Let \(\mathbb{R}^\infty\) be the subset of \(\mathbb{R}^\omega\) consisting of all sequences that are eventually zero. What is the closure of \(\mathbb{R}^\infty\) in \(\mathbb{R}^\omega\) in the uniform topology? Justify your answer.
Solution
Let \(c_0\) denote the set of sequences converging to \(0\). We claim \[ \overline{\mathbb R^\infty}^{\,\bar\rho}=c_0. \]
First take \(x=(x_i)\in c_0\). Given \(0<\varepsilon<1\), choose \(N\) such that \[ |x_i|<\varepsilon\qquad(i>N). \] Let \[ y=(x_1,\dots,x_N,0,0,\dots)\in\mathbb R^\infty. \] Then \[ \bar\rho(x,y)=\sup_{i>N}|x_i|<\varepsilon, \] so every uniform neighborhood of \(x\) meets \(\mathbb R^\infty\).
Conversely, suppose \(x\) does not converge to \(0\). Then there is \(\varepsilon\in(0,1)\) such that \(|x_i|\ge\varepsilon\) for infinitely many \(i\). For any eventually-zero \(y\), choose such an \(i\) beyond the last nonzero coordinate of \(y\). Then \[ \bar\rho(x,y)\ge\min\{|x_i|,1\}\ge\varepsilon. \] Hence the uniform ball \(B_{\bar\rho}(x,\varepsilon)\) contains no eventually-zero sequence. Thus \(x\) is not in the closure.