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?

  • 6
    $\begingroup$ Looks okay to me. $\endgroup$ Nov 1 '16 at 15:21
  • $\begingroup$ @BrianM.Scott can we connect using either email or WhatsApp? Being in contact with someone seasoned like you will give me pleasure and inspire me to make an effort to do better in topology. $\endgroup$ Mar 17 '19 at 7:49

Looks great! The only improvement I can see is that in the converse direction, you might as well just use $\varepsilon$ instead of $\frac{\varepsilon}{2}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.