Abstract Nonsense

Crushing one theorem at a time

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).

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June 9, 2010 - Posted by | Fun Problems, Munkres, Topology | , , , , , , ,

1 Comment »

  1. Cool Story bro.

    Comment by Jeremy Janson | November 3, 2011 | Reply


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