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In the notes of the current course I'm taking, the infinite-dimensional unit ball, sphere, and disk are defined to be $$B^\infty:=\{x\in\mathbb{R}^\infty\colon\|x\|<1\},$$ $$D^\infty:=\{x\in\mathbb{R}^\infty\colon\|x\|\leq 1\},$$ $$S^\infty:=\{x\in\mathbb{R}^\infty\colon\|x\|=1\},$$ respectively (using the Euclidean norm). Here $\mathbb{R}^\infty$ is the space of all sequences with finitely many non-zero terms. There is an exercise to prove $$B^\infty=\bigcup_{n=1}^\infty B^n,\ D^\infty=\bigcup_{n=1}^\infty D^n,\ S^\infty=\bigcup_{n=1}^\infty S^n,$$ where $B^n$, $D^n$, and $S^n$ are the $n$-dimensional variants of the unit ball, disc, and sphere, respectively. This confuses me, since elements of $\mathbb{R}^n$ aren't even elements of $\mathbb{R}^\infty$ to begin with, right? If so, how can these two possibly be equal? I wonder if I'm supposed to consider a function such as $\mathbb{R}\to\mathbb{R}^\infty$ given by $x\mapsto (x,0,...)$. Anyway, any clarity on this problem would be helpful, thank you!

Jose Avilez
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Kazutoshi Ko
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    As you point out, there is a natural inclusion of $\mathbb{R}^n$ into $\mathbb{R}^m$ whenever $m > n$, including $m = \infty$ (though if you want to be careful about it, you can consider the inclusions $$\mathbb{R} \hookrightarrow \mathbb{R}^2 \hookrightarrow \mathbb{R}^3 \hookrightarrow \dotsb $$ and take an appropriate kind of limit). – Xander Henderson Jun 01 '21 at 04:42
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    Could you please define the norm? This is a bit confusing, what about the element (1, 1/2, 1/4, ....)? – Adrián Hinojosa Calleja Jun 01 '21 at 05:01
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    This is a great question. It shows you are being very careful in your thinking, which is good. If you wanted to you could "stretch" the definition of $\mathbb{R}^3$ for example so that $B^3\subseteq\mathbb{R}^3:=(x_1, x_2, x_3, 0, 0, 0, \ldots) \subseteq \mathbb{R}^\infty$. This $\mathbb{R}^3$ (a subspace of $\mathbb{R}^\infty$) and $(x_1, x_2, x_3)$ (not a subspace of $\mathbb{R}^\infty$) both have all the same properties (i.e. they are "isomorphic" to throw out a loaded term you may or may not be familiar with). – Selrach Dunbar Jun 01 '21 at 05:06
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    $\mathbb R^{\infty}$ with its usual product topology is (metrizable but not) normable. The question doesn't makes sense since you did not define $|x|$. – Kavi Rama Murthy Jun 01 '21 at 05:06
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    Following Munkres and other authors, I have interpreted this question to mean $\mathbb{R}^\infty$ is the set of all sequences that are eventually zero. (So this is a subspace of $\ell^2$). – Selrach Dunbar Jun 01 '21 at 06:50
  • @KaviRamaMurthy Apologies, I should've defined that in the question. As noted by Selrach, $\mathbb{R}^\infty$ is the set of all sequences such that all but finitely many terms are zero. I think my professor defined the norm to just be the Euclidean norm applied to the nonzero terms (which are finite). – Kazutoshi Ko Jun 01 '21 at 13:11

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