- Dense set: set whose closure is full space
- Baire Category theorem: Intersection of countably many dense sets is dense (countable interesction of chonk is chonk)

- Let $D[i]$ be family of dense sets.
- Denote $C(\cdot)$ for the closure of a set.
- Let $p \in W$, we need to show that $p \in C(\cap_i D[i])$.
- So for any $\epsilon$, we need to show that there is a point $w$ (for witness) that is $\epsilon$ close to $p$ in $\cap_i D[i]$.
- So we need to show that $w$ is in each of the $D[i]$.
- We will create a sequence of points that will converge to $w$

```
def find_witness(Ds, p, eps):
"""returns a point 'w' such that `w ∈ Ds[i] ∀i` and that `|p - w| < eps`."""
seq = []
cur = D[0].get_close_pt(p, eps/3)) # cur is 'eps/3' close to p.
d_scale = 1 # current distance is eps / 3^d
yield cur
# loop invariant: 'cur' is eps/3^d from p
for k in range(1, infty):
for i in range(0, k):
d += 1
# 1. next is closer to to point than 'cur' is by (1/3)
next = D[i].get_close_pt(cur, eps/3**d)
# 2. (cur, next) distance is a monotonic decreasing function of 'd',
# so d(cur, next) > d(next, next')
# Proof:
# - dist(cur, next) <= dist(cur, p) + dist(p, next)
# - dist(cur, next) <= eps/3**(d) + eps/3**(d+1) <= 4 eps / 3**d
# - this dist(cur, next) is monotone decreasing in d, and thus
# sequence is cauchy.
yield cur
cur = next
```

- A nowhere dense set is a set whose closure is empty.
- A meagre set/ a set of first category is a set which can be written as a countable union of nowhere dense sets.
- A non meagre set/a set of second category is a set which cannot be written in this way
- Baire Category: A complete metric space is non-meagre / second category in itself.

- By contradiction, assume that $X$ is the union of nowhere dense sets $N_i$.
- complement the set $N_i$ to get $D_i \equiv X - N_i$.
- We claim that the sets $D_i$ are dense.
- By baire category theorem, $\cap D_i$ is dense.
- But this means that $(\cap D_i)^C$ is nowhere dense, that is, $\cup D_i^C = \cup C_i$ is nowhere dense.
- If $\cup C_i$ is nowhere dense, then $(\cup C_i) \neq X$.

- Baire Category, stmt 2: If $X$ is the union of a countable family of closed subsets, then at least one of the closed subsets contains an open set

- TODO

- Let $D$ be an enumerable set and $X$ a complete metric space.
- We have some statement $\forall x \in X, \exists d \in D, P_d(x)$.
- We get a uniform statement: $\exists d \in D, \forall x \in X, P_d(x)$
- To do this, we first define $F_d \equiv \{ x \in X : P_d(x) \}$ (filter $x$ by $P_d$).
- If we are lucky, then each of the $F_d$ are closed. Since $d \in D$ is enumerable, and we have that $X = \cup_d F_d$, we apply baire category.
- Baire category tells us that there is a $\mathcal d \in D$ such that $F_{\mathcal d}$ has non empty interior.
- By setup the situation such that if $int(F_D) \neq \emptyset$, then $F_D = X$, which gives us the uniform statement.

- Let $P$ be infinitely differentiable, such that for each $x \in \mathbb R$, there is a number $n(x) \in \mathbb N$ such that $\partial^{n(w)} P /\partial x^{n(w)}|_w = 0$.
- This is again a case where we switch a pointwise fact --- is locally like a poly as nth order derivative vanishes, into a global fact --- is actually a polynomial.
- Let us try the above proof sketch.
- Proof by contradiction, assuming $P$ is not a polynomial.
- Define $X \equiv \{ x : \forall x \in (a, b), P|_{(a, b)}~\text{is not a polynomial} \}$
- Define $F_d \equiv \{ v \in [0, 1] : (\partial^d f / partial x^d)(v) = 0 \}$
- Clearly, $\cup F_d \equiv [0, 1]$, and each of th $F_d$ are closed (zero set of fn).
- By baire category, one of the $F_d$ has an open set inside it.
- This means that for some open set, there is some natural $D$ such that the $D$th derivative vanishes.
- From this, it is "clear" that $f$ is a polynomial?

- Assume $[0, 1]$ is countable.
- So $[0, 1] = \cup_i \{x[i]\}$ for a countable number of points $x[i]$.
- See that each of the sets $\{x[i]\}$.
- But we know that a nonempty set $X$ is not a countable union of nowhere dense sets.

- Let $X$ be a banach space, $Y$ a normed vector space, $B(X, Y)$ be all bounded linear operators from $X$ to $Y$. Let $F$ be a collection of bounded linear operators from $X$ to $Y$. If $\forall x_0 \in X, \sup_{T \in F} ||T(x_0)||_Y < \infty$ (set of operators is pointwise bounded), then $sup_{T \in F} ||T||< \infty$ (set of operators is uniformly bounded)