Topology, Chapter 2 , Section 19Mason Keller

1.(Theorem 19.2) In order to prove Theorem 19.2, one can use the results of Lemma 13.3to show that the topology generated by the basis given in Theorem 19.2 (called the ”basis-basis” Bb) is the same as the topology generated by the basis given in Theorem 19.1 (calledthe ”open set-basis” BU).Now, let x be a point in

∏α∈J

Xα with the box topology. Let Bb be some element of Bb that

contains x. We desire to construct BU in BU that contains x and is a subset of Bb. Notethat Bb and BU are products over α, so for the subset relation to hold, it must hold in eachcoordinate. That is to say, that it must be that πα(BU) ⊂ πα(Bb) for all α ∈ J . Further,recall that both πα(BU) and πα(Bb) are sets in the same topology (Xα). Since basis elementsare open in the topology they generate and since πα(Bb) is a basis for Xα, a valid choice forπα(BU) is πα(Bb), which as per the previous argument makes the relation x ∈ BU ⊂ Bb holdfor any Bb containing some x.To prove the other inclusion, one proceeds as before but instead of choosing πα(BU) = πα(Bb),one chooses πα(Bb) to be one of the basis elements that makes up πα(BU) and contains x.Such a basis element is guaranteed to exist, since the topology on Xα is generated by thebasis of which πα(Bb) is an element of. This choice, then, satisfies the relation x ∈ Bb ⊂ BU

for any BU containing some x.Because these two relations hold, the bases generate equivalent topologies.

Now, to prove Theorem 19.2’s results for the product topology, one can realize that allthe sets in these two bases are also elements of the basis for the box topology, and so theconstruction proceeds in the exact same way. If one is not so clever, one can write out someextraneous proof as such:Let us use the notation of the previous argument and show that the relation x ∈ BU ⊂ Bb

holds for any Bb containing some x. As stated, the argument above holds for finitely manyπα. Let us consider α for which it does not. This means thatπα(Bb) = Xα, which sug-gests that a suitable choice for πα(BU) is Xα itself. To proceed in the other direction (thex ∈ Bb ⊂ BU direction), for any ”non-special” α, one notes that πα(BU) = Xα. Conse-quently a choice for πα(Bb) is any basis element in Xα that contains x.

2. (Theorem 19.3) It should be clear that the box topology on just∏Aα differs at most

from the subspace topology it inherits from∏Xα in that, in the subspace topology, U∩

∏Aα

has to be open for all U ∈ T . However, each U can be written as the union of productsof open sets in each α. This implies that the subspace topology inherited from

∏Xα has

that ∀(α ∈ J)(πα(U) ∩ Aα is open in Aα. Aα, however, has the subspace topology put onit in either instance, so this set is open in both cases. Since what could differ does not, thetopologies on

∏Aα are equivalent in the box topology. Because the box topology is finer

than the product, the theorem also holds for the product topology.

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3.(Theorem 19.4) Let Xα be a set of Hausdorff topologies for α ∈ J . Then, let∏α∈J

Xα

have the box topology put on it. Next, let x1 and x2 be arbitrary distinct points in thisproduct. This implies that πα(x1)! = πα(x2) for at least one α ∈ J . Let choose on of theseparticular α for which they differ, and denote it as i. Now, πi(x1) and πi(x2) are distinctpoints in Xi, which is by hypothesis Hausdorff. This means there are disjoint neighborhoodsof each point. Now, basis elements in the product space, which are indeed open sets in theproduct space, have the form

∏α∈J

Uα. Note that two of these products are disjoint if any

two Uα are disjoint. I choose as a neighborhood of x1∏

α∈J∧α<iXα ×D1 ×

∏α∈J∧α>i

Xα where

D1 is the disjoint neighborhood of πi(x1) that exists by hypothesis. Similarly, i choose as aneighborhood of x2

∏α∈J∧α<i

Xα × D2 ×∏

α∈J∧α>iXα where D2 is the neighborhood of πi(x2)

that is disjoint to πi(x1). These two neighborhoods are disjoint, and have been constructedfor arbitrary distinct points x1 and x2. This proves that the box topological product space isHausdorff if the coordinate spaces are Hausdorff. The result holds in the product topologybecause it is coarser than the box topology.

