We don't know any continuous functions on compact Haussdorf spaces; Let $X$ be a topological space. What functions $X \rightarrow \mathbb R$ are continuous? We only have the constant functions! If $X$ is a metric space, can we get more continuous functions? We can probably do something like $f_p(x) \equiv d(p, x)$. 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 $X$ into $\mathbb R^n$, I then get so many continuous functions! I can take the polynomials on $\mathbb R^n$ and restrict to $X$. 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 $X$ is normal and $A, B$ are closed disjoint subsets of $X$, there exists a continuous function $f: X \rightarrow [0, 1]$ such that $f(A) = 0$ and $f(B) = 1$. 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 $X$ is normal and $C \subseteq O \subseteq X$ with $C$ closed, $O$ open, then there exists a set $U \subseteq X$ that is open such that $C \subseteq U \subseteq \overline{U} \subseteq O$. (In fact, this is iff!) See that we "reverse the direction"; We started with closed-open, we end with closed-(open-closed)-open.

• Consider $C$ and $O^c$. These are two closed sets. Since $C$ is contained in $O$, $C$ does not meet $O^c$ ( $C \subseteq O$, we have $C \cap O^c = \emptyset$).
• By normality, we have two opens $P, Q$ such that $C \subseteq P$, $O^c \subseteq Q$, and $P \cap Q = \emptyset$.
• So we have $C \subseteq P$, and $P \subseteq Q^c$. This gives us $C \subseteq P \subseteq Q^c \subseteq O$.
• We have $\overline{P} \subseteq Q^c$ as $P \subseteq Q^c$ and $Q^c$ is closed, and thus contains all of its limit points.
• This together gives $C \subseteq P \subseteq \overline{P} \subseteq Q^c \subseteq O$.

§ 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 $X$ to build the continuous function using the pre-image characterization.
• Suppose we succeeded. Then $[0, p) \subseteq [0, 1]$ is open in the subspace topology.
• Define $U_p \equiv f^{-1}([0, p)) \subseteq X$. Each of these $U_p$ are included in one another as we make $p$larger.
• In fact, we have

• We have that $\overline{U_p} = f^{-1}([0, p]) \subseteq U_q = f^{-1}([0, q))$ for $p < q$.
• If we now think of the original sets, we needed $f(A) = 0$, $f(B) = 1$. So we must have that $A \subseteq f^{-1}([0, p))$for all $p > 0$.
• Similarly, we have that $U_p \subseteq B^c$ for $p < 1$, as till $p$ reache $1$, we cannot get to $B$.
• This gives us $A \subseteq U_p \subseteq B^c$.
• This is the only properties we will use to reconstruct $f$!
• Really, I only need a dense subset of $U_p$. So let's say I pick $\{ U_p : p \in \mathbb Q \}$.
• I can reconstruct $f(x)$ by first thinking of $f(x)$. There are sets $U_p$ that reach towards $f(x)$. Consider the closest such
• So take $y = sup(\{ p \in \mathbb Q : p \leq f(x) \} \cup \{ 0 \})$. Because $\mathbb Q$is dense in $\mathbb R$, this works out and we get $y = f(x)$
• But if $p \leq f(x)$, this means $x \not \in U_p$ since $U_p$ covers $[0, p)$. So we write this as: $y = sup(\{ p \in \mathbb Q : x \not \in U_p \} \cup \{ 0 \})$. Because $\mathbb Q$is dense in $\mathbb R$, this works out and reconstructs for us us $y = f(x)$(see that we did not use $f$ in the definition of $y$).

§ Claim 1

Claim: If $P \subseteq (0, 1)$ is dense, and $\{ U_p \}_{p \in P}$ is a collection of open subsets of $X$ indexed by $p$, such that:

1. $A \subseteq U_p \subseteq B^c$.
2. $\overline{U_p} \subseteq U_q$ whenever $p < q$

Then $f(x) = sup(\{ p \in P : x \not \in U_p \} \cup \{ 0 \})$ is continuous and $f(A) = 0$, $f(B) = 1$. See that we don't even need normality! (Time: 29:30 in video)

• Proof: It's clearly well defined based on $sup$. If it's continuous, then it obeys the properties based on the containment assumptions (1) and (2).
• For $p \in P$, We claim $x \not \in U_p \implies f(x) \geq p$. If $x \not \in U_p$, then $p$ is in the set of points we take a $sup$ over. Hence, we have that $f(x) \geq p$since $f(x)$ is the $sup$.
• The contrapositive is that $f(x) < p \implies x \in U_p$.
• To show $f: X \rightarrow [0, 1]$ is continuous, it suffices to show that preimages of open sets are open for a basis. We know a basis consisting of intervals $\{ (p, q) : p, q \in P \}$. We need to show that $f^{-1}((r_0, s_0))$ is open for $r_0, s_0 \in P$ with $0 < r_0 < s_0 < 1$. [The cases where $r_0 = 0$ or $s_0 = 1$ are easy modifications ].
• Choose a $x \in f^{-1}((r_0, s_0))$. Since $P$ is dense, we can find $r_x, s_x$ such that $r_0 < r_x < x < s_x < s_0$.
• We claim that $x \in \overline{U_{r_x}}^c \cap U_{s_x}$. $\overline{U_{r_x}}^c$ is open as it is the complement of a closed set. Hence, we have shown that $x$ is in this open nbhd.
• Since $f(x) < U_s$, we must have $x \in U_s$ by our contrapositive.
• We claim that $x \not \in \overline{U_r}$. Proof by contradiction; suppose $x \in \overline{U_r}$, hence $r < f(x)$. Then, for any $r' \in P$ such that $r < r' < f(x)$, we would have $x \in U_r'$ (by 2). We claim that this is A CONTRADICTION.
• Since $f(x) = sup(\cdot)$, and $r < f(x)$ we have something in the $sup$ that is bigger than $r$. Let this thing be $r'$ such that $x \not \in U_{r'}$. But this is a contradiction to $x \in \overline{U_r}$.
• TODO: there is more to proof!

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

With $P \equiv \mathbb Q \cap (0, 1)$ there exists a collection $\{ U_p \}_{p \in P}$ as we need.

We will prove this by induction. Choose an bijection $\mathbb Q \cap (0, 1) \equiv \{ p_1, p_2, \dots \}$. Let $P_n \equiv \{ p_1, p_2, \dots, p_n \}$. To define $U_{p_i}$ choose an open set such that:

Such a set exists by our characterization of normality. Now suppose $U_{p_1}, \dots U_{p_n}$ have been constructed such that $\{ U_p \}$ satisfies the claims (i), (ii). To construct $U_{p_{n+1}}$. Recall that the $p_i$ can be in arbitrary order, since we choose an arbitrary bijection. So let $r$ be the index such that $p_r$ comes right before $p_{n+1}$, and $s$ comes right after $p_{n+1}$. This gives us $r = \max \{ p_i | 1 \leq i \leq n \land p_i < p_{r+1} \}$, $s = \min \{ p_i | 1 \leq i \leq n \land p_i < p_{r+1} \}$. Then $r \leq s$ and $\overline U_r \subseteq U_s$. Thus by our characterization of normality, there exists an open $V$ such that:

§ Tietze Extension theorem

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

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

§ References

• Urhyson's Lemma video
• Urhyson's Lemma lecture notes