Analysis

21.1 General Topological Spaces

It turns out that metric spaces are not sufficiently general for some applications. This section is a brief introduction to general topology. In making this generalization, the properties of balls in a metric space are stated as axioms for a subset of the power set of a given set. This subset of the power set (set of all subsets) will be known as a basis for a topology. The properties of balls which are of importance are that the intersection of finitely many is the union of balls and that the union of all of them give the whole space. Recall that with a metric space, an open set was just one in which every point was an interior point. This simply meant that every point is contained in a ball which is contained in the given set. All that is being done here is to make these simple properties into axioms.

Note that this is simply the analog of saying a set is open exactly when every point is an interior point.

Proof: If p ∈∅ then there exists B ∈ℬ such that p ∈ B ⊆∅ because there are no points in ∅. Therefore, ∅∈ τ. Now if p ∈ X, then by part 2.) of Definition 21.1.1 p ∈ B ⊆ X for some B ∈ℬ and so X ∈ τ.

If C⊆ τ, and if p ∈∪C, then there exists a set, B ∈C such that p ∈ B. However, B is itself a union of sets from ℬ and so there exists C ∈ℬ such that p ∈ C ⊆ B ⊆∪C. This verifies 21.2.

Finally, if A,B ∈ τ and p ∈ A∩B, then since A and B are themselves unions of sets of ℬ, it follows there exists A₁,B₁ ∈ℬ such that A₁ ⊆ A,B₁ ⊆ B, and p ∈ A₁ ∩B₁. Therefore, by 1.) of Definition 21.1.1 there exists C ∈ℬ such that p ∈ C ⊆ A₁ ∩B₁ ⊆ A∩B, showing that A ∩ B ∈ τ as claimed. Of course if A ∩ B = ∅, then A ∩ B ∈ τ. ■

Note that if the topological space is Hausdorff, then this definition is equivalent to requiring that every open set containing p contains infinitely many points from E. Why?

Proof: Suppose first that E is closed and let x be a limit point of E. Is x ∈ E? If x

E, then E^C is an open set containing x which contains no points of E, a contradiction. Thus x ∈ E.

Now suppose E contains all its limit points. Is the complement of E open? If x ∈ E^C, then x is not a limit point of E because E has all its limit points and so there exists an open set, U containing x such that U contains no point of E other than x. Since x

E, it follows that x ∈ U ⊆ E^C which implies E^C is an open set because this shows E^C is the union of open sets. ■

Proof: If x≠p, there exist open sets U and V such that x ∈ U,p ∈ V and U ∩ V = ∅. Therefore,

^C is an open set so is closed. ■

Note that the Hausdorff axiom was stronger than needed in order to draw the conclusion of the last theorem. In fact it would have been enough to assume that if x≠y, then there exists an open set containing x which does not intersect y.

Proof: Let C denote all the closed sets which contain E. Then C is nonempty because X ∈C.

{ } (∩ {A : A ∈ C})C = ∪ AC : A ∈ C ,

an open set which shows that ∩C is a closed set and is the smallest closed set which contains E.

Proof: Let x ∈E and suppose that x

E. If x is not a limit point either, then there exists an open set, U,containing x which does not intersect E. But then U^C is a closed set which contains E which does not contain x, contrary to the definition that E is the intersection of all closed sets containing E. Therefore, x must be a limit point of E after all.

Now E ⊆E so suppose x is a limit point of E. Is x ∈E? If H is a closed set containing E, which does not contain x, then H^C is an open set containing x which contains no points of E other than x negating the assumption that x is a limit point of E. ■

The following is the definition of continuity in terms of general topological spaces. It is really just a generalization of the ε - δ definition of continuity given in calculus.

You should prove the following.

Proof: Suppose x ∈∏ _i=1ⁿA_i∩∏ _i=1ⁿB_i where A_i and B_i are open sets. Say

x =(x1,⋅⋅⋅,xn).

Then x_i ∈ A_i ∩ B_i for each i. Therefore, x ∈∏ _i=1ⁿA_i ∩ B_i ∈ℬ and ∏ _i=1ⁿA_i ∩ B_i ⊆ ∏ _i=1ⁿA_i. ■

The definition of compactness is also considered for a general topological space. This is given next.

