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Here is Prob. 7, Sec. 19, 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," that is, $\left( x_1, x_2, \ldots \right)$ such that $x_i \neq 0$ for only finitely many values of $i$. What is the closure of $\mathbb{R}^\infty$ in $\mathbb{R}^\omega$ in the box and product topologies? Justify your answer.

My Attempt:

First, we note that, for any real number $\alpha \neq 0$, we can find an open interval $( \beta, \gamma)$ such that $\alpha \in (\beta, \gamma)$ and such that $0 \not\in (\beta, \gamma)$: in fact we can take $$ ( \beta, \gamma) \colon= \begin{cases} \left( \frac{\alpha}{2}, 2 \alpha \right) \ & \mbox{ if } \ \alpha > 0, \\ \left( 2 \alpha, \frac{\alpha}{2} \right) \ & \mbox{ if } \ \alpha < 0. \end{cases} \tag{0} $$

For the box topology:

Let $\mathbf{a} \colon= \left( \alpha_1, \alpha_2, \ldots \right)$ be an arbitrary element of $\mathbb{R}^\omega \setminus \mathbb{R}^\infty$. Then there are infinitely many natural numbers $n$ for which $\alpha_n \neq 0$, and for each one of these $n$ we can find an open interval $\left( \beta_n, \gamma_n \right)$ containing $\alpha_n$ and not containing $0$, as in (0) above.

Let $ \left( n_1, n_2, \ldots \right)$ be the strictly increasing sequence of all those natural numbers $n$ for which $\alpha_n \neq 0$. Let us consider the box topology basis element $$ B \colon= \prod_{n \in \mathbb{N} } \left( a_n, b_n \right) = \left( a_1, b_1 \right) \times \left( a_2, b_2 \right) \times \cdots, $$ where $$ \left( a_n, b_n \right) \colon= \begin{cases} \left( \beta_n, \gamma_n \right) \ & \mbox{ if } \ n = n_k \mbox{ for some } k \in \mathbb{N}, \\ (-1, 1) \ & \mbox{ otherwise}. \end{cases} \tag{1} $$ Then $$\mathbf{a} \in B \subset \mathbb{R}^\omega \setminus \mathbb{R}^\infty, \tag{2}$$ for if $\mathbf{x} \colon= \left( x_1, x_2, \ldots \right) \in B$, then $x_{n_k} \in \left( \beta_{n_k}, \gamma_{n_k} \right)$, which implies that $x_{n_k} \neq 0$ for all $k \in \mathbb{N}$; that is, $x_n \neq 0$ for infinitely many natural numbers $n$, and so $\mathbf{x} \in \mathbb{R}^\omega \setminus \mathbb{R}^\infty$.

Thus we have shown that for each point $\mathbf{a} \in \mathbb{R}^\omega \setminus \mathbb{R}^\infty$, we can find a box topology basis element $B$ such that (2) holds. Therefore $\mathbb{R}^\omega \setminus \mathbb{R}^\infty$ is open, and so $\mathbb{R}^\infty$ is closed in $\mathbb{R}^\omega$ in the box topology, which implies that $$ \overline{\mathbb{R}^\infty} = \mathbb{R}^\infty. $$

Now for the product topology:

Let $\mathbf{x} \colon= \left( x_1, x_2, \ldots \right)$ be any element of $\mathbb{R}^\omega$, and let $U \colon= \prod_{n \in \mathbb{N} } U_n$ be a product topology basis element containing $\mathbf{x}$. Then $U_n$ is open in $\mathbb{R}$ for each $n \in \mathbb{N}$, and $U_n \neq \mathbb{R}$ for at most finitely many $n \in \mathbb{N}$; let $n_1, \ldots, n_k$ be all the natural numbers $n$ for which $U_n \neq \mathbb{R}$, and let $$N \colon= \max \left\{ \ n_1, \ldots, n_k \ \right\}. $$ Then $U_n = \mathbb{R}$ for every natural number $n > N$. Now let $\mathbf{x}^\prime \colon= \left( x_1^\prime, x_2^\prime, \ldots \right)$ such that, for each $n \in \mathbb{N}$, $$ x_n^\prime \colon= \begin{cases} x_n \ & \mbox{ if } n \leq N, \\ 0 \ & \mbox{ if } n > N. \end{cases} $$ Then obviously this $\mathbf{x}^\prime \in U \cap \mathbb{R}^\infty$ so that $U \cap \mathbb{R}^\infty$ is non-empty.

