}\subseteq\mathfrak{J}$.

Conversely, suppose that $\mathcal{S}\subseteq\mathfrak{J}$. It suffices to prove that $f_\alpha:A\to X_\alpha$  for a fixed but arbitrary $\alpha\in\mathcal{A}$. So, to do this let $U$ be open in $X_\alpha$ then $f_{\alpha}^{-1}\left(U\right)\in\mathfrak{J}$ and thus by assumption $f_\alpha^{-1}\left(U\right)\in\mathfrak{J}$; but this precisely says that $f_\alpha^{-1}\left(U\right)$ is open in $A$. By prior comment the conclusion follows. $\blacksquare$

So, let

$\mathcal{C}=\left\{\mathfrak{T}:\mathfrak{J}\text{ is a topology on }A\text{ and }\mathcal{S}\subseteq\mathfrak{J}\right\}$

and let

$\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{I}\in\mathcal{C}}\mathfrak{T}$

By previous problem $\mathfrak{J}$ is in fact a topology on $A$, and by our lemma we also know that all the mappings $f_\alpha:A\to X_\alpha$ are continuous since $\mathcal{S}\subseteq\mathfrak{J}$. To see that it’s the coarsest such topology let $\mathfrak{U}$ be a topology for which all of the $f_\alpha:A\to X_\alpha$ are continuous. Then, by the other part of our lemma we know that $\mathcal{S}\subseteq\mathfrak{U}$ and thus $\mathfrak{U}\in\mathcal{C}$. So,

$\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}\subseteq\mathfrak{U}$

And thus $\mathfrak{J}$ is coarser than $\mathfrak{U}$.

The uniqueness is immediate.

b) It follows from the previous problem that we must merely show that $\mathcal{S}$ is a subbasis for the topology $\mathfrak{J}$. The conclusion will follow from the following lemma (which was actually an earlier problem, but we reprove here for referential reasons):

Lemma: Let $X$ be a set and $\Omega$ be a subbasis for a topology on $X$. Then, the topology generated by $\Omega$ equals the intersection of all topologies which contain $\Omega$.

Proof: Let

$\mathcal{C}\left\{\mathfrak{T}:\mathfrak{T}\text{ is a topology on }X\text{ and }\Omega\subseteq\mathfrak{T}\right\}$

and

$\displaystyle \mathfrak{J}=\bigcap_{\mathfrak{T}\in\mathcal{C}}\mathfrak{T}$

Also, let $\mathfrak{G}$ be the topology generated by the subbasis $\Omega$.

Clearly since $\Omega\subseteq\mathfrak{G}$ we have that $\mathfrak{J}\subseteq\mathfrak{G}$.

Conversely, let $U\in\mathfrak{G}$. Then, by definition to show that $U\in\mathfrak{J}$ it suffices to show that $U\in\mathfrak{T}$ for a fixed but arbitrary $\mathfrak{T}\in\mathcal{C}$. To do this we first note that by definition that

$\displaystyle U=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$

where each

$U_\alpha=O_1\cap\cdots\cap O_{m_\alpha}$

for some $O_1,\cdots,O_{m_\alpha}\in\Omega$. Now, if $\mathfrak{T}\in\mathcal{C}$ we know (since $\Omega\subseteq\mathfrak{T}$) that $O_1,\cdots,O_{m_\alpha}\in\mathfrak{T}$ and thus

$O_1\cap\cdots\cap O_{m_\alpha}=U_\alpha\in\mathfrak{T}$

for each $\alpha\in\mathcal{A}$. It follows that $U$ is the union of sets in $\mathfrak{T}$ and thus $U\in\mathfrak{T}$. It follows from previous comment that $\mathfrak{G}\subseteq\mathfrak{J}$.

The conclusion follows. $\blacksquare$

The actual problem follows immediately from this.

c) So, let $g:Y\to A$ be some mapping and suppose that $f_\alpha\circ g:Y\to X_\alpha$ is continuous for each $\alpha\in\mathcal{A}$. Then, given a subbasic open set $U$ in $A$ we have that

$U=f_{\alpha_1}^{-1}\left(U_{\alpha_1}\right)\cap\cdots\cap f_{\alpha_n}^{-1}\left(U_{\alpha_n}\right)$

for some $\alpha_1,\cdots,\alpha_n$ and for some open sets $U_{\alpha_1},\cdots,U_{\alpha_n}$ in $X_{\alpha_1},\cdots,X_{\alpha_n}$ respectively. Thus $g^{-1}(U)$ may be written as

$\displaystyle g^{-1}\left(\bigcup_{j=1}^{n}f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}g^{-1}\left(f_{\alpha_j}^{-1}\left(U_{\alpha_j}\right)\right)=\bigcup_{j=1}^{n}\left(f_{\alpha_j}\circ g\right)^{-1}\left(U_{\alpha_j}\right)$

but since each $f_{\alpha_j}\circ g:Y\to X_{\alpha_j}$ we see that $g^{-1}\left(U\right)$ is the finite union of open sets in $Y$ and thus open in $Y$. It follows that $g$ is continuous.

Conversely, suppose that $g$ is continuous then $f_\alpha\circ g:Y\to X_{\alpha}$ is continuous since it’s the composition of continuous maps.

d) First note that

$\displaystyle f^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\bigcap_{\alpha\in\mathcal{A}}f\alpha^{-1}\left(U_\alpha\right)$

from where it follows that the initial topology under the class of maps $\{f_\alpha\}$ on $A$ is the same as the initial topology given by the single map $f$. So, in general we note that if $X$ is given the initial topology determined by $f:X\to Y$ then given an open set $f^{-1}(U)$ in $X$ we have that $f\left(f^{-1}(U)\right)=U\cap f(X)$ which is open in the subspace $f(X)$.

June 9, 2010 -

## 1 Comment »

1. Cool Story bro.

Comment by Jeremy Janson | November 3, 2011 | Reply