In general topological spaces there may be no concept of “bounded”. Even if there is, closed and bounded is not necessarily the same as compactness. However, in any Hausdorff space every compact set must be a closed set.

Proof: Suppose p

K. For each x ∈ X, there exist open sets, U_x and V _x such that

x ∈ Ux,p ∈ Vx,

and

Ux ∩ Vx = ∅.

If K is assumed to be compact, there are finitely many of these sets, U_x1,

,U_xm which cover K. Then let V ≡∩_i=1^mV _xi. It follows that V is an open set containing p which has empty intersection with each of the U_xi. Consequently, V contains no points of K and is therefore not a limit point of K. ■

Proof: Suppose to the contrary that ∅ = ∩K. Then consider

{ } C ≡ KC : K ∈ K .

It follows C is an open cover of K₀ where K₀ is any particular element of K. But then there are finitely many K ∈K, K₁,

,K_r such that K₀ ⊆∪_i=1^rK_i^C implying that ∩_i=0^rK_i = ∅, contradicting the finite intersection property. ■

There is a fundamental theorem for locally compact Hausdorff spaces which is presented next.

Proof: Since X is locally compact, there exists a basis of open sets whose closures are compact U. Denote by C the set of all U ∈U which contain k and let C^′ denote the set of all closures of these sets of C intersected with the closed set V ^C.  Thus C^′ is a collection of compact sets. There are finitely many of the sets of C^′ which have empty intersection. If not, then C^′ has the finite intersection property and so there exists a point p in all of them. Since X is a Hausdorff space, there exist disjoint basic open sets from U, A,B such that k ∈ A and p ∈ B. Therefore, p

A contrary to the above requirement that p be in all such sets. It follows there are sets A₁,,A_m in C such that

--- --- VC ∩ A1 ∩⋅⋅⋅∩ Am = ∅

Let U_k = A₁ ∩

∩A_m. Then U_k ⊆A₁ ∩∩A_m and so it has empty intersection with V ^C. Thus it is contained in V . Also U_k is a closed subset of the compact set A₁ so it is compact and k ∈ U_k.

For the second part, consider all such U_k. Since K is compact, there are finitely many which cover K U_k1,

,U_kn. Then let U ≡∪_i=1ⁿU_ki. It follows that

U-= ∪n Uk- i=1 i

and each of these is compact so this set works. ■

The following is Urysohn’s lemma for locally compact Hausdorff spaces.

Proof: This involves using Lemma 21.1.21 repeatedly. First use this lemma to obtain V open such that its closure is compact and contained in U. Let D ≡{r_n}_n=1^∞ be the rational numbers in (0,1]. Using Lemma 21.1.21, let V _r1 be an open set such that

-- -- H ⊆ Vr1 ⊆ Vr1 ⊆ V, Vr1 is compact

Suppose V _r1,

,V _rk have been chosen and list the rational numbers r₁,,r_k in order,

rl1 < rl2 < ⋅⋅⋅ < rlk for {l1,⋅⋅⋅,lk} = {1,⋅⋅⋅,k}.

If r_k+1 > r_lk then letting p = r_lk, let V _rk+1 satisfy

V- ⊆ V ⊆ V- ⊆ V, V- compact p rk+1 rk+1 rk+1

If r_k+1 ∈ (r_li,r_li+1), let p = r_li and let q = r_li+1. Then let V _rk+1 satisfy

-- -- -- Vp ⊆ Vrk+1 ⊆ Vrk+1 ⊆ Vq, Vrk+1 compact

If r_k+1 < r_l1, let p = r_l1 and let V _rk+1 satisfy

H ⊆ Vr ⊆ Vr ⊆ Vp, V-r compact k+1 k+1 k+1

Thus there exist open sets V _r for each r ∈ ℚ ∩

with the property that if r < s,

-- -- H ⊆ Vr ⊆ V r ⊆ Vs ⊆ Vs ⊆ V.

Now let

⋃ f (x) = min(inf{t ∈ D : x ∈ Vt},1),f (x) ≡ 1 if x ∕∈ Vt. t∈D

(Recall D = ℚ ∩ (0,1].) I claim f is continuous.

