Munkres Chapter 2 Section 19 (Part I)
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
Noting that and is a collection of open subsets of finishes the argument.
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.
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.
Problem: Prove that .
Clearly this is continuous since
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 . #
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 .
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.
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
And by definition of an -tuple this implies that
for each . But, since each is injective it follows that
For each . Thus,
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
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
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.