I have a question about the proof I am looking at. The question is as following:
Let $\mathbf{w}_i=(0, 0, \cdots 0, i, i, i, \cdots)\in \mathbb{R}^\omega$(where there are $i-1$ zeros). Does this sequence converge in uniform topology?
and a part of the proof goes like this:
... For the uniform topology, note that, given $i ≥ 1$, $\bar{\rho}(\mathbf{w}_i, \mathbf{0})>1$, so $\mathbf{w}_i \notin B_{\bar{\rho}}(\mathbf{0}, 1)$ for all $i$... and so the sequence does not converge in uniform topology of $\mathbb{R}^\omega$.
My question is, how can $\bar{\rho}(\mathbf{w}_i, \mathbf{0})>1$ happen? The uniform metric I learned is defined as $$\bar{\rho}(\mathbf{x, y})=\sup\{\bar{d}(x_\alpha, y_\alpha) |\alpha \in J\}$$ where $J$ is the index set(in this case, $\mathbb{N}$), and $\bar{d}$ is the standard bounded metric so it is always less than or equal to $1$. Then, the supremum also must be less than or equal to $1$. So, $\bar{\rho}(\mathbf{a, b})\leq 1$ for any two points. Where am I wrong?