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 topologies? Justify your answer.
Solution
1. In the product topology, \(\overline{\RR^\infty}=\boxed{\RR^\omega}\).
A nonempty basic open set is \(\prod_iU_i\) with \(U_i\) open and nonempty, and \(U_i=\RR\) except for \(i\) in a finite set \(F\). Choose \(z_i\in U_i\) for \(i\in F\) and \(z_i=0\) for \(i\notin F\). Then \(z=(z_i)\) is eventually zero and lies in \(\prod_iU_i\), so every nonempty open set meets \(\RR^\infty\).
2. In the box topology, \(\overline{\RR^\infty}=\boxed{\RR^\infty}\).
Let \(x\notin\RR^\infty\), so the set \(S=\{i:x_i\ne0\}\) is infinite. Put \(U_i=\RR-\{0\}\) for \(i\in S\) and \(U_i=\RR\) for \(i\notin S\). Then \(\prod_iU_i\) is a box neighborhood of \(x\). An eventually-zero sequence \(z\) has \(z_i=0\) for some \(i\in S\), because \(S\) is infinite, so \(z\notin\prod_iU_i\). Hence the complement of \(\RR^\infty\) is box-open.
3. Q.E.D.
Steps 1 and 2 give the closure in the two topologies.