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I am aware that this question has been asked repeatedly on this site. I am asking this question myself because I do not want to know the answer to problem yet. If I am wrong, please explain, and I would be grateful for a hint. Please DO NOT blatantly give the answer. Thank you.

Let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all sequences that are "eventually zero," that is, all sequences $(x_{1},x_{2},...)$ such that $x_{i}\neq0$ 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?

In the product topology I have that all constant sequences $(a,a,...)$ for $a \in \mathbb{R}$ lie in the closure of $\mathbb{R}^{\infty}$.

Then the more I thought of it, I thought the closure of $\mathbb{R}^{\infty}$ was $\mathbb{R}^{\omega}$.

In the box topology, I am fairly confident that the only sequence that lies in the closure of $\mathbb{R}^{\infty }$is the constant zero sequence.

user404735
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Your hunch about the product topology is correct. Go and prove it, the proof will be the same for constant sequences or any other one...

In the box topology, the $0$-sequence is already in $\Bbb R^\infty$, so certainly in its closure, as all of $\Bbb R^\infty$ is. Show no new point will be, in the box topology. What does it mean that $(x_n) \notin \Bbb R^\infty$?

Henno Brandsma
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