# Closure in box and product topology

Let $\mathbb R^\infty$ be the subset of $\mathbb R^\omega$ consisting of all sequences that are eventually zero that is all sequences $(x_1,x_2,\ldots)$ 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 topology?

How do you even start to think about this problem? It seems like it is one of the problems from Munkres (but not sure).

• Hint: Look at $x = (1,1,1,\ldots)$; and take a basic open set around it. Does this basic open set intersect $\mathbb{R}^{\infty}$ in the two topologies you are looking at? – Prahlad Vaidyanathan Oct 13 '13 at 4:46
• @PrahladVaidyanathan: I am having difficulty in grasping the concept of open sets around sequences? Could you please elaborate on that more? Right now, I don't know what your hint means – Rutherford Mark Oct 13 '13 at 4:52

1. The product topology: A basic open set of $\mathbb{R}^{\omega}$ is of the form $$U = U_1\times U_2\times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots$$ where $U_i$ are open sets in $\mathbb{R}$. Now take any $x\in \mathbb{R}^{\omega}$, and any basic open set $U$ as above containing $x$. Let $$y = (x_1, x_2, \ldots, x_n, 0,0,0,\ldots)$$ Now note that $y \in U$ (since $U$ only really cares about the first $n$ components). Also, $y \in \mathbb{R}^{\infty}$.

Hence $\mathbb{R}^{\infty}$ is dense in $\mathbb{R}^{\omega}$ in the product topology.
2. The box topology : Every basic open set is of the form $$W = W_1\times W_2\times \cdots \times W_n\times W_{n+1}\times \cdots$$ Now take any $x \notin \mathbb{R}^{\infty}$; then $x_n \neq 0$ for infinitely many $n$. In particular, if $$W_n = \begin{cases} (x_n - |x_n|/2, x_n + |x_n|/2) \quad & \text{if x_n\neq 0} \\ (-1,1) &\text{if x_n = 0} \end{cases}$$ then if $W = \prod W_i$ as above, then $x\in W$ and $$W \cap \mathbb{R}^{\infty} = \emptyset$$ Do you see why?

Hence $\mathbb{R}^{\infty}$ is closed in the box topology.
• Actually, $x_n$ can be $0$ for some $n$ when $x \notin \mathbb R^\infty$. – dfeuer Oct 13 '13 at 5:06
• @dfeuer : Thanks, edited. – Prahlad Vaidyanathan Oct 13 '13 at 5:22
• You can even use $\Bbb R$ instead of $(-1,1)$. – dfeuer Oct 13 '13 at 5:23
• @Prahlad Nice explanation.Thanks – Vishesh Oct 13 '13 at 5:33
• VERY NICE SIR..BTW I KNOW U PERSONALLY@PrahladVaidyanathan – sani Sep 18 '17 at 14:52

In the box topology: Let $(x_n) \notin \Bbb R^\infty$. Can you find an open set about $(x_n)$ that doesn't contain any element of $\Bbb R^\infty$? Think about the values of $n$ for which $x_n=0$ and the ones for which $x_n\ne 0$—you will have to handle these separately.

In the product topology: Let $(x_n) \in \Bbb R^\omega$. Let $U$ be a basic open set about $(x_n)$. What sorts of things are in $U$?

• But we don't know when is $x_n=1$ or $x_n=0$ So how can we proceed further? They can take on those values any time as long as it is finitely many times – Rutherford Mark Oct 13 '13 at 4:54
• Sorry, I wrote $=1$ when I meant $\ne 0$. Think about building a basic open set around the value not in $\Bbb R^\infty$ that won't intersect $\Bbb R^\infty$, and see if you're able to do it. – dfeuer Oct 13 '13 at 4:59

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, $$\overline{ \mathbb{R}^\infty } = \mathbb{R}^\omega.$$