−1 f ([0,a)) = ∪{Vt : t < a,t ∈ D},

an open set.

Next consider x ∈ f⁻¹

so f ≤ a. If t > a, then x ∈ V _t because if not, then

f (x) ≡ inf{t ∈ D : x ∈ Vt} > a.

Thus

f−1([0,a]) ⊆ ∩ {Vt : t > a} = ∩{Vt : t > a}

which is a closed set. If x ∈∩{V _t : t > a}, then x ∈∩{V _t : t > a} and so f

≤ a. Thus f⁻¹ is closed. This is also true if a = 1. In this case, f⁻¹ = X. Therefore, for a ≥ 0,

f−1((a,1])∪ f−1([0,a]) = X

and so f⁻¹

is an open set. Next consider f⁻¹. This was shown to be open above. It follows that f is continuous because f⁻¹ = f⁻¹ ∩ f⁻¹ the intersection of two open sets. Since this is so for every interval, it follows that the inverse image of any open set is open and so f is continuous. Clearly f = 0 on H. If x ∈ V ^C, then xV _t for any t ∈ D so f = 1 on V ^C. Let g = 1 − f. ■

In any metric space there is a much easier proof of the conclusion of Urysohn’s lemma which applies.

Proof: Consider

and suppose without loss of generality that f ≥ f. Then choose y ∈ S such that f + ε > d. Then

Since ε is arbitrary, it follows that ≤ d. ■

Proof: Let δ > 0 be such that for all h ∈ H,dist

> δ. If no such δ exists, then you could get a contradiction by obtaining a sequence such that dist≤ 1∕n. Then there is a convergence subsequence, still denoted as h_n such that h_n → h ∈ H. Then from Lemma 21.1.23, dist = lim_n→∞dist = 0 and since U^C is closed, it follows that h ∈ U^C which contradicts h ∈ H. Now consider the balls B. These cover the compact set and so there are finitely many which do so. B,,B where the closure of each of these is compact. Also B

(hj,δ)

⊆ U. Because if x ∈B

(hj,δ)

, then d ≤ δ < dist. Let V = ∪_j=1^mB. Thus V = ∪_j=1^mB

(hj,δ)

because there are only finitely many sets. Also V is compact because it is a finite union of compact sets. Now define

( C) g (x) ≡ -----distx,V--------- dist(x,H )+ dist(x,VC )

This is continuous, equals 1 on H and equals 0 off V because the denominator is always positive since both H,V ^C are closed. ■

A useful construction when dealing with locally compact Hausdorff spaces is the notion of the one point compactification of the space.

The reason this is called a compactification is contained in the next lemma.

Proof: Since

is a locally compact Hausdorff space, it follows is a Hausdorff topological space. The only case which needs checking is the one of p ∈ X and ∞. Since is locally compact, there exists an open set of τ, U having compact closure which contains p. Then p ∈ U and ∞∈U^C and these are disjoint open sets containing the points, p and ∞ respectively. Now let C be an open cover of with sets from . Then ∞ must be in some set, U_∞ from C, which must contain a set of the form K^C where K is a compact subset of X. Then there exist sets from C, U₁,,U_r which cover K. Therefore, a finite subcover of is U₁,,U_r,U_∞.

To see the last claim, suppose U contains ∞ since otherwise there is nothing to show. Notice that if C is a compact set, then X ∖C is an open set. Therefore, if x ∈ U ∖

, and if ∖C is a basic open set contained in U containing ∞, then if x is in this basic open set of , it is also in the open set X ∖C ⊆ U ∖. If x is not in any basic open set of the form ∖C then x is contained in an open set of τ which is contained in U ∖. Thus U ∖ is indeed open in τ. ■

Proof: Suppose first that X is compact. Then if C is an open cover consisting of basic open sets, it follows it admits a finite subcover because these are open sets in C.

Next suppose that every basic open cover admits a finite subcover and let C be an open cover of X. Then define

to be the collection of basic open sets which are contained in some set of C. It follows is a basic open cover of X and so it admits a finite subcover, . Now each U_i is contained in an open set of C. Let O_i be a set of C which contains U_i. Then is an open cover of X. ■

Actually, there is a profound generalization of this lemma.