ADDENDUM TO PROBLEM 22.12Abstract. Consider reading these notes to be an alternate solution to (a),which states to “reread discussion 1 3.7. ” We will also provide a rigorous re-formulation of the question in (b), as in lecture. None of these considerationsis required to solve the problem, but I think it makes it easier.Let X be a set, and let 2X:= {E : E ⊆ X}. That is, 2Xis a set, the elementsof which are the subsets of X. Call a collection of sets B ⊆ 2Xa base for X if thefollowing two requirements hold:(i) ∀x ∈ X, ∃B ∈ B such that x ∈ B(ii) B1, ..., Bk∈ B and y ∈ ∩ki=1Bi⇒ ∃B ∈ B such that y ∈ B ⊆ ∩ki=1BiNote: In class we used the notation {B} for B.Now call a set of functions B ⊆ C[0, 1] a basic pointed ε - ball centered at f0if itcan be written in the following manner:B = {f ∈ C[0, 1] : |f (x) − f0(x)| < ε for all x ∈ {x1, ..., xk}}for some fixed f0∈ C[0, 1], ε > 0, and finite set of points F = {x1, ..., xk} ⊆ [0, 1].We abbreviate such a set B = B(f0, ε, F ).Now we are in a position to state problem 22.12(b) more clearly:Problem 22.12 (b): Show that the collectionB := {B(f0, ε, F ) : f0∈ C[0, 1], ε > 0, F a finite subset of [0,1]}forms a base for C[0, 1] = {f : [0, 1] → R | f is continuous}.Note: Of course, the book uses S instead of [0, 1], but the argument is identicalfor any S. So you may assume S = [0, 1], but afterwards try to understand whythis is a ‘WLOG’ assumption.Proof: property (i) is trivial. One could immediately reduce condition (ii) tothe case of when only two sets are intersected, by a simple induction. So writeBi= B(fi, εi, Fi) for i = 1, 2 (or i = 1, 2, ..., k if you don’t believe the inductionthing.)Let g ∈ ∩Bi. Rewriting, we have, ∀i and ∀x ∈ Fi, that g(x) ∈ (fi(x)−εi, fi(x)+εi).Then for each such x we may choose a δxi> 0 so that (using metric notation forthe standard distance in R):Bδxi(g(x)) ⊆ Bεi(fi(x)),12since open balls are open sets. Let δ = min{δxi: x ∈ Fi; i = 1, 2, ..., k} > 0, whichis strictly positive, being the minimum of a finite list of positive numbers. Thenobviously we have g ∈ B(g, δ, (∪Fi)) ⊆ ∩B(fi, εi, Fi) owing to the fact that thechoices were made so as to hold for all i. The set ∪Fiis finite, since it is a finiteunion of finite sets, and hence we are done.1. A Rant About General TopologyAbstract sets X do not come with a set of instructions for how to make bases forthem. The same set may have many different useful bases. Think of this set of sets,{B} = B, as our primary investigative tool, when confronted with an arbitrary setE ⊆ X.To determine if E is open, we select a random point x ∈ E and see if we can fita B around it, so that the B is still entirely within E. Sets for which this attemptis inevitably successful we call open. Upon verification that this will always work,we may ‘upgrade’ an E to an O. Note that requirement (i) for B simply states thatthis upgrade is valid when the set under consideration, E, is all of X.Any specific point x ∈ E for which the procedure is successful; that is, such that∃B ∈ B with x ∈ B ⊆ E, we call an interior point of E. The above condition canbe restatedE is open ⇐⇒ all elements (points) of E are interior to E.Now we will restate the above in a more precise fashion. Assume we have a baseB for X. Define a collection of subsets of X, to be called open sets, as follows:O := {E ∈ 2X: ∀x ∈ E ∃B ∈ B such that x ∈ B ⊆ E}We will show that O satisfies the following three requirements:(i) ∅, X ∈ O(ii) If P ⊆ O is an arbitrary collection of open sets, then their union[O∈PO ∈ Ois open.(iii) If O1, ..., Ok∈ O is a finite collection of open sets, then their intersection\i∈[k]Oi∈ Ois open.Any set of sets that satisfies (i), (ii), (iii) is called a topology on X. In otherwords, if one says the phrase ‘X has a topology,’ one really means that ‘X hasa distinguished class of subsets, which we call open. Considered as a whole theysatisfy the above requirements. We term this distinguished class a topology.’Now, a general collection of open sets possibly may not be presented using abase. Most of the ones people have invented are, however. Think of ‘calling a setopen if all of its elements are interior’ as a guideline for identifying open sets. As itturns out, thes e open sets form a topology. The only thing that must be verified isthat our even further distinguished collection of basis sets must in fact be a base.3That is because we may argue as follows, using only the above axiomatic pro-gram. Assume we have a base B, and define a collection of sets O as above . Clearly∅ ∈ O, since the hypothesis x ∈ ∅ is vacuous. The first requirement, (i), in thedefinition of a base, exactly states that every x ∈ X is interior to X; that is,X ∈ O.Let P ⊆ O, and consider an arbitrary x ∈[O∈PO. (We must show this x is aninterior point.) In particular, we have x ∈ O1for some O1∈ P, by definition ofunion. O1is open, so we may choose B ∈ B such thatx ∈ B ⊆ O1⊆[O∈PO.In particular, this B demonstrates that x is interior to the union of the sets in P.Now let O1, ..., Ok∈ O be open, and consider an arbitrary x ∈ ∩Oi. (Again wemust show x is interior. We suppress the intersection index notation; it is alwaysi ∈ [k] := {1, ..., k}.) For each i we may choose Bisuch that x ∈ Bi⊆ Oi, since xis interior to e ach Oi(by simple virtue of its presence in Oi). So x ∈ ∩Bi⊆ ∩Oi.Using requirement (ii) for B, select B ∈ B so thatx ∈ B ⊆ ∩Bi⊆ ∩Oi.Hence, x is interior to ∩Oi, so it is an open set. We can collect the ideas above andstate:Theorem: The collection of open sets as defined by a base forms a topology.Exercise: Let X, d be a metric space. DefineB = {Bε(x) : ε > 0, x ∈ X},the set of all open ε balls centered at points of X. Show B is a base for X.Exercise: Find several examples of different bases, B16= B2, for the same set X,such that they generate the same topology. (Exclude trivial examples like ‘excludingall balls of radius exactly 1’ or other dumb answers you might think are clever.)Try different metrics on the same set, or an order