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I was trying to show this result. I have seen proofs that show that the dimension is at least $\mathfrak{c}$, however, I'm unable to prove that it is exactly $\mathfrak{c}$. The last line of this article is unclear. How does separability imply that dimension is $\mathfrak{c}$? enter image description here

Sahiba Arora
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  • You can also use the Baire category theorem to prove that any infinite dimensional Banach space has an uncountable hamel basis, hint: proof by contradiction. – NotaChoice Dec 23 '21 at 03:43

1 Answers1

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If $X$ is separable, it has a countable dense subset $A$. Every point of $X$ is a limit of some sequence in $A$, and there are only $|A|^{|\mathbb{N}|}=\mathfrak{c}$ different sequences in $A$. So $|X|\leq\mathfrak{c}$. A Hamel basis for $X$ is a subset of $X$, and so any Hamel basis has at most $\mathfrak{c}$ elements.

Eric Wofsey
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  • I want to ask that do we assume x_n’s to be unit or smt? Since otherwise how do we know that the series of x_t converges? – boyler Feb 20 '22 at 05:41