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I have a question concerning Example 1.8.6 in the book Algebraische Topologie by Stöcker and Zieschang.

They consider $\mathbb{R}^\infty = \lim_n \mathbb{R}^n$ as the space of sequences of real numbers $(x_n)$ where $x_k = 0$ for all $k$ greater than some $N \in \mathbb{N}$. By their definition of the weak topology, a subset $X\subseteq \mathbb{R}^\infty$ is open iff all intersections $X \cap \mathbb{R}^n$ are open in the standard topology on $\mathbb{R}^n$.

In their example, they claim that for any sequence of positive real numbers $(\varepsilon_n)$, the set $U := \bigcup_{n=1}^\infty \{y\in \mathbb{R}^n \mid \lvert y \rvert < \varepsilon_n \}$ is a neighborhood of $(0,\dotsc)$.

What I fail to understand: How can we find an open subset of $U$ if we take $\varepsilon_n = n^{-1}$, or more generally, any sequence that converges to 0?

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3 Answers 3

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It is false. We have $$U = \bigcup_m B_m(\epsilon_m)$$ where $B_m(\varepsilon_m) = \{y\in \mathbb{R}^m \mid \lvert y \rvert < \varepsilon_m \}$ is the open ball in $\mathbb R^m$ with radius $\varepsilon_m$ and center $0$. Clearly $$U \cap \mathbb R^n = \bigcup_m B_m(\varepsilon_m) \cap \mathbb R^n .$$ For $m \ge n$ we have $B_m(\epsilon_m) \cap \mathbb R^n = B_n(\varepsilon_m)$ which is open in $\mathbb R^n$, but for $m < n$ the set $B^n_m(\varepsilon_m) = B_m(\varepsilon_m) \cap \mathbb R^n$ is contained in the $m$-dimensional linear subspace $\mathbb R^m \subset \mathbb R^n$ and thus does not contain any interior points. Therefore, if there exists $N$ such that $\sup_{m > N}\varepsilon_m < \varepsilon_N$, then $U$ is not open in $\mathbb R^\infty$. To see this oberserve that $$U \cap \mathbb R^{N+1} = \bigcup_{m \le N}B^{N+1}_m(\varepsilon_m) \cup \bigcup_{m > N} B_{N+1}(\varepsilon_m) .$$ The set $V = \bigcup_{m > N} B_{N+1}(\varepsilon_m)$ is the open ball with radius $\sup_{m > N}\varepsilon_m$. The set $W = \bigcup_{m \le N}B^{N+1}_m(\varepsilon_m)$ does not contain any interior points, but due to $\sup_{m > N}\varepsilon_m < \varepsilon_N$ it contains points outside of $V$ (surely all points $y \in B_{N+1}(\varepsilon_N)$ with $\lvert y \rvert > \sup_{m > N}\varepsilon_m$). Thus $W \cup V$ cannot be open in $\mathbb R^{N+1}$.

As an example take $\varepsilon_1 = 2, \varepsilon_m = 1$ for $m > 1$.

Remark:

We have shown that in general $U$ is not an open neigborhood of $0$. But could it be that it is always a not necessarily open neigborhood of $0$, i.e. contains an open neigborhood of $0$? To prove that it is impossible would require much more work.

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For each $n$, $U \cap \Bbb R^n$ is essentially the open ball with radius $\varepsilon_n$ and hence open in $\Bbb R^n$. I see no issues with the convergence to $0$ at all. The definition just applies.

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  • $\begingroup$ Maybe im missing something here, but the intersection $U\cap \mathbb{R}^2$ should be $(-\epsilon_1,\epsilon_1)\times \{0 \} \cup \{(x,y)\mid \sqrt{x^2 + y^2} < \epsilon_2 \}$. But this does not contain the box $(-\epsilon_1,\epsilon_1) \times (-\epsilon_2,\epsilon_2)$. $\endgroup$
    – lboeke
    Commented Nov 11, 2021 at 12:58
  • $\begingroup$ @lboeke How do you see $\Bbb R^1$ inside $\Bbb R^2$? Aren't all members of $\Bbb R^\infty$ just real sequences? $\endgroup$ Commented Nov 11, 2021 at 13:01
  • $\begingroup$ By the inclusion $x \mapsto (x,0)$. So $\mathbb{R}^\infty$ is a quotient of the disjoint union of the $\mathbb{R}^n$, where the inclusions give us the equivalence relation. $\endgroup$
    – lboeke
    Commented Nov 11, 2021 at 13:33
  • $\begingroup$ @lboeke Another more convenient way to see the space is $\Bbb R^\infty = \{y=(y)_n \mid \exists N(y) \in \Bbb N: \forall n > N: y_n =0\}$ and define $\Bbb R^n = \{y \in \Bbb R^\infty: N(y)=n\}$ we give these subspaces $\Bbb R^n$ the standard topology from the norm on $\Bbb R^n$ and the open sets are as defined. Then we can even say $U \cap \Bbb R^n$is just an open ball so open by definition. So $\Bbb R^\infty$ is just an increasing union and we have an induced topology from the cover. $\endgroup$ Commented Nov 11, 2021 at 13:45
  • $\begingroup$ @lboeke I first read your $U$ as $\{(y) \in \Bbb R^\infty\mid |y_n| < \varepsilon_n\}$ instead. $\endgroup$ Commented Nov 11, 2021 at 13:49
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Let's assume $V$ is an open set contained in $U$. We then have that $V\cap \mathbb{R}^m$ is open for any $m$. Fix some $m_1$ and find $r_{m_1}$ such that an open ball with radius $r_{m_1}$ around $0$ lies in $V\cap \mathbb{R}^{m_1}$. Now, we look at $V\cap \mathbb{R}^{m_2}$ for some $m_2>m_1$. This is again an open set, and therefore contains an open ball with radius $r_{m_2}$. If now $r_{m_2} < r_{m_1}$, then the union $B^{m_1}_{r_{m_1}}(0) \cup B^{m_2}_{r_{m_2}}(0) \subset \mathbb{R}^{m_2}$ is not open. This means that while $V\cap \mathbb{R}^{m_2}$ contains such a union of cells, the converse is not true as long as $V$ is open. In the case that $r_{m_2} \geq r_{m_1}$, everything is fine, but this is incompatible with a sequence like $\varepsilon_n$ in the question.

What I'm trying to get to: There should be some lower bound on the $\varepsilon_n$ in the definition of $U$.

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