I recently read that the cardinality of $\mathbb{R}^n$ (the number of points in n-dimensional Euclidean space) is the same as $\beth_1$ just like the cardinality of the set of real numbers, however, I wonder if This is true for $\mathbb{R}^\infty$ too? That's all. [Note: For those who don't know $\beth_1=2^{\aleph_0}$.]
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2Does this answer your question? cardinality of all real sequences – PrincessEev Jul 21 '22 at 21:53
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1What's $\Bbb R^\infty$? – Sassatelli Giulio Jul 21 '22 at 21:56
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@SassatelliGiulio It's the direct limit $\mathbb{R}^\infty:=\varinjlim_{n\in\mathbb{N}}\mathbb{R}^n$. – Samuel Adrian Antz Jul 21 '22 at 22:01
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@SamuelAdrianAntz Could be, but I think it is the infinite product. – Paul Frost Jul 21 '22 at 23:13
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@PaulFrost I was gonna comment, it's the same, but then read a comment under the answer below, that $\mathbb{R}^\infty$ is only the set of sequences with finite support, which is indeed something I didn't consider, but is true concerning the colimit. – Samuel Adrian Antz Jul 22 '22 at 16:43
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3@SamuelAdrianAntz To make precise what I meant: Notation is often ambiguous and depends on the "taste" of the author. Thus $\mathbb R^\infty$ could denote the direct limit or could denote the infinite product. These objects are not the same. But they have the same cardinality ;-) – Paul Frost Jul 22 '22 at 16:51
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$\mathbb R^\infty=\mathbb R^\mathbb N=2^{\mathbb N\times \mathbb N}$ so $|\mathbb R^\infty|=|2^\mathbb N|=\frak c.$
To see that $\mathbb R^\mathbb N=2^{\mathbb N\times \mathbb N}$, note that each $x\in \mathbb R$ may be expressed as a sequence of $1$'s and $0$'s; i.e. its binary expansion. Then, $\mathbb R^\mathbb N=\left(2^\mathbb N\right )^{\mathbb N}.$ Now, $2^{\mathbb N\times \mathbb N}$ is isomorphic to $\left(2^\mathbb N\right )^{\mathbb N}$ via the (Currying) map $g\mapsto f:f(n)(m)=g(n,m).$
Matematleta
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Thanks this really helped, it becuase I was trying to figure out a universal homomorphism but needed to know this little fact. – Rafael Lazo Jul 21 '22 at 22:19
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$\lim\limits_{\stackrel{\longrightarrow}{n\in\Bbb N}} \Bbb R^n$ with morphisms $(\phi_{n,m}(v))_j=\begin{cases}v_j&\text{if }j\le n\ 0&\text{if }n<j\le m\end{cases}$ is not (at least, not in the categories Set, $\Bbb R$-Vec or Top and with the obvious morphisms) $\Bbb R^{\Bbb N}$, but rather $\Bbb R^{(\Bbb N)}={x\in\Bbb R^{\Bbb N},:, \operatorname{card}{n\in\Bbb N,:, x_n\ne 0}<\aleph_0}$. – Sassatelli Giulio Jul 21 '22 at 22:58
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$\mathbb R^\infty$ is the set of sequences with finite support whereas $\mathbb R^\mathbb N$ is the set of sequences. But the cardinality of the former is no larger than that of the latter and is certainly larger than that of any $\mathbb R^n$ so it must be $\frak c$ assuming the continuum hypothesis. – Matematleta Jul 21 '22 at 23:12
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@Matematleta You don't need to use CH and, truth be told, this aspect should be addressed in the answer. – Sassatelli Giulio Jul 21 '22 at 23:49