Let $C = \{(x_n) \in \mathbb{R}^\omega ~|~ \lim x_n = 0 \}$. I am trying to show that this set is closed in the uniform topology, which is the topology generated by the metric $\rho (x,y) = \sup d(x_i,y_i)$, where $d(x_i,y_i) = \min \{|x_i-y_i|,1\}$, the standard bounded metric on $\mathbb{R}$. Here is a partial attempt:
Let $(x_n) \in \overline{C}-C$. Then $\lim x_n = \ell \neq 0$, and without loss of generality suppose $\ell > 0$. Since $x_n- \ell < x_n < x_n + \ell$ or $x_n \in (x_n - \ell, x_n + \ell)$ for every $n$, clearly the set $\prod (x_n - \ell, x_n + \ell)$ contains $(x_n)$. Moreover, if $(z_n) \in \prod (x_n - \ell, x_n + \ell)$, then $x_n- \ell < z_n < x_n + \ell$ and by the squeeze theorem $0 < \lim z_n < 2 \ell$. Now, if this is open in the uniform topology, then we have an open neighborhood of $(x_n)$ that doesn't intersect $C$, giving us the desired contradiction.
At this point, I mistakenly thought that this was a basis element of the product topology, and therefore open in the uniform topology This is obviously a fallacious inference, but it still could be open. My question is, is this in fact open in the uniform topology? If so, I could use some hints on how to show this. If not, I could still some some hints!
Note the definition of closure I am working with is as follows: the closure of $A$ is defined as the intersection of all closed sets containing $A$. I do have the following characterization though: $x \in \overline{A}$ if and only if every open neighborhood of $x$ intersects $A$. I do not have any notion of closure in terms of convergent sequences.