# Weak topology on $\mathbb{R}^\infty$: Constructing neighborhoods from sequences

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?

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.

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.

• 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 $(-\epsilon_1,\epsilon_1) \times (-\epsilon_2,\epsilon_2)$. Commented Nov 11, 2021 at 12:58
• @lboeke How do you see $\Bbb R^1$ inside $\Bbb R^2$? Aren't all members of $\Bbb R^\infty$ just real sequences? Commented Nov 11, 2021 at 13:01
• 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. Commented Nov 11, 2021 at 13:33
• @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. Commented Nov 11, 2021 at 13:45
• @lboeke I first read your $U$ as $\{(y) \in \Bbb R^\infty\mid |y_n| < \varepsilon_n\}$ instead. Commented Nov 11, 2021 at 13:49

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 $$U$$.