© | Dror Bar-Natan: Classes: 2004-05: Math 1300Y - Topology:

(10)

Next: Class Notes for Tuesday September 28, 2004 Previous: Homework Assignment 1

Agenda for September 28, 2004

this document in PDF: Agenda-040928.pdf

Comment. $f$ is continuous at $x$ iff for every neighborhood $V$ of $f(x)$, its inverse image $f^{-1}(V)$ contains a neighborhood of $x$.

Agenda. We will discuss two primary notions and the interaction between them and along the way also learn about sequences....

First notion -- the product topology. (The naive definition and the box topology), definition by listing our requirements, uniqueness and existence, interaction with the trivial topology, the subspace topology, $T_2$ and the discrete topology.

Second notion -- metric spaces and metrizability Definition, examples, the metric topology, $T_2$-ness, metrizability.

The interaction We'll prove three theorems:

Theorem 1. (good) $\emptyset\neq\prod_{k=1}^\infty X_k$ is metrizable iff every $X_k$ is metrizable.

Theorem 2. (who cares?) ${\mathbb{R}}^{\mathbb{N}}_{\text{box}}$ is not metrizable.

Theorem 3. (bad) ${\mathbb{R}}^{\mathbb{R}}$ is not metrizable.

In order to prove Theorems 2 and 3 we will need to know about sequences, and these are quite interested by themselves:

Sequences. Convergence, sequential closure.

Proposition 1. The sequential closure is always a subset of the closure, and in a metrizable space, they are equal.

Proposition 2. If $F:X\to Y$ and $X$ is metric, then $f$ is continuous iff for every sequence in $X$, the convergence $x_k\to x$ implies the convergence $f(x_k)\to f(x)$.

The generation of this document was assisted by L^ATEX2HTML.