• The GSI will be Nick Brody. Office hours and office
• Midterm by vote will be Monday, Oct 29
• Course webpage has been updated. Homework is recommended to be written in Latex, submitted on paper
• Shoutout to arXiv.org - preprints, a wonderful way to see the current state of a subject

# Seminorm

As an example of a seminorm, let $$V$$ be a vector space over $$\mathbb{R}$$, and take $$\varphi \in V'$$, a linear functional in the dual space. Define

$\| \nu \|_\varphi = p_\varphi(\nu)$

More than just $$0$$ can be in the kernel of $$\varphi$$ (trivially take $$\varphi = 0$$), so this is a seminorm.

# Sequences

Let $$X$$ be a set. A sequence in $$X$$ is a function $$f \colon \mathbb{N} \to X$$. We notate it as $$\{ x_n \}$$ or $$\{ x_n \}_{n \in \mathbb{N}}$$.

Let $$(X, d)$$ be a metric space, and let $$\{x_n\} \subset X$$. We say that $$\{ x_n \}$$ converges to $$x_n$$, written $$\{ x_n \} \to x$$, if

$\forall \epsilon > 0 \ \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, d(x, x_n) < \epsilon$

Proposition. Let $$(X, d_x), (Y, d_y)$$ be metric spaces, and let $$f \colon X \to Y$$ be continuous. Let $$\{ x_n \} \to x \in X$$. Then $$\{ f(x_n) \} \to f(x)$$.

This proposition also has a converse. If we apply $$f$$ to all sequences converging at a point $$x \in X$$ and the resultant sequences also converge, then $$f$$ is continuous at that point.

Heading towards completions, let us define some notions in metric spaces.

Let $$(X, d)$$ be a metric space. Let $$S \subset X$$, we say that $$S$$ is dense in $$X$$ if

$\forall x \in X \ \forall \epsilon > 0 \ \exists s \in S \text{ with } d(x, s) < \epsilon$

For $$x \in X$$, the open ball about $$x$$ of radius $$\epsilon > 0$$ is

$B(x, \epsilon) = \{ x' \in X \colon d(x, x') < \epsilon \}$

Proposition. Let $$(X, d_x), (Y, d_y)$$ be metric spaces. Let $$S \subset X$$, with $$S$$ dense in $$X$$. Let $$f, g \colon X \to Y$$ be continuous. Then if $$f|_S = g|_S$$, we have $$f = g$$.
Proof. Omitted.

For $$(X, d)$$ a sequence $$\{ x_n \} \subset X$$ is a Cauchy sequence if

$\forall \epsilon > 0 \ \exists N \in \mathbb{N} \text{ s.t. for } m, n \geq N, d(x_m, x_n) < \epsilon$

Cauchy sequences are preserved under uniformly continuous maps:

Proposition. Let $$(X, d_x), (Y, d_y)$$ be metric spaces, and let $$f \colon X \to Y$$. If $$f$$ is uniformly continuous, and if $$\{ x_n \}$$ is a Cauchy sequence in $$X$$, then $$\{ f(x_n) \}$$ is Cauchy in $$Y$$.
Proof. Let $$\epsilon > 0$$ be given. Then there is $$\delta > 0$$ such that if $$x, x' \in X$$ with $$d_x(x, x') < \delta$$ then $$d_y(f(x), f(x')) < \epsilon$$. Since $$\{ x_n \}$$ is Cauchy, there exists an $$N$$ s.t. if $$m, n \geq N$$, then $$d_x(x_m, x_n) < \delta$$. But then $$d_y(f(x_n), f(x_m)) < \epsilon$$.

# Toward Completions

In an ideal metric space, we want all Cauchy sequences to converge:

Let $$X$$ be a metric space. If every Cauchy sequence in $$X$$ converges, then $$(X, d)$$ is said to be complete.

Let $$\mathbb{Q}$$ be the set of rational numbers. We can define a metric $$d$$ on $$\mathbb{Q}$$ by $$d(r, s) = |r - s|$$. We intuitively know that $$\mathbb{Q}$$ is not complete, as for example $$p(t) = t^2 - 2$$ has no roots in $$\mathbb{Q}$$.

However, $$\mathbb{R}$$ with the same metric is complete, and $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$.

Inspired by the relationship by $$\mathbb{Q}$$ and $$\mathbb{R}$$, let’s define a formal notion of completion (draw diagram):

Let $$(X, d)$$ be a metric space. By a completion of $$(X, d)$$ we mean a complete metric space $$(\overline{X}, d)$$ with an isometric map $$f \colon X \to \overline{X}$$ with dense range.

Existence of completions is important, but it turns out we can say something about uniqueness of completions by constructing isomorphisms:

Proposition. If $$(Y_1, d_{y_1}), f_1 \colon X \to Y_1$$ and $$(Y_2, d_{y_2}), f_2 \colon X \to Y_2$$ are completions of $$(X, d)$$, then there is a $$g \colon Y_1 \to Y_2$$ isometric onto with $$f_2 = g \circ f_1$$.