# discrete topology, product topology

• Dec 6th 2008, 11:03 AM
GenoaTopologist
discrete topology, product topology
My friend and I are still stuck on:

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

Background:

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 $f(n) \in \{ 0 , 1\}$.

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

Progress:

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.
• Dec 6th 2008, 02:31 PM
Opalg
Quote:

Originally Posted by GenoaTopologist
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.

More accurately, those sets form a base for the product topology. It is not true that every open set is of that form. For a set U to be open in the product topology, it is necessary that every point of U should contain a basic neighbourhood that is contained in U.

For example. 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 think that illustrates the main techniques you will need to answer the rest of the question.
• Dec 6th 2008, 10:47 PM
GenoaTopologist
Thank you so much!
• Dec 7th 2008, 01:34 PM
xianghu21
(d) $\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|=5 \}$
(e) $\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|\leq5 \}$

So, now for parts (d.) and (e.) , these are tricky to see if these will be open, closed, both, or neither by looking at the complements and arguing similarly. Is (d.) closed? Because the complement of (d.) would be open (at least I am thinking). And then part (e.) is very similar to part (d.) but it looks like this set is open. So should (e.) be open? This is a hard question that we are struggling on. Thanks for your help. This is much more understandable now.
• Dec 8th 2008, 05:10 AM
Opalg
Quote:

Originally Posted by xianghu21
(d) $\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|=5 \}$
(e) $\{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|\leq5 \}$

So, now for parts (d.) and (e.) , these are tricky to see if these will be open, closed, both, or neither by looking at the complements and arguing similarly. Is (d.) closed? Because the complement of (d.) would be open (at least I am thinking). And then part (e.) is very similar to part (d.) but it looks like this set is open. So should (e.) be open?

Let $D = \{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|=5 \}$, $E = \{f \in \prod_{n \in \omega} X_n | \text{ }|\{ n \in \omega | f(n)=0 \}|\leq5 \}$. If $f\notin E$ then there must be (at least) six values of n for which f(n)=1. The set of all elements of the product space that take the value 1 at those six points is an open neighbourhood of f that is disjoint from E. So E is closed.

But the set D is different. For each integer m>4, define $f_m$ by $f_m(1) = f_m(2) = f_m(3) = f_m(4) = f_m(m) = 0$, and $f_m(n) = 1$ for all other values of n. Then $f_m\in D$, but $\lim_{m\to\infty}f_m = g$, where g(n) = 0 for n=1,2,3,4, but g(n) =1 for all other values of n. So $g\notin D$. That shows that D is not closed.

I'll leave you to think about whether D and E are open or not.
• Dec 13th 2008, 03:19 PM
GaloisTheory1
So, now how does one show that $D$ and $E$ are open or not open? I never had a course in graduate level (or upper level) topology/set theory, but I think this is a very interesting problem.