A Universe of Sorts

§ Urhyson's lemma

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

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

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

U_p = f^{-1}([0, p)) \subseteq f^{-1}([0, p]) \subseteq f^{-1}([0, q)) \subseteq f^{-1}([0, q]])

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:

$A \subseteq U_p \subseteq B^c$ .
$\overline{U_p} \subseteq U_q$ 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:

A \subseteq U_{p_1} \subseteq U_{p_1}^c \subseteq B^c

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:

\overline U_r \subseteq V \subset \overline V \subseteq U_s

§ Tietze Extension theorem

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

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.