## Munkres Chapter 2 Section 19 (Part I)

**1.**

**Problem: **Suppose that for each the topology on is given by a basis . The collection of all sets of the form

Such that is a basis for with the box topology, denote this collection by . Also, the collection of all sets of the form

Where for finitely many and otherwise is a basis for the product topology on .

**Proof:** To prove the first part we let be open. Then, by construction of the box topology for each we may find some such that is open in and . So, then for each we may find some such that and thus

Noticing that and every element of is open finishes the argument.

Next, we let be open with respect to the product topology. Once again for each we may find some such that is open in for each and for all but finitely many , call them . So, for each we may find some such that and so

Where

Noting that and is a collection of open subsets of finishes the argument.

**2.**

**Problem: **Let be a collection of topological spaces such that is a subspace of for each . Then, is a subspace of if both are given the product or box topology.

**Proof**:** **Let denote the topologies inherits as a subspace of and as a product space respectively. Note that are generated by the bases (where is the basis on with the product topology), and

So, let where then

Where is open in , and thus is open in . Also, since for all but finitely many it follows that for all but finitely many . And so . Similarly, if then

Where is open in , but this means that for some open set in and so

From where it follows that and thus are equal.

The case for the box topology is completely analgous.

**3.**

**Problem:** Let be a collection of Hausdorff spaces, then is Hausdorff with either the box or product topologies

**Proof**: It suffices to prove this for the product topology since the box topology is finer.

So, let be distinct. Then, for some . Now, since is Hausdorff there exists disjoint open neighborhoods of respectively. So, are disjoint open neighborhoods of respectively. The conclusion follows.

**4.**

**Problem:** Prove that .

**Proof:** Define

Clearly this is continuous since

**5.**

**Problem:** One of the implications holds for theorem 19.6 even in the box topology, which is it?

**Proof:** If where the latter is given the box topology then we have that each is continuous and thus so is each . #

**6.**

**Problem: **Let be a sequence of points in the product space . Prove that converges to if and only if the sequences coverge to for each . Is this fact true if one uses the box topology?

**Proof:** Suppose that is a neighborhood of such that

is infinite. Notice then that if that from where it follows that is a neighborhood of which does not contain all but finitely many values of contradicting the fact that in .

Conversely, suppose that for each and let be a basic open neighborhood of . Then, letting be the finitely many indices such that . Since each there exists some such that . So, let . Now, note that if then for some . But, since clearly we must have that and thus . It follows that for every we have that . Then, since every neighborhood of contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows.

Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider . Clearly each coordinate converges to zero, but is a neighborhood of in the product topology. But, if one claimed that for every (for some that they’d be wrong. To see this merely note and so and thus .

**7.**

**Problem:** Let be the subset of consisting of all eventually zero sequences. What is in the box and product topology?

**Proof:** We claim that in the product topology . To see this let be a basic non-empty open set in with the product topology. Since we are working with the product topology we know there are finitely many indices such that . So, for each select some and consider where

Clearly then and thus every non-empty open set in intersects and the conclusion follows.

Now, we claim that with the box topology that . To see this let . Then, there exists some subsequence of the sequence which is non-zero. For each form an interval such that . Then, consider

Clearly then is a neighborhood of and since each clearly has an infinite subsequence of non-zero values it is disjoint from . It follows that in with the box topology that is closed and thus as desired.

**8.**

**Problem:** Given sequences and of real numbers with define by the equation

Show that if is given the product topology that is a homeomorphism. What happens if is given the box topology?

**Proof: **Let us first prove that is a bijection. To do this we prove something more general…

**Lemma:** Let and be two classes of untopologized sets and a collection of bijections . Then, if

we have that is a bijection.

**Proof:** To prove injectivity we note that if

Then,

And by definition of an -tuple this implies that

for each . But, since each is injective it follows that

For each . Thus,

as desired.

To prove surjectivity we let be arbitrary. We then note that for each fixed we have there is some such that . So, if is the corresponding -tuple of these values we have that

from where surjectivity follows. Combining these two shows that is indeed a bijection.

**Lemma:** Let and be two classes of non-empty topological spaces and a corresponding class of continuous functions such that . Then, if and are given the product topologies the mapping

is continuous.

**Proof:** Since the codomain is a product space it suffices to show that

is continuous for each . We claim though that the diagram

commutes where and denote the canonical projections from and to and respectively. To see this we merely note that

and

which confirms the commutativity of the diagram. But, the conclusion follows since is the composition of two continuous maps (the projection being continuous since is a product space).

The lemma follows by previous comment.

We come to our last lemma before the actual conclusion of the problem.

**Lemma:** Let and be two classes of non-empty topological spaces and a set of homeomorphisms with . Then,

is a homeomorphism if and are given the product topology.

**Proof:** Our last two lemmas show that is bijective and continuous. To prove that it’s inverse is continuous we note that

And similarly for the other side. Thus,

Which is continuous since each is continuous and appealing to our last lemma again. Thus, is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows.

Thus, getting back to the actual problem we note that if we denote that each is a homeomorphism. Thus, since it is easy to see that

we may conclude by our last lemma (since we are assuming that we are giving in both the domain and codomain the product topology) that is a homeomorphism.

This is also continuous if we give the box topology. To see this we merely need to note that and thus if all of the are open then so are (since each is continuous) is each and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function.

Why no love for section 19? (I’m working through the book to study for the math subject GRE.)

Comment by Pat | June 7, 2010 |

n/m. Looking at the problems they pretty much seem to be the same. You working out of the first edition?

Comment by Pat | June 7, 2010 |

Haha, yeah recheck the title, I just mislabeled it. And no, I am working out of the second edition.

Also, is there really that much top. on the GREs?

Comment by drexel28 | June 7, 2010 |

I don’t think there’s too much, which is why I only plan to work my way through the first three chapters. Afterwards will be a crash course in algebra (Fraleigh’s book which seems good). Keep it up, really cool what you’re doing!

Comment by Pat | June 8, 2010 |

Thank you! I appreciate that someone actually read this. You’ll notice that this blog mostly used to be theorems, but I just really like doing problems. So, I figured over the summer I’d “try” to do all the problems in the “trifecta”: Topology-Munkres, Algebra-Dummit and Foote (over Herstein,yes), and Analysis-Rudin (baby Rudin). To be fair, I know all of these subjects fairly well. In fact, I’ve been through most of Munkres and Rudin’s book themselves. That said, it’ll still be hard.

P.S. I would not suggest Fraleigh. That book is kind of dumbed down and if you intend to pursue a graduate degree in mathematics (as the GRE reference suggests) it won’t do.

Comment by drexel28 | June 8, 2010 |