## 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 of non-empty sets, with , the product

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

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

So, then if we consider as just a class of sets, the fact that we have indexed them implies there exists a surjective “indexing function$

where clearly since we have already indexed out set we have that . So, consider

This is clearly a well-defined mapping and and thus

from where it follows that

Conversely, let be a class of sets and let be an indexing function. We may then index by . Then, by assumption

Thus there exists some

Such that

Thus, we have that

is a well-defined mapping with

For each . It follows that we have produced a choice function for and the conclusion follows.

*Remark:* We have assumed the existence of a bijective indexing function , but this is either A) a matter for descriptive set theory or B) obvious since satisfies the conditions. This depends on your level of rigor.

**10.**

**Problem: **Let be a set; let be an indexed family of spaces; and let be an indexed family of functions

**a) **Prove there is a unique coarsest topology on relative to whish each of the functions is continuous.

**b) **Let

and let . Prove that is a subbasis for .

**c)** Show that the map is continuous relative to if and only if each map is continuous.

**d)** Let be defined by the equation

Let denote the subspace of of the product space . Prove taht the image under of each element of is an open set in .

**Proof:**

**a)** We first prove a lemma

**Lemma:** Let be a topology on , then all the mappings are continuous if and only if where is defined in part b).

**Proof:**Suppose that all the mappings are continuous. Then, given any open set we have that is continuous and so is open and thus from where it follows that .

Conversely, suppose that . It suffices to prove that for a fixed but arbitrary . So, to do this let be open in then and thus by assumption ; but this precisely says that is open in . By prior comment the conclusion follows.

So, let

and let

By previous problem is in fact a topology on , and by our lemma we also know that all the mappings are continuous since . To see that it’s the coarsest such topology let be a topology for which all of the are continuous. Then, by the other part of our lemma we know that and thus . So,

And thus is coarser than .

The uniqueness is immediate.

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

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

**Proof: **Let

and

Also, let be the topology generated by the subbasis .

Clearly since we have that .

Conversely, let . Then, by definition to show that it suffices to show that for a fixed but arbitrary . To do this we first note that by definition that

where each

for some . Now, if we know (since ) that and thus

for each . It follows that is the union of sets in and thus . It follows from previous comment that .

The conclusion follows.

The actual problem follows immediately from this.

**c) **So, let be some mapping and suppose that is continuous for each . Then, given a subbasic open set in we have that

for some and for some open sets in respectively. Thus may be written as

but since each we see that is the finite union of open sets in and thus open in . It follows that is continuous.

Conversely, suppose that is continuous then is continuous since it’s the composition of continuous maps.

**d) **First note that

from where it follows that the initial topology under the class of maps on is the same as the initial topology given by the single map . So, in general we note that if is given the initial topology determined by then given an open set in we have that which is open in the subspace .

Cool Story bro.

Comment by Jeremy Janson | November 3, 2011 |