Definition: Let $(\Omega, \tau)$ be a topological space. The indicator function $1_A: \Omega \to \mathbb{R}$ on a subset $A \subset \Omega$ is defined by $1_A(x) = 1$ if $x \in A$ and $1_A(x) = 0$ if $x \notin A$.
Let $\Omega = \mathbb{R}^{\mathbb{R}}$ and $M = \{1_A: A \subset \mathbb{R} \ \vert \ \text{$A$ is finite}\}$. Show that $1_\mathbb{R} \in \overline{M}$ and prove that there exists no sequence $\{1_{A_n}\}_{n \in \mathbb{N}} \subset M$ such that $1_{A_n} \to \mathbb{1_\mathbb{R}}$. Here we are using the product topology.
We know that if $U = \displaystyle{\prod_{\alpha \in \mathbb{R}} U_\alpha} $ is a basic open set of $\mathbb{R}^\mathbb{R}$ in the product topology containing $1_\mathbb{R}$, then $U_\alpha = \mathbb{R}$ for all but finitely many $\alpha$. Let $F \subset \mathbb{R}$ be the set such that $U_\alpha \neq \mathbb{R} \ \forall \alpha \in F$. Since $1_\mathbb{R} \in U$, then $1 \in U_{\alpha_i} \ \forall \alpha_i \in F$. Then $1_F$ witnesses $U \cap M \neq \emptyset$, therefore $1_{\mathbb{R}} \in \overline{M}$.
The second part is where I got stuck. My only ideas were:
-- construct an open set that contains $1_{\mathbb{R}}$ but doesn't contain any tail of any sequence $\{1_{A_n}\}_{n \in \mathbb{N}}$
-- somehow use the fact that $1_\mathbb{R}$ is a limit point of $M$?
But I haven't had any success with either of those. Can anyone help?
All you need for this is the fact that convergence in $\mathbb R^{\mathbb R}$ implies convergence of each coordinate. $\cup_n A_n$ is a countable subset of $\mathbb R$ so there is point $x$ in $\mathbb R\setminus \cup_n A_n$. If $I_{A_n} \to I_{\mathbb R}$ then Then $I_{A_n}(x) \to 0 \neq I_{\mathbb R} (x)$.