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Let $R^\infty = \{ (x_i)_{i=1}^\infty | \exists n : x_j = 0, \forall j \geq n \}$. Then the topology on $R^\infty$ is such that a subset $U$ is open if and only if $U \cap \mathbb{R}^n$ is open for each $n$.

I think that if a sequence $x_n \rightarrow x$ in this topology then there is some $m$ such that $x, x_i \in \mathbb{R}^m$ for large enough enough $n$. Here I'm thinking of $\mathbb{R}^k \subset \mathbb{R}^j$ if $k \leq j$. Then we would have $x_n \rightarrow x$ in the topology of $\mathbb{R}^m$. But I cannot see why this is true. Can someone give me a hint about how to show it?

user390884
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2 Answers2

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Hint: try a contrario.

For $x \in \mathbb{R}^{\infty}$ let $h(x) = \min \{ m \in \mathbb{N} : (\forall n > m) \, x(n) = 0 \}$ so it is the smallest number $m \in \mathbb{N}$ such that $x \in \mathbb{R}^m$.

Now suppose that $x_n \to x$ but $h(x_n)$ is unbounded. Without loss of generality $x = 0$ and (by passing to a subsequence) $h(x_n)$ strictly increases. Construct a neighborhood of $0$ such that all terms of $x_n$ lie outside.

Adayah
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  • And how do you propose to do that? – Henno Brandsma Sep 10 '19 at 21:34
  • @HennoBrandsma Diagonally: for each $k \in \mathbb{N}$ if $k = h(x_n)$ for some (unique) $n$, then set $U_k = (-\varepsilon, \varepsilon)$ where $\varepsilon > 0$ is such that $x_n(k) \notin (-\varepsilon, \varepsilon)$ (which can be done since $x_n(k) \neq 0$). Otherwise put $U_k = \mathbb{R}$. Then $$U = \mathbb{R}^{\infty} \cap \prod_{k=1}^{\infty} U_k$$ works. – Adayah Sep 24 '19 at 06:28
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I apologise for earlier, misread this as the product topology and not the box topology!

Suppose $x_n \rightarrow x$ in the box topology. I will assume $x=(0,0, \ldots)$ as we may always shift the set that we use later. Suppose that for all $k \in \mathbb{N}$ and $N \in \mathbb{N}$ there exists $n \geq N$ such that $x_n(k) \neq 0$. Define the strictly increasing sequence $(n_k)_{k = 1}^\infty \in \mathbb{N}^\mathbb{N}$ where $x_{n_k}(k) \neq 0$. We now define the open $U = \prod_{i=1}^\infty I_i$ where $I_i = (-1,1)$ if $i \neq n_k$ for all $k \in \mathbb{N}$ and $I_i = (-|x_{n_k}(k)|, |x_{n_k}(k)|)$ if $i=n_k$ and note that for all $N \in \mathbb{N}$ there exists $n_k \geq N$ such that $x_{n_k} \notin U$.

S. Dewar
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