22M:132Fall 07J. SimonMetrization Theorems(Relates to text Sec. 34)Introduction. What properties of a topolo gical space (X, T ) are enough toguarantee that the topology actually is given by some metric? The space has to benormal, since we know metric spaces are normal. And the topology has to have acountable local basis at each point, since metric spaces have that property. InChapter 6 of the text, there a r e theorems saying, “(X, T ) is metric if and only if ithas the following topological properties . . . ”. The conditions are (1) the space isregular, and (2) there is some countability property stronger than saying there is acountable local basis at each point, but a little weaker than 2nd-countable (but stillstrong enough that regular + the property =⇒ normal ) . In the current textsection, the theorem is less general: we characterize s eparable metric spaces; but thisis a good introduction to the ideas.How does one prove that some topolo gy on a space is given by a metric? There aretwo choices: either explicitly define the metric and prove the metric topology is thesame as T , or show that X is homeomorphic to a subspace of a known metric space.We used the first approach when we defined a metric on Rωthat generates theproduct topology; and now we will see a good example of the other approach, whichis also how the most general metrization theorems are proven.Theorem (Urysohn metrization theorem). If (X, T ) is a regular space with acountable basis for the topology, then X is homeomorphi c to a subspace of the metricspace Rω.The way I stated the a bove theorem, it is ambiguous: we have studied two(inequivalent) metrics for Rω: the product space metric and the uniform metric.The theorem is true with either metric, but it is an “if and only if” for the productmetric. Recall that in the product topo lo gy, Rωhas a countable dense subset: theset S = all vectors (q1, q2, . . .) where each qi∈ Q and all but finitely many qiare 0.Since Rωin the product topology is metrizable a nd has a countable dense subset, itmust be 2nd-countable. Each subspace of space with a countable basis also has acountable basis. And, of course, each subspace of a metric space is metric. Weconclude that each subspace of (Rω, product metric) is metric and has a countablebasis.The above paragraph combines ideas from various parts of our course. So let us takethis as one (bunch of) of our sample problems for the Final Exam. Your task s toorganize the various f acts, ideas, etc. into a coherent proof (and to be able to fill indetails if asked); as always, an exam problem might involve filling in the details ofone particular part of a longer argument).cJ. Simon, all rights reserved page 1Problem. [sample fo r final exam]If (X, T ) is homeomorphic to a subspace of (Rω, product topology), then (X, T ) isregular and h as a countable basis.Key steps:a. There is a metric o n Rωthat gives the product topology.b. Rωin the product topology has a countable den se subset.c. A metric space with a countable dense subset has a countable basis for thetopology.d. Each subspace of a 2nd-countable space is 2nd-countable.e. Each subspace of a metric space is metrizable.e. Each metric space i s regular.f. Put the pieces together.On t he other hand, the metric space (Rω, uniform metric) is not second-count able(so not separable) since it has an uncountable discrete subspace K = the set o f allvectors (t1, t2, . . .) where each ti= 0 or 1. The set of all sequences of 0’s and 1’s isuncountable, and the distance between any two elements of K is 1. So eachsubspace of (Rω, uniform metric topology) is a metric space, but it need not beseparable. We really should state the Urysohn metrization theorem as two theorems:Theorem. (X, T ) is regular with a countable basis ⇐⇒ (X, T ) is homeomorphicto a subspace of (Rω, product topology me tric).Theorem. (X, T ) is regular with a countable basis =⇒ (X, T ) is homeomorphic toa subspace of (Rω, uniform metric topology).Proving the metrization theorem[s]. The text gives the details, so I will focuson the gestalt and some highlights. Our goal is to define an embedding of X intoRω. We wa nt to assign to each point x ∈ X a point F (x) ∈ Rω, that is a (countablyinfinite) list of “coordinates”: F (x) = (x1, x2, . . .). How can we find numbers thatmeasure how a point x ∈ X is related topologically to all the other points of X?This is the bit o f magic in this theorem. We will use Urysohn’s lemma infinitelymany times to define a sequence of functions fn: X → [0, 1]; these will be thecoordinate functions.The space (X, T ) has a countable basis B and it it regular, so it is normal. Givenany closed set A and open neighborhood U(A), t here exists a Urysohn function forthe disjoint closed sets X − U and A. That is, there exists f : X → [0, 1] such t hatf(x) = 0 for all x /∈ U and f( a) = 1 for all a ∈ A. In particular, for any pair Bn, Bmof elements of B that happen to have¯Bn⊆ Bm, t here exists a functionf : X → [0, 1] with f = 1 on¯Bnand f = 0 outside Bm.cJ. Simon, all rights reserved page 2Since B is countable, the set of such pairs Bn, Bmis countable. Number these pairsin any order, and let f1, f2, . . . be the Urysohn functions defined in the precedingparagraph. Then define F : X → [0 , 1] byF (x) = (f1(x), f2(x), . . .).We need to prove that the function F is 1-1, continuous, and has a continuousinverse (From F (X) → X). The questions of continuity have to depend on whattopology we use for Rω. But we can check injectivity befor e worrying about thetopology.Proposition. T he function F : X → Rωis injectiveProof. Suppose x, y ∈ X with x 6= y. Since X is Hausdorff, there exist disjointneighborhoods U(x), V (y). Since B is a basis, there exists some Bmwithx ∈ Bm⊆ U. Since X is regular, there exists a neighbor hood U′(x) such that¯U′⊆ Bm. And, again since B is a basis, there exists a basis set Bnwithx ∈ Bn⊆ U′. Since¯U′⊆ Bm, we thus have¯Bn⊆ Bm. The Urysohn function fjassociated to the pair Bn, Bmhas x → 1 and y → 0; so F (x) 6= F (y). The text goes on to show that , in the product topology, F is continuous and has acontinuous inverse. The proof that F is cont inuous is easy because each coordinatefunction is continuous; the proof that F is an open map takes more wo rk; see thetext for the details.To use the uniform topology, we need to chang e F . Recall that in the producttopology, if we are