Exercise 19.7

Question

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}}

ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}}

Let ${\mathbb{R}}^\infty$ be the subset of ${\mathbb{R}}^{\omega}$ consisting of all sequences that are ``eventually zero,'' that is all sequences $\seqinf{x}$ such that $x_i 
eq 0$ 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.

Dan Whitman · Accepted Answer

ewcommand\parens[1]{\left( #1 ight)} ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}} ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}} First, we claim that ${\mathbb{R}}^\infty$ is dense in ${\mathbb{R}}^{\omega}$ in the product topology in the sense that its closure is all of ${\mathbb{R}}^{\omega}$. \begin{proof} We show that any point of ${\mathbb{R}}^{\omega}$ is in $\overline{{\mathbb{R}}^{\omega}}$. So consider any point $x = \seqinf{x} \in {\mathbb{R}}^{\omega}$ and any neighborhood $U$ of $x$. Then there is a basis element $B = \prod U_n$ containing $x$ where $B \subset U$. By the definition of the product topology each $U_n$ is open and $U_n = {\mathbb{R}}$ for all but finitely many values of $n$. So let $I$ be a finite subset of ${{\mathbb{Z}}_+}$ such that $U_n = {\mathbb{R}}$ for all $n otin I$ and $U_n$ is merely just open for $n \in I$. Consider now the sequence $y = \seqinf{y}$ defined by \begin{gather*} y_n = \begin{cases} x_n & n \in I \ 0 & n otin I \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$. Since $I$ is finite clearly $y \in {\mathbb{R}}^\infty$. Also $y_n = x_n \in U_n$ when $n \in I$ since $B = \prod U_n $ contains $x$. We also have $y_n = 0 \in {\mathbb{R}} = U_n$ when $n otin I$ so that either way $y_n \in U_n$ and hence $y \in \prod U_n = B$. Thus also $y \in U$ since $B \subset U$. Since $U$ was an arbitrary neighborhood and $U$ intersects ${\mathbb{R}}^\infty$ (with $y$ being a point in the intersection), this shows that $x \in \overline{{\mathbb{R}}^\infty}$ by Theorem~17.5. This of course shows the desired result since $x$ was any element of ${\mathbb{R}}^{\omega}$. \end{proof} For the box topology, we claim that ${\mathbb{R}}^\infty$ is already closed. \begin{proof} We show this by showing that any point not in ${\mathbb{R}}^\infty$ is not a limit point of ${\mathbb{R}}^\infty$ so that ${\mathbb{R}}^\infty$ must already contain all its limit points. So consider any $x = \seqinf{x} otin {\mathbb{R}}^\infty$ so that $x_n eq 0$ for infinitely many values of $n$. Now define the sets \begin{gather*} U_n = \begin{cases} (-1, 1) & x_n = 0 \ (x_n/2, 2x_n) & x_n > 0 \ (2x_n, x_n/2) & x_n < 0 \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$. Clearly, each $U_n$ is a basis element of ${\mathbb{R}}$ and is therefore open. Also clearly each $x_n \in U_n$. It, therefore, follows that $B = \prod U_n$ is a basis element of ${\mathbb{R}}^{\omega}$ and is therefore open, and that $x \in B$. Hence $B$ is a neighborhood of $x$. Then, for any $y = \seqinf{y} \in B$ we have that each $y_n \in U_n$. For infinitely many $n \in {{\mathbb{Z}}_+}$ we then have that $x_n eq 0$ and hence $x_n > 0$ or $x_n < 0$. In the former case $y_n \in U_n = (x_n/2, 2x_n)$ so that $0 < x_n/2 < y_n$. In the latter case $y_n \in U_n = (2x_n, x_n/2)$ so that $y_n < x_n / 2 < 0$. Hence either way $y_n eq 0$ so that $y otin {\mathbb{R}}^\infty$ since this is true for infinitely many $n$. Since $y \in B$ was arbitrary, this shows that $B$ cannot not intersect ${\mathbb{R}}^\infty$. Therefore $x$ is not a limit point of ${\mathbb{R}}^\infty$ since $B$ is a neighborhood of $x$. \end{proof}

Exercise 19.7

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Exercise 19.7

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