§ Urhyson's lemma

We don't know any continuous functions on compact Haussdorf spaces; Let XX be a topological space. What functions XRX \rightarrow \mathbb R are continuous? We only have the constant functions! If XX is a metric space, can we get more continuous functions? We can probably do something like fp(x)d(p,x)f_p(x) \equiv d(p, x). But the compact Haussdorf spaces are very nice, we should know something about them! we often asumme we have an embedding (a homeomorphism) of a compact Haussdorff space XX into Rn\mathbb R^n, I then get so many continuous functions! I can take the polynomials on Rn\mathbb R^n and restrict to XX. Polynomials are dense in the compact-open topology! (Stone-Weirstrass)

§ Normal space (T4)

A space where points are closed, and two disjoint closed subsets can be separated by open neighbourhoods. It's stronger than haussdorf, because we can separate subsets, not just points.

§ Urhyson lemma

If XX is normal and A,BA, B are closed disjoint subsets of XX, there exists a continuous function f:X[0,1]f: X \rightarrow [0, 1] such that f(A)=0f(A) = 0 and f(B)=1f(B) = 1. This gives us "interesting continuous function" on an arbitrary topological space. We ask for normal, because compact Hausfdorff spaces are normal. So "good smooth manifolds" for example, are normal.

§ Lemma: Re-characterization of normality

If XX is normal and COXC \subseteq O \subseteq X with CC closed, OO open, then there exists a set UXU \subseteq X that is open such that CUUOC \subseteq U \subseteq \overline{U} \subseteq O. (In fact, this is iff!) See that we "reverse the direction"; We started with closed-open, we end with closed-(open-closed)-open.
  • Consider CC and OcO^c. These are two closed sets. Since CC is contained in OO, CC does not meet OcO^c ( COC \subseteq O, we have COc=C \cap O^c = \emptyset).
  • By normality, we have two opens P,QP, Q such that CPC \subseteq P, OcQO^c \subseteq Q, and PQ=P \cap Q = \emptyset.
  • So we have CPC \subseteq P, and PQcP \subseteq Q^c. This gives us CPQcOC \subseteq P \subseteq Q^c \subseteq O.
  • We have PQc\overline{P} \subseteq Q^c as PQcP \subseteq Q^c and QcQ^c is closed, and thus contains all of its limit points.
  • This together gives CPPQcOC \subseteq P \subseteq \overline{P} \subseteq Q^c \subseteq O.

§ Proof: Intuition

  • Suppse we succeeded. Then the only thing we know is that the space is normal, so it is rich in open sets. We're going to convert the existence of a continuous function into properties of open pre-images. We will then show that we have "enough opens" in XX to build the continuous function using the pre-image characterization.
  • Suppose we succeeded. Then [0,p)[0,1][0, p) \subseteq [0, 1] is open in the subspace topology.
  • Define Upf1([0,p))XU_p \equiv f^{-1}([0, p)) \subseteq X. Each of these UpU_p are included in one another as we make pplarger.
  • In fact, we have
Up=f1([0,p))f1([0,p])f1([0,q))f1([0,q]])U_p = f^{-1}([0, p)) \subseteq f^{-1}([0, p]) \subseteq f^{-1}([0, q)) \subseteq f^{-1}([0, q]])
  • We have that Up=f1([0,p])Uq=f1([0,q))\overline{U_p} = f^{-1}([0, p]) \subseteq U_q = f^{-1}([0, q)) for p<qp < q.
  • If we now think of the original sets, we needed f(A)=0f(A) = 0, f(B)=1f(B) = 1. So we must have that Af1([0,p))A \subseteq f^{-1}([0, p))for all p>0p > 0.
  • Similarly, we have that UpBcU_p \subseteq B^c for p<1p < 1, as till pp reache 11, we cannot get to BB.
  • This gives us AUpBcA \subseteq U_p \subseteq B^c.
  • This is the only properties we will use to reconstruct ff!
  • Really, I only need a dense subset of UpU_p. So let's say I pick {Up:pQ}\{ U_p : p \in \mathbb Q \}.
  • I can reconstruct f(x)f(x) by first thinking of f(x)f(x). There are sets UpU_p that reach towards f(x)f(x). Consider the closest such
  • So take y=sup({pQ:pf(x)}{0})y = sup(\{ p \in \mathbb Q : p \leq f(x) \} \cup \{ 0 \}). Because Q\mathbb Qis dense in R\mathbb R, this works out and we get y=f(x)y = f(x)
  • But if pf(x)p \leq f(x), this means x∉Upx \not \in U_p since UpU_p covers [0,p)[0, p). So we write this as: y=sup({pQ:x∉Up}{0})y = sup(\{ p \in \mathbb Q : x \not \in U_p \} \cup \{ 0 \}). Because Q\mathbb Qis dense in R\mathbb R, this works out and reconstructs for us us y=f(x)y = f(x)(see that we did not use ff in the definition of yy).

