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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)$.

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Dror Bar-Natan 2004-09-28