Discrete topology, product topology

[mathsss2]

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)$$\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n) = 0 \}|\leq5 \}$$

 

[mathsss2]

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

 

[morphism]

And what are your thoughts on the problem?

 

[mathsss2]

Here is what I know:

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.$$f$$ is a function. Here $$\omega = \mathbb{N} \cup \{ 0 \}$$ (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 $$f(n) \in \{ 0 , 1\}.$$

This is as far as I've gotten.

 

[morphism]

There's a nice graphical representation of the product topology on Y^X (i.e. the product of the space Y |X| times). Namely, if we draw X as an "x-axis" and Y as a "y-axis", then elements in X^Y are "graphs of functions" in the X-Y "plane". An open nbhd of an element f is the set of all functions g whose graphs are close to the graph of f at finitely points. We get different nbhds by varying the closeness to f and/or the set of finite points.

In our case the product space is 2^w=2^N, whose "plane" looks like two copies of the naturals N. In other words, if you were to imagine this as a 'subset' of R^2, it's just the set $\{(n,i) \colon n \in \bN, i \in \{0,1\}\}$.

Maybe this will help you.

 

[mathsss2]

Progress:

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 $$C = \{f \in \prod_{n \in \omega} X_n | \text{ }\forall n \in \omega \text{ }f(n) = 0 \Rightarrow f(n + 1) = 1 \}$$. 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.)