Thus we have shown that every product topology basis element $U$ in $\mathbb{R}^\omega$ containing $\mathbf{x}$ intersects $\mathbb{R}^\infty$. So $\mathbf{x} \in \overline{\mathbb{R}^\infty}$.

But $\mathbf{x}$ was an arbitrary element of $\mathbb{R}^\omega$. Thus we can conclude that every element of $\mathbb{R}^\omega$ is in the closure of $\mathbb{R}^\infty$. That is, $$ \overline{ \mathbb{R}^\infty } = \mathbb{R}^\omega. $$

Is my logic correct? If so, then have I also managed to obtain the correct results? I don't see where I have erred!

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    $\begingroup$ Why for a > 0, is a and 0 in (a,2a)? How about (-a,2a)? $\endgroup$ Sep 24, 2017 at 2:52
  • $\begingroup$ @WilliamElliot thank you so much for pointing out the issue. I've edited my post to remove it. See how you like it now. $\endgroup$ Sep 24, 2017 at 4:59
  • $\begingroup$ I haven't checked your work..... A suggestion: For $x=(x_n)_n\in \mathbb R^{\omega}$ let $f_x(n)=1$ if $x_n=0$ and let $f_x(n)=|x_n|/2$ if $x_n\ne 0.$ ...... If $x \not \in \Bbb R^{\infty}$ then $V(x)=\{ (y_n)_n\in \Bbb R^{\omega} : \forall n\; (|x_n-y_n|<f_x(n)\;)\}$ is a nbhd of $x$ in the box topology which is disjoint from $\Bbb R^{\infty}$ because if $(y_n)_n\in V(x)$ then $\forall n\; (x_n\ne 0\implies y_n \ne 0)$ $\endgroup$ Sep 24, 2017 at 6:01
  • $\begingroup$ @DanielWainfleet I would be grateful if you could take time reviewing my proof. $\endgroup$ Sep 24, 2017 at 6:30
  • $\begingroup$ Your work is correct but overly complicated.Let $B=\prod_nU_n$ be a member of the (canonical) base for the (Tychonoff) product topology. Each $U_n$ is a non-empty open subset of $\Bbb R,$ and the set $S=\{n: U_n\ne \Bbb R\}$ is finite. Take $y_n\in U_n$ such that $(U_n=\Bbb R\implies y_n=0).$ Then $(y_n)_n\in B,$ and the set $\{n: y_n\ne 0\}$ is a subset of the finite set $S,$ so $(y_n)_n\in \Bbb R^{\infty}.$... So $\Bbb R^{\infty}$ is a dense set because it has non-empty intersection with every member of the base. $\endgroup$ Sep 24, 2017 at 6:45

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Why use $(\beta, \gamma)$? Just note that $E_0 = \mathbb{R}\setminus\{0\}$ is open $\mathbb{R}$. Then if $x \in \mathbb{R}^\omega$ and $x \notin \mathbb{R}^\infty$, then the set $N_x := \{ n \in \mathbb{N}: x_n \neq 0\}$ is indeed infinite, and then define the box open set $U = \prod U_n$ where $U_n = E_0$ for $n \in N_x$, $U_n = \mathbb{R}$ for $n \notin N_x$, contains $x$ and if $y \in U$ then $N_y \subseteq N_x$, so that $U \subseteq \mathbb{R}^\omega\setminus \mathbb{R}^\infty$. So indeed $\mathbb{R}^\infty$ is closed.

The denseness of $\mathbb{R}^\infty$ in the product topology I have no comment on.

Essentially the proofs are fine.

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