Determine whether or not \(\mathbb{R}^\omega\) is connected in the uniform topology.
Solution
The uniform metric is \(\bar\rho(\mathbf x,\mathbf y)=\sup_n\min\{\abs{x_n-y_n},1\}\). Let \(\mathcal B\) be the set of bounded sequences.
1. If \(\bar\rho(\mathbf x,\mathbf y)<\frac12\), then \(\mathbf x\) is bounded if and only if \(\mathbf y\) is bounded.
\(\bar\rho(\mathbf x,\mathbf y)<\frac12\) forces \(\abs{x_n-y_n}<\frac12\) for every \(n\), so \(\sup_n\abs{y_n}\le\sup_n\abs{x_n}+\frac12\) and symmetrically.
2. \(\mathcal B\) is open and closed in the uniform topology, and \(\mathcal B\ne\varnothing,\RR^\omega\).
By step 1, the ball \(B_{\bar\rho}(\mathbf x,\frac12)\) lies in \(\mathcal B\) when \(\mathbf x\in\mathcal B\) and in \(\RR^\omega-\mathcal B\) when \(\mathbf x\notin\mathcal B\), so both sets are open. \(\mathbf 0\in\mathcal B\) and \((1,2,3,\ldots)\notin\mathcal B\).
3. Q.E.D.
By step 2, \(\mathcal B\) and its complement separate \(\RR^\omega\), so \(\RR^\omega\) is \(\boxed{\text{not connected}}\) in the uniform topology.