5. (Theorem 19.6) (metaproof) In example two, Munkres gives an example of a functionfor which each coordinate-function is continuous, but the function itself is not. This meansthat coordinate-function continuity does not imply function continuity in the box topology.By the language of problem five, this implies that the other implication, namely that func-tion continuity implies coordinate-function continuity, must hold in the box topology.

6.(Coordinate convergence is logically equivalent to point convergence) First, I choose toanswer the question in the box topology first. Let us consider Rω. Let χn ∈ Rω, πα(χn) = 1/nfor all α ∈ Z+. Clearly each coordinate of this series converges to 0. However, note that∏i∈Z+

Ui, Ui = (−πi, πi) is a neighborhood of the 0 ”vector” that the coordinates would imply

χn converges to. If it were that χn did converge to this point, then we would have some nsuch that 1

n∈ (−π

i, πi)∀i, however, as i approaches infinity, the set is 0, and there is no n

such that 1n∈ {0}.

Let us consider the product topology on∏α∈J

Xα, and the statement that point convergence

implies coordinatewise convergence. Let χn be a series of points that converges. Let χend bean element of the set of convergence points. Now, for any neighborhood U of χend, there issome N such that ∀(n > N)(χn ∈ U). This neighborhood U , however, can be expressed inthe form of a basis element from the basis given in theorem 19.2 (due to the definition of ba-sis). Let this suitable basis element be denoted by B, and recall that its form is

∏α∈J

Bα such

that Bα is a basis element from Xα, and that finitely many of these Bi are different from Xi.For the point χn to be an element of this set means that πα(χn) ∈ πα(

∏α∈J

Bα) = Bα. Now, all

χn after some arbitraryN , by hypothesis, lay in this basis element (it is a neighborhood aboutthe supposed convergence point χend). This implies that ∀∀(α ∈ J)(n > N)(πα(χn) ∈ Bα).

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Each (πα(χn) is a series of points in the topology Xα. Because of how we came to Bα (thatis, by U), a different choice of U could generate any and all Bα, each with an N such that∀(n > N)(πα(χn) ∈ Bα). Open sets in Xα can be composed of basis elements, so that thetruth of the above proposition is a sufficient proof that each ”coordinate series” convergesin the same way that that the entire series does.

Now, for all α ∈ J , let χα,n denote some series in πα(∏α∈J

Xα). Further, let each of these

χα,n converge to at least some χα,end. We desire to show that∏α∈J

χα,end = χend is a conver-

gence point of∏α∈J

χα,n = χn. Equivalently, we wish to construct some N for an arbitrary

open set U about χend in the product space such that for all (n > N)χn ∈ U . Any U canbe expressed as the union of basis elements of the form

∏α∈J

Uα where Uα ∈ Tα and only for

finitely many α does Uα differ from Xα. Any N that ”works” for this basis element works forU , since it is, by definition, a subset of U . Now, let I be the necessarily finite set (producttopology) of α ∈ J such that Uα! = Xα. However, these Uα are still neighborhoods aboutχα,end. This coordinate series, by definition has some Nα for which it does not escape thisparticular Uα. For each coordinate in I, we have a particular Ni. I assert that N = max(Ni)is a valid choice for the entire series χn. Further, I assert that this maximum must existbecause the set of Ni is finite. Clearly it works for all α ∈ I. For the other coordinatesthough, since they are the whole Xα, N = 0 is a choice. Note that the if N ”works”, anyinteger greater than N also works. So N = max(Ni) is a valid choice for all α ∈ J , whichcompletes the proof that coordinate-wise convergence implies series convergence.

Because these two implications hold in the product topology, it true that a series convergesif and only if it’s ”coordinate series” converge.

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