# Friday, August 26

:::{.remark}
Recall that a function $f\in \smooth\Mfd(M, N)$ if its local coordinate representations \( \psi_ \beta\circ f \circ \varphi_{ \alpha}\inv \in C^\infty(\RR^n, \RR^n) \).
Smoothness is local: if $\mcu\covers M$ then $f$ is smooth iff $\ro{f}{U_\alpha}$ are all smooth.
Next goal: prove that there are lots of smooth functions.
The Urysohn analog will take some work, but we can easily prove a more modest result:
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:::{.proposition title="Smooth $T_{3\, {1\over 2}}$ "}
Given $p\in M, A \subseteq M$ closed with $p\not\in A$, there exists a smooth function $f:M\to\RR$ with $f(p) = 1$ and $\ro{f}{A} = 0$.
Note that this is essentially regularity with separation by a function instead. 
:::

:::{.proof title="Sketch"}
Let \( \varphi_{ \alpha}: U_{ \alpha} \to \RR^d \) be a coordinate chart where without loss of generality (composing with a translation) \( \varphi_{ \alpha}(p) = \vector 0\).
Note $U_\alpha \sm A$ is open in both $U_{ \alpha}$ and $M$, so \( \varphi_{ \alpha}(U_{ \alpha} \sm A) \) is open in \( \varphi_{ \alpha}(U_{ \alpha}) \).
One can find an $r$ such that $\cl B_r(\vector 0) \subset \phi_{ \alpha}(U_{ \alpha \sm A})$ since we are in $\RR^d$.
Claim to show in HW: its pullback \( \varphi_{ \alpha}(\cl B_r(\vector 0)) \) is closed in $M$.[^subtle_hausd]
The pullback is disjoint from $A$.
Making the function $f$: take a smooth bump function $\RR^d \to [0, 1]$ where $g(\vector 0)=1$ and is zero for $\norm{\vector x} \geq r$.
Then define
\[
f(z) \da
\begin{cases}
g( \varphi_{ \alpha}(z) ) &  z\in U_{ \alpha}
\\
0 & z\in M\sm \varphi_{ \alpha}(\cl B_r(\vector 0)).
\end{cases}
.\]
Then $\ro{f}{A} =0$ since $A \intersect \varphi_{ \alpha}\inv(\cl B_r(\vector 0)) = \emptyset$.
Constructing such a $g$: reduce to a 1-dimensional problem by taking it to be the form $g(\vector x) = h(\norm{\vector x}^2)$.
So we want $h: \RR\to I$ with $h(0) = 1, h(s) = 0$ for all $s\geq r^2$.
Modify the basic function 
\[
\rho(s) = 
\begin{cases}
e^{-{1\over s}} & s > 0 
\\
0 & s \leq 0.
\end{cases}
,\]
which is smooth on $\RR$ since one can show that $\rho^{(k)}(s) = p\qty{1\over 2} e^{-{1\over s}}$ where $p$ is a polynomial in $1/s$.
So take $h(s) = {p(r^2-s) \over p(r^2)}$.

[^subtle_hausd]: 
This is subtle and necessitates the Hausdorff condition, and is in fact the only place it is used.

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:::{.definition title="(Regular) coordinate balls"}
If $B \subseteq M$ is of the form $B = \varphi_{ \alpha}\inv(B_r(p))$ for some $p\in \RR^d$ is a **coordinate ball**.
It is a **regular coordinate ball** if $\cl B = \varphi_{ \alpha}\inv (\cl B_r(p))$ and $\cl B_r(p) \subseteq \varphi_{ \alpha}( U_ \alpha)$.
:::

:::{.proposition title="?"}
For $M\in\smooth\Mfd$ there is a stable base $\mcb = \ts{B_n}_{n=1}^\infty$ for the topology on $M$ consisting of regular coordinate balls.
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:::{.proof title="?"}
Let $\ts{ \varphi_k: U_k\to \RR^d}$ be a countable subatlas of the atlas for $M$.
Define the countable set
\[
\mcb = \ts{ \varphi_k\inv(B_q(\vector a)) \st \vector a \in \QQ^d \intersect \phi_k(U_k), q\in \QQ, B_{2q}(\vector a) \subseteq \varphi_k(U_k) }
.\]
One needs that for all $x\in V$ open in $M$, $x\in B \subseteq V$ for some $B\in \mcb$.
For this, take $q$ small enough such that \( \varphi_k\inv(B_{2q}(\vector x) ) \subseteq V \intersect U_k \) and $\vector a\in \QQ^d$ with $\vector a \in B_q(\phi_k(\vector x))$.
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:::{.remark}
Note that in fact the closure of $B$ is still contained in $V$, which we'll use later for partitions of unity.
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