discrete topology, product topology

**Erdos32212** · December 2nd 2008, 12:58 PM

For each $n \in \omega$ , let $X_n$ be the set $\{0, 1\}$ , and let $\tau_n$ be the discrete topology on $X_n$ . For each of the following subsets of $\prod_{n \in \omega} X_n$ , say whether it is open or closed (or neither or both) in the product topology.

(a) $\{f \in \prod_{n \in \omega} X_n | f(10)=0 \}$
(b) $\{f \in \prod_{n \in \omega} X_n | \text{ }\exists n \in \omega \text{ }f(n)=0 \}$
(c) $\{f \in \prod_{n \in \omega} X_n | \text{ }\forall n \in \omega \text{ }f(n)=0 \Rightarrow f(n+1)=1 \}$
(d) $\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|=5 \}$
(e) $\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|\leq5 \}$

Recall that $\omega=\mathbb{N} \cup \{ 0 \}$

**Moo** · December 3rd 2008, 11:01 AM

Hello,

I'm not sure I understand correctly... Is f a function ? Then what does it mean for a function to be in $\prod_{n \in \omega} X_n$ ?

**ThePerfectHacker** · December 3rd 2008, 04:11 PM

Originally Posted by Moo

I'm not sure I understand correctly... Is f a function ? Then what does it mean for a function to be in $\prod_{n \in \omega} X_n$ ?

Yes f is a function. Here $\omega = \mathbb{N}$ (the reason for using $\omega$ is because he is using it to refer to the natural as an ordinal, but whatever that is not important). If it helps you can think of $\prod_{n\in \omega} X_n$ as $\prod_{n=0}^{\infty} X_n$ . We define $\prod_{n=0}^{\infty}X_n$ to be the set of all functions $f: \mathbb{N} \to \{ 0 , 1\}$ that satisfies $f(n) \in \{ 0 , 1\}$ .

(I would try doing the problem, it is just that I am not sure how "open" sets are defined here. I know there is a way to define "open" for an infinite Cartesian product but I just cannot find a good explanation anywhere on the internet. And I never learned topology

. Perhaps you, Moo, know how "open" is defined for infinite Cartesian products).

**Erdos32212** · December 3rd 2008, 08:48 PM

ThePerfectHacker & Moo, thank you both for helping me (and trying to help). Here is what I understand about this problem now:
So remember that open sets in the infinite product topology is really just having all but finitely many the whole space and the rest are open. Since the individual factors are discrete, you only need to check that all but finitely many are the whole space.

e.g. in (a) the 10th coordinate has a specific value, but all other coordinates can be whatever, so this is certain open.

However, I am still struggling with this problem. Do you have any ideas on the other parts?

**Erdos32212** · December 7th 2008, 01:01 PM

Thanks to Opalg [Opalg], we know this:

Take set (b). Let $B = \{f \in \prod_{n \in \omega} X_n | \;\exists n \in \omega \; f(n) = 0 \}$ . If $f\in B$ then there exists m such that f(m)=0. Then the set $\{g \in \prod_{n \in \omega} X_n |\; g(m) = 0\}$ is an open neighbourhood of f contained in B. Therefore B is open.

It's usually more difficult to check when a set is closed. You have to look at its complement and decide whether that is open. Sometimes this is straightforward. For example, the complement of set (a) is the set of all f such that f(10)=1. That is open, so set (a) is closed as well as open.

For a slightly less easy example, look at set (c). Let $C = \{f \in \prod_{n \in \omega} X_n | \matht{ }\forall n \in \omega \matht{ }f(n) = 0 \Rightarrow f(n + 1) = 1 \}$ . If $f\notin C$ then there exists m such that f(m)=f(m+1)=0. Then $\{g \in \prod_{n \in \omega} X_n |\; g(m) = g(m + 1) = 0\}$ is an open neighbourhood of f containing no points of C. Therefore the complement of C is open and so C is closed.

I still do not know how to do parts (d.) and (e.)

Thread: discrete topology, product topology

LinkBack

Thread Tools

Display

discrete topology, product topology

Similar Math Help Forum Discussions

Order topology = discrete topology on a set?

[SOLVED] Discrete topology on an uncountable set

Discrete Topology

Discrete Topology Question

discrete topology, product topology

Search Tags