Here's Prob. 5, Sec. 20 in the book Topology by James R. Munkres, 2nd edition:
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.
My effort:
Let $x = \left( x_1, x_2, x_3, \ldots \right)$ be an element of the closure of $\mathbb{R}^\infty$ in the uniform metric topology on $\mathbb{R}^\omega$. Then, for any real number $\varepsilon \in (0, 1)$, we can find a point $y = \left( y_1, y_2, y_3, \ldots \right)$ in $\mathbb{R}^\infty$ such that $\tilde{\rho}(x,y) < \varepsilon$, where
$$\tilde{\rho}(x,y) = \sup \left\{ \ \min \left\{ \ \left\vert x_n – y_n \right\vert, \ 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\}.$$
So, for each $n \in \mathbb{N}$, we have $\left\vert x_n – y_n \right\vert < \varepsilon$.Now as $ y \in \mathbb{R}^\infty$, so there exists a natural number $N$ such that $y_n = 0$ for all $n > N$. So we can conclude that $\left\vert x_n \right\vert < \varepsilon$ for all $n > N$, form which it follows that the sequence $x$ converges to the real number $0$.
Conversely, if $x = \left( x_1, x_2, x_3, \ldots \right)$ is a sequence of real numbers converging to $0$, then, for any given real number $\varepsilon \in (0, 1)$, we can find a natural number $N$ such that $\left\vert x_n \right\vert < \frac{\varepsilon}{2}$ for all $n > N$.
Now let $y = \left( x_1, \ldots, x_N, 0, 0, \ldots \right)$. Then clearly $y \in \mathbb{R}^\infty$ and $\tilde{\rho}(x,y) \leq \frac{\varepsilon}{2}$, thus showing that $x$ is in the closure of $\mathbb{R}^\infty$.
Thus, the closure of $\mathbb{R}^\infty$ in the uniform topology on $\mathbb{R}^\omega$ equals the set $c_0$ of all the sequences of real numbers which converge to $0$, in the standard metric on $\mathbb{R}$.
Am I right?
Best Answer
Looks great! The only improvement I can see is that in the converse direction, you might as well just use $\varepsilon$ instead of $\frac{\varepsilon}{2}$.