[Math] Closure in box and product topology

general-topology

Let $\mathbb R^\infty$ be the subset of $\mathbb R^\omega$ consisting of all sequences that are eventually zero that is all sequences $(x_1,x_2,\ldots)$ such that $x_i\neq 0$ for only finitely many values of $i$. What is the closure of $\mathbb R^\infty$ in $\mathbb R^\omega$ in the box and product topology?

How do you even start to think about this problem? It seems like it is one of the problems from Munkres (but not sure).

Best Answer

  1. The product topology: A basic open set of $\mathbb{R}^{\omega}$ is of the form $$ U = U_1\times U_2\times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots $$ where $U_i$ are open sets in $\mathbb{R}$. Now take any $x\in \mathbb{R}^{\omega}$, and any basic open set $U$ as above containing $x$. Let $$ y = (x_1, x_2, \ldots, x_n, 0,0,0,\ldots) $$ Now note that $y \in U$ (since $U$ only really cares about the first $n$ components). Also, $y \in \mathbb{R}^{\infty}$.

    Hence $\mathbb{R}^{\infty}$ is dense in $\mathbb{R}^{\omega}$ in the product topology.
  2. The box topology : Every basic open set is of the form $$ W = W_1\times W_2\times \cdots \times W_n\times W_{n+1}\times \cdots $$ Now take any $x \notin \mathbb{R}^{\infty}$; then $x_n \neq 0$ for infinitely many $n$. In particular, if $$ W_n = \begin{cases} (x_n - |x_n|/2, x_n + |x_n|/2) \quad & \text{if $x_n\neq 0$} \\ (-1,1) &\text{if $x_n = 0$} \end{cases} $$ then if $W = \prod W_i$ as above, then $x\in W$ and $$ W \cap \mathbb{R}^{\infty} = \emptyset $$ Do you see why?

    Hence $\mathbb{R}^{\infty}$ is closed in the box topology.
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