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I think I can prove that closure of every countable set in any metric space has cardinality at most $\mathcal c=2^{\aleph _0}$ . So if a metric space is separable i.e. has a countable dense subset $A$ then obviously $X=\bar A$, so $|X|=|\bar A| \le 2^{\aleph_0 }$ . Now suppose $X$ is an infinite dimensional Banach space and $B$ be a Hamel basis for $X$ , then $B$ is uncountable (see https://matthewhr.files.wordpress.com/2012/08/uncountability-of-banach-hamel-basis1.pdf ) ; so I think if $X$ is an infinite dimensional banach space then we can write , as the underlying field is $\mathbb R$ , if $\aleph _1$ is the smallest uncountable cardinal , that $|X| \ge |\mathbb R \times ...\aleph_1 {times}|$ ; now assuming the existence of an infinite dimensional separable Banach space , we see that $|\mathbb R \times ...\aleph_1 {times}| \le |X| \le 2^{\aleph_0}$ ; so am I correct ? Is it true that $|\mathbb R \times ...\aleph_1 {times}| \le 2^{\aleph_0}$ ? Is $|\mathbb R \times ...2^{\aleph_0} \space {times}| \le 2^{\aleph_0}$ also true ?

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    $\begingroup$ BTW you can use something like this: $\underset{\aleph_1\text{-times}}{\underbrace{\mathbb R\times\cdots\times\mathbb R}}$ $\underset{\aleph_1\text{-times}}{\underbrace{\mathbb R\times\cdots\times\mathbb R}}$. (Although it is probably better not to use something like that in the title, it would look too big.) $\endgroup$ Mar 12, 2015 at 12:13

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If $B$ is a basis for a vector space $V$ over $F$, then $V$ is not a product of $|B|$ many copies of $F$, as a set. Instead it is all the sequences in this product which are finitely non-zero. It is the direct sum.

And the difference is huge, in this case.

In general, $A\times A\times\ldots$ "$\aleph_1$ times" is the same as $A^{\omega_1}$, which, if $A$ has at least two elements, is at least as large as $2^{\aleph_0}$ and consistently larger; if we take "$2^{\aleph_0}$ copies", then we get something whose cardinality is $2^{2^{\aleph_0}}$ which is in fact larger.

But with a vector space this is different, because we only take very few of the function in this product. So in fact the cardinality is equal to that of $|\operatorname{Fin}(B\times F)|$, which under the axiom of choice and the assumption that at least one of these is infinite, means that it is either $|F|$ or $|B|$.

In the case where you look at $F=\Bbb R$, this means that if you take any vector space of dimension $\leq2^{\aleph_0}$, its cardinality remains $2^{\aleph_0}$ as well.


What does it mean about Banach spaces?

If a Banach space is infinite dimensional, then its dimension cannot be countable. This is a nice application of Baire's category theorem. So its Hamel basis has to be uncountable. But we can in fact prove that its Hamel basis has to have cardinality $2^{\aleph_0}$, which is a stronger statement, since it is consistent that $\aleph_1<2^{\aleph_0}$, in which case a normed space of dimension $\aleph_1$ is necessarily not a Banach space, and therefore not complete.

(And finally, "assuming the existence of an infinite dimensional separable Banach space" is a strange statement. $\ell_2$ is an infinite dimensional, separable Banach space. In fact, any $\ell_p$ for $p\in[1,\infty)$ is an infinite dimensional separable Banach space.)

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