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 December 6th, 2008, 11:09 AM #1 Newbie   Joined: Dec 2008 Posts: 4 Thanks: 0 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 $\omega=\mathbb{N} \cup \{ 0 \}$ 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 $C= \{f \in \prod_{n \in \omega} X_n | \latext{ }\forall n \in \omega \latext{ }f(n) = 0 \Rightarrow f(n + 1) = 1 \}$. 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. [color=red]I still do not know how to do parts (d.) and (e.)[/color]

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