Functional Analysis 7 -- Baire's Category Theorem 1

This is course note from Siran Li-Shanghai Jiao Tong University.

Consider \(X\) as nonempty, complete metric space.

Philosophy: Select \(E \subset X.\) Is \(E\) big or small?

Topologically: small -- meagre,

Measure: volume, length.

Primary Example: \(Q \subset \mathbb{R}.\) \(\bar{Q} = \mathbb{R}\) but it is a null set.

We can enumerate \(Q = \{q_1, q_2, ...\}\) Then \[ Q \subset \cup I_i, I_i = ]q_i-\frac{\epsilon}{2^{i+1}}, q_i + \frac{\epsilon}{2^{i+1}}[ \] Then we have \[ \sum length( I_i ) = \sum \frac{\epsilon}{2^i} = \epsilon \] where \(\epsilon > 0\) is arbitrary.

Definition: A set \(E \subset X\) is nowhere dense iff \(\bar{E}\) has no interior.

Write \(A^o = int(A) := \{x \in A: \exists \delta > 0, s.t. B^o(x, \delta) \subset A\}.\) Note that \(A^o\) is open. \(A^o \subset A \subset \bar{A}, A \backslash A^o = \partial A.\)

Note that \(\bar{A^o} \neq \bar{A}^o.\)

Eg: \(Q = \cup \{q\}\) is the union of countable nowhere dense set. But \(Q\) is not nowhere dense.

Definition: Meagre

\(E \subset X\) is meagre iff \(E\) can be written as a countable union of nowhere dense set. One may also say that \(E\) is of first category.

Definition: Residual

\(E \subset X\) is residual iff \(X \backslash E\) is meagre. One may also say that \(E\) is comeagre.

Definition:

\(E \subset X\) is of the 2nd category if it is not of the 1st category.

Properties:

1. \(E \subset X\) is meagre, \(F \subset E\) then \(F\) is meagre.

2. Countable union(intersection) of meagre(residual) sets is meagre(residual).

3. A countable intersection of dense open sets is residual

Lemma: \(X\backslash A^o = \bar{X \backslash A}.\)

Baire Category Theorem:

Let \(X\) be a nonempty compact metric space. Let \(\{U_n\}\) be a sequence of open, dense subset of \(X.\) Then \(\cap U_n\) is dense.

Let \(\{U_n\}\) be a sequence of open, dense subsets of \(X.\) Then \(\underset{n}\cap U_n\) is dense.

Definition:

A countable \(\cap\) of open sets is called a \(G_{\delta}\)-set.

A countable \(\cup\) of closed sets is a \(F_{\delta}\)-set.

Corollary:

Residual \(\Leftrightarrow\) contains dense \(G_{\delta}\)

Proof: \(\Leftarrow\) If \(X\cap O_n \subset E \subset X\) where \(O_n\) is open and dense. Then $E# is residual.

\(\Rightarrow\) If \(E\) is residual, then \(X\backslash E\) = \(\cup M_n\) for \(int \bar{M}_n = \phi.\) \(\Rightarrow E = X \backslash (\cup M_n) = \cap (X \backslash M_n).\) where \(\cap (X \backslash \bar{M}_n) \subset \cap (X \backslash M_n).\) Hence \(X \backslash \bar{M}_n\) is dense since, \(\bar{X\backslash \bar{M}_n} = X\backslash(int(\bar{M}_n)) = X,\) given \(M_n\) is nowhere dense. \(\Box\)

Corollary:

1. A countable \(\cap\) of dense \(G_{\delta}\)-set is dense \(G_{\delta}\)-set.

2. If \(M \subset X\) is meagre, then \(x \backslash M\) is dense.

3. Any meagre set in \(X\) has empty interior.

4. \(X\) is no meagre in itself.

5. Let \(X\) be a Banach space. Its Hamel basis are either finite or uncountable.

proof: (3) Let \(M \subset X\) be meagre. Then \(M = \cup E_n\) for \(int \bar{E}_n =\phi.\) \(\Rightarrow X \backslash int M = \bar{X \backslash M} = \bar{X \backslash E_n}.\) But \(E_n\) is nowhere dense, leading to that \(X\backslash E_n\) is residual. \(X\backslash E_n\) contains dense \(G_{\delta}.\) Then \(\cap(X\backslash E_n)\) contains dense \(G_{\delta}.\) Then \(\bar{\cap(X\backslash E_n)} = X.\) \(int M = \phi.\)

5. Consider dim \(X = \infty.\) If \(\{e_1, ...\}\) is a countable Hamel basis, then \(X = \cup_n span\{e_1,...,e_n\}.\) Then \(span\{e_1,...,e_n\}\) is nowhere dense, leads to \(X\) is meagre. This leads to contradiction to (4).

Fact: There is a "Berstein set" \(F\subset [0,1]\) s.t. \(\forall [a,b] \subset [0,1], F\cap[a,b]\) is neither meagre nor residual.

Applications:

Definition:

\(x \in \mathbb{R}\backslash Q\) is Diophantine of exponent \(\alpha > 0\) if \(\forall p/q \in Q, |x - p/q|>C/q^{\alpha}.\)

proof: 考虑代数数，作为整系数多项式系数的根，距离一个有理数的距离有下界。

Proposition:

Any degree-d-algebraic number is Diophantine pf exponent \(d.\)

On the other hand,

Proposition:

\(\forall a \in \mathbb{R}\backslash Q, \exists p/q \in Q\) s.t. \(|\alpha - p/q| \le 1/q^{a}.\)

proof: \(\forall \alpha \in \mathbb{R}\backslash Q, \forall n \in \mathbb{N},\) consider \(\alpha, 2\alpha,...,n\alpha \mod 1.\) For \(0 \le n_1 < n_2 \le n, 0 \le (n_2 - n_1)\alpha \mod 1 \le 1/n \le 1/(n_2-n_1).\) So for \(q = n_2-n_1,\) \(q\alpha \mod 1 \le q^{-1} \leftrightarrow \exists p \in \mathbb{Z}, |q\alpha-p| \le 1/q.\)

Definition:

\(x \in \mathbb{R}\backslash Q\) is Liouville iff \(\forall n \in \mathbb{N}, \exists p/1 \in Q,\) s.t. \(|x-p/q| < q^{-n}.\)

eg: Liouville number \[ x = \sum\frac{1}{10^{n!}}. \] Liouville numbers are transcendental(超越数).

Theorem：

A generic("residually many") irrational number is Liouville.

proof: Consider set {Liouville numbers} = \(\cap O_n.\) Here \[ O_n = \{x \in [0,1]: \exists p,q \in \mathbb{Z}_+, |x - p/q| < q^{-n}.\} \] So, \(Q \cap O_n\) is dense in \(O_n.\) This leads to that {Liouville numbers} is residual and dense by BCI(Borel-C-Lemma).

Theorem:

A random irrational is Diophantine (2+\(\epsilon\)) \(\forall \epsilon > 0.\)

Theorem:

A generic continuous function on \([0,1]\) is nowhere differentiable.

John Nash: Rapid Oscillation (振幅很小,频率增加的扰动) -> Convex Integration