Let \(\bar{\rho}\) be the uniform metric on \(\mathbb{R}^\omega\). Given \(\mathbf{x} = (x_1, x_2, \ldots) \in \mathbb{R}^\omega\) and given \(0 < \epsilon < 1\), let
\[ U(\mathbf{x}, \epsilon) = (x_1 - \epsilon, x_1 + \epsilon) \times \dots \times (x_n - \epsilon, x_n + \epsilon) \times \dots. \]
Show that \(U(\mathbf{x}, \epsilon)\) is not equal to the \(\epsilon\)-ball \(B_{\bar{\rho}}(\mathbf{x}, \epsilon)\).
Show that \(U(\mathbf{x}, \epsilon)\) is not even open in the uniform topology.
Show that
\[ B_{\bar{\rho}}(\mathbf{x}, \epsilon) = \bigcup_{\delta < \epsilon} U(\mathbf{x}, \delta). \]
Solution
The uniform metric is \[ \bar\rho(\mathbf x,\mathbf y)=\sup_i\min\{|x_i-y_i|,1\}, \] and throughout \(0<\varepsilon<1\).
Define \[ y_i=x_i+\varepsilon\left(1-\frac1{i+1}\right). \] Then \(|y_i-x_i|<\varepsilon\) for every \(i\), so \(\mathbf y\in U(\mathbf x,\varepsilon)\). But \[ \bar\rho(\mathbf x,\mathbf y)=\sup_i|x_i-y_i|=\varepsilon, \] so \(\mathbf y\notin B_{\bar\rho}(\mathbf x,\varepsilon)\). Thus the two sets are not equal.
Use the same \(\mathbf y\in U(\mathbf x,\varepsilon)\). Given any \(r>0\), choose \(i\) so large that \[ \varepsilon-|y_i-x_i|<r/2. \] Alter only the \(i\)th coordinate, increasing it by some \(s<r\) large enough that the new coordinate lies outside \((x_i-\varepsilon,x_i+\varepsilon)\). The resulting point \(\mathbf z\) satisfies \[ \bar\rho(\mathbf y,\mathbf z)<r \] but \(\mathbf z\notin U(\mathbf x,\varepsilon)\). Hence no uniform ball about \(\mathbf y\) lies inside \(U(\mathbf x,\varepsilon)\), so this set is not uniform-open.
If \(\mathbf y\in U(\mathbf x,\delta)\) for some \(\delta<\varepsilon\), then \[ \bar\rho(\mathbf x,\mathbf y)\le\delta<\varepsilon, \] so \(\mathbf y\) lies in the \(\varepsilon\)-ball. Conversely, if \[ s=\bar\rho(\mathbf x,\mathbf y)<\varepsilon, \] choose \(\delta\) with \(s<\delta<\varepsilon\). Since \(\delta<1\), the inequality defining \(s\) implies \[ |x_i-y_i|<\delta \] for every \(i\). Thus \(\mathbf y\in U(\mathbf x,\delta)\). Therefore \[ B_{\bar\rho}(\mathbf x,\varepsilon)=\bigcup_{\delta<\varepsilon}U(\mathbf x,\delta). \]