Exercise 20.5

Question

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.

Amit Awasthi · Accepted Answer

\begin{proof}[Solution]
  We claim that $A = \overline{\mathbb{R}^\infty}$ is the set of all sequences that converge to zero; we denote this latter set by $X$. It suffices to show by Theorem $17.5$ that $x \in X$ if and only if every basis element $U 
i x$ intersects $A$. First suppose $x \in X$ and let $U 
i x$ be a basis element in the uniform topology; we then see that we can find an open ball $B(x,\epsilon) \subseteq U$. We know we can find $N \in \mathbb{N}$ such that $|x_n| < \epsilon$ for all $n \ge N$ by the definition of convergence. Then, define $y$ such that $y_n = x_n$ for all $n < N$, and zero otherwise; this means $y \in A$. Then, $\overline{ho}(x,y) < \epsilon$, and so $y \in B(x,\epsilon) \cap A$.
  \par Now suppose $x 
otin X$; it suffices to find a basis element containing $x$ that does not intersect $A$. Since $x 
otin X$, there exists a ball $B(0,\epsilon) \subseteq \mathbb{R}$ such that $\{x_n\}_{n \ge N} 
ot\subseteq B(0,\epsilon)$ for any $N$. The ball $B(x,\epsilon/2)$, then, does not intersect $A$, since for any $y \in B(x,\epsilon/2)$, it is not the case that $y_n = 0$ for all $n \ge N$ for some $N$.
\end{proof}

Dan Whitman · Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} ewcommand{\vect}[1]{\mathbf{#1}} \def\va{\vect{a}} \def\vb{\vect{b}} ewcommand\parens[1]{\left( #1 ight)} ewcommand\seqfin[2]{\parens{#1_1, \ldots, #1_{#2}}} ewcommand\seqinf[1]{\parens{#1_1, #1_2, \ldots}} \def\br{\bar{ }} \def eals{{\mathbb{R}}} \def inf{ eals^\infty} Let ${\mathbb{R}}^0$ denote the subset of ${\mathbb{R}}^{\omega}$ consisting of all sequences that converge to zero. We then claim that $\overline{ inf} = {\mathbb{R}}^0$, i.e. the closure of $ inf$ is ${\mathbb{R}}^0$. \begin{proof} $(\subset)$ We show this by contrapositive. So suppose that $\mathbf{x} otin {\mathbb{R}}^0$ so that the sequence $\mathbf{x} $ does \emph{not} converge to zero. Then there is a neighborhood $V$ of $0$ in the standard topology on ${\mathbb{R}}$ such that, for any $N \in {{\mathbb{Z}}_+}$, there is an $n \geq N$ where $x_n otin V$. It also follows from \href{./exercise_20-4}{Lemma~20.4.1} that there is an $\e > 0$ such that $B_d(0, \e) \subset V$. So let $\d = \min\left\{\e, 1 ight\}$ and consider the set $B_{\br}(\mathbf{x} , \d)$, which is clearly a neighborhood of $\mathbf{x} $ in the uniform topology. Also suppose that $\mathbf{y} $ is any element of $ inf$ so that $\mathbf{y} $ is eventually zero. Then there must be an $N$ where $y_n = 0$ for all $n \geq N$. From before we have that there is a specific $n \geq N$ where $x_n otin V$ so that also $x_n otin B_d(0, \e)$, and thus \begin{gather*} d(y_n, x_n) = d(0, x_n) = d(x_n, 0) \geq \e \geq \d \,. \end{gather*} Then we have that both $d(y_n, x_n) \geq \d$ and $1 \geq \d$ so that \begin{gather*} \bar{d}(y_n, x_n) = \min\left\{d(y_n, x_n), 1 ight\} \geq \d \,. \end{gather*} From this it clearly follows that \begin{gather*} \br(\mathbf{y} , \mathbf{x} ) = \sup\left\{\bar{d}(y_n, x_n) \mid n \in {{\mathbb{Z}}_+} ight\} \geq \d \end{gather*} so that $\mathbf{y} otin B_{\br}(\mathbf{x} , \d)$. Since $\mathbf{y} $ was arbitrary, this shows that $B_{\br}(\mathbf{x} , \d)$ does not intersect $ inf$. This in turn shows that $\mathbf{x} $ is not a limit point of $ inf$. Now, clearly also $\mathbf{x} $ cannot be eventually zero since then it would converge to zero, hence $\mathbf{x} otin inf$ either. Therefore $\mathbf{x} $ cannot be in the closure of $ inf$. Hence by the contrapositive, we have that $\overline{ inf} \subset {\mathbb{R}}^0$. $(\supset)$ Now consider any $\mathbf{x} \in {\mathbb{R}}^0$ so that the sequence $\mathbf{x} $ converges to zero. Consider any neighborhood $U$ of $\mathbf{x} $ in the uniform topology so that, by \href{./exercise_20-4}{Lemma~20.4.1}, there is an $\e > 0$ such that $B_{\br}(\mathbf{x} , \e) \subset U$. Now, since $B_d(0, \e/2)$ is a neighborhood of $0$, it follows that there is an $N \in {{\mathbb{Z}}_+}$ where $x_n \in B_d(0, \e/2)$ for all $n \geq N$ since $\mathbf{x} $ converges to $0$. Now define the sequence $\mathbf{y} $ where \begin{gather*} y_n = \begin{cases} x_n & n < N \ 0 & n \geq N \end{cases} \end{gather*} for $n \in {{\mathbb{Z}}_+}$. Clearly $\mathbf{y} $ is eventually zero so that $\mathbf{y} \in inf$. We also claim that $\mathbf{y} \in U$. To see this consider any $n \in {{\mathbb{Z}}_+}$. If $n < N$ then clearly $y_n = x_n$ so that \begin{gather*} \bar{d}(y_n, x_n) \leq d(y_n, x_n) = d(x_n, x_n) = 0 < \e/2 \,. \end{gather*} If $n \geq N$ then $y_n = 0$ and we have from before that $x_n \in B_d(0, \e/2)$ and hence \begin{gather*} \bar{d}(y_n, x_n) \leq d(y_n, x_n) = d(0, x_n) = d(x_n, 0) < \e/2 \,. \end{gather*} Hence it follows that \begin{gather*} \br(\mathbf{y} , \mathbf{x} ) = \sup\left\{\bar{d}(y_n, x_n) \mid n \in {{\mathbb{Z}}_+} ight\} \leq \e/2 < \e \end{gather*} so that $\mathbf{y} \in B_{\br}(\mathbf{x} , \e) \subset U$. This shows that $\mathbf{y} \in inf \cap U$. If $\mathbf{y} = \mathbf{x} $ then of course $\mathbf{x} = \mathbf{y} \in inf$ itself. If $\mathbf{y} eq \mathbf{x} $ then we have shown that $\mathbf{x} $ is a limit point of $ inf$ since $U$ was an arbitrary neighborhood. Thus either way $\mathbf{y} \in \overline{ inf}$ so that $\overline{ inf} \supset {\mathbb{R}}^0$ since $\mathbf{x} $ was arbitrary. \end{proof} Lastly, we note that $ inf$ is a proper subset of its closure $\overline{ inf} = {\mathbb{R}}^0$ since, for example, the sequence $\mathbf{x} $ defined by $x_n = 1/n$ for all $n \in {{\mathbb{Z}}_+}$ clearly converges to zero so is in ${\mathbb{R}}^0$ but is not eventually zero so is not in $ inf$.

Exercise 20.5

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Exercise 20.5

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