§ Claim 1

Claim: If P(0,1)P \subseteq (0, 1) is dense, and {Up}pP\{ U_p \}_{p \in P} is a collection of open subsets of XX indexed by pp, such that:
  1. AUpBcA \subseteq U_p \subseteq B^c.
  2. UpUq\overline{U_p} \subseteq U_q whenever p<qp < q
Then f(x)=sup({pP:x∉Up}{0})f(x) = sup(\{ p \in P : x \not \in U_p \} \cup \{ 0 \}) is continuous and f(A)=0f(A) = 0, f(B)=1f(B) = 1. See that we don't even need normality! (Time: 29:30 in video)
  • Proof: It's clearly well defined based on supsup. If it's continuous, then it obeys the properties based on the containment assumptions (1) and (2).
  • For pPp \in P, We claim x∉Up    f(x)px \not \in U_p \implies f(x) \geq p. If x∉Upx \not \in U_p, then pp is in the set of points we take a supsup over. Hence, we have that f(x)pf(x) \geq psince f(x)f(x) is the supsup.
  • The contrapositive is that f(x)<p    xUpf(x) < p \implies x \in U_p.
  • To show f:X[0,1]f: X \rightarrow [0, 1] is continuous, it suffices to show that preimages of open sets are open for a basis. We know a basis consisting of intervals {(p,q):p,qP}\{ (p, q) : p, q \in P \}. We need to show that f1((r0,s0))f^{-1}((r_0, s_0)) is open for r0,s0Pr_0, s_0 \in P with 0<r0<s0<10 < r_0 < s_0 < 1. [The cases where r0=0r_0 = 0 or s0=1s_0 = 1 are easy modifications ].
  • Choose a xf1((r0,s0))x \in f^{-1}((r_0, s_0)). Since PP is dense, we can find rx,sxr_x, s_x such that r0<rx<x<sx<s0r_0 < r_x < x < s_x < s_0.
  • We claim that xUrxcUsxx \in \overline{U_{r_x}}^c \cap U_{s_x}. Urxc\overline{U_{r_x}}^c is open as it is the complement of a closed set. Hence, we have shown that xx is in this open nbhd.
  • Since f(x)<Usf(x) < U_s, we must have xUsx \in U_s by our contrapositive.
  • We claim that x∉Urx \not \in \overline{U_r}. Proof by contradiction; suppose xUrx \in \overline{U_r}, hence r<f(x)r < f(x). Then, for any rPr' \in P such that r<r<f(x)r < r' < f(x), we would have xUrx \in U_r' (by 2). We claim that this is A CONTRADICTION.
  • Since f(x)=sup()f(x) = sup(\cdot), and r<f(x)r < f(x) we have something in the supsup that is bigger than rr. Let this thing be rr' such that x∉Urx \not \in U_{r'}. But this is a contradiction to xUrx \in \overline{U_r}.
  • TODO: there is more to proof!

§ Claim: PQ(0,1)P \equiv \mathbb Q \cap (0, 1)

With PQ(0,1)P \equiv \mathbb Q \cap (0, 1) there exists a collection {Up}pP\{ U_p \}_{p \in P} as we need. We will prove this by induction. Choose an bijection Q(0,1){p1,p2,}\mathbb Q \cap (0, 1) \equiv \{ p_1, p_2, \dots \}. Let Pn{p1,p2,,pn}P_n \equiv \{ p_1, p_2, \dots, p_n \}. To define UpiU_{p_i} choose an open set such that:
AUp1Up1cBcA \subseteq U_{p_1} \subseteq U_{p_1}^c \subseteq B^c
Such a set exists by our characterization of normality. Now suppose Up1,UpnU_{p_1}, \dots U_{p_n} have been constructed such that {Up}\{ U_p \} satisfies the claims (i), (ii). To construct Upn+1U_{p_{n+1}}. Recall that the pip_i can be in arbitrary order, since we choose an arbitrary bijection. So let rr be the index such that prp_r comes right before pn+1p_{n+1}, and ss comes right after pn+1p_{n+1}. This gives us r=max{pi1inpi<pr+1}r = \max \{ p_i | 1 \leq i \leq n \land p_i < p_{r+1} \}, s=min{pi1inpi<pr+1}s = \min \{ p_i | 1 \leq i \leq n \land p_i < p_{r+1} \}. Then rsr \leq s and UrUs\overline U_r \subseteq U_s. Thus by our characterization of normality, there exists an open VV such that:
UrVVUs \overline U_r \subseteq V \subset \overline V \subseteq U_s

§ Tietze Extension theorem

If XX is normal, AXA \subseteq X is closed, given a continuous bounded function f:ARf: A \subseteq \mathbb R then there exists and continuous and bounded function F:XRF: X \rightarrow \mathbb R such that FA=fF|A = f.

§ Urhyson's Metrization theorem

A normal space with a countable basis is metrizable. We know that metrizable is normal. This says that normal is not so far away from metrizable.

§ Embedding of topological manifolds

If XX is a compact topological nn-manifold, then there exists an embedding into some Rm\mathbb R^m. That's saying that there are "m" very interesting continuous functions, the coordinate functions! So it makes sense Urhyson's is involved.

§ References