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Let $X$ be an infinite-dimensional Separable Banach Space. Prove that $\dim X=\mathfrak{c}$.

On the direction of $\dim X\ge \mathfrak{c}$, I thought taking the subset of all elements $x\in X$ which can be represented as an convergent expansion $$X\ni x = \sum_{n=1}^\infty c_ne_n$$ where $$e_n^i=\cases{0\quad i\neq n \\1\quad i=n,}$$ which will be mapped to $\ell^\infty$, a space of cardinality $\mathfrak c$ by the isomorphism $x\mapsto(c_1,c_2,\dots)$.

About the other direction I don't have any clue. We know that a Banach space with a basis must be separable but how can we use that fact to find out that $\dim X\le\mathfrak{c}$?

Andrés E. Caicedo
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    What is $\aleph$? If you mean $|\mathbb R|$, please use $\mathfrak c$ instead. – Andrés E. Caicedo Oct 09 '14 at 05:37
  • To show that $\dim X\le 2^{\aleph_0}$, show that $|X|=2^{\aleph_0}$. – Jonas Meyer Oct 09 '14 at 05:41
  • That was exactly what I meant. changed it. –  Oct 09 '14 at 05:41
  • @JonasMeyer but that's exactly what I need to prove (that $|X|=2^{\aleph_0}$). so as I said I can't prove only the $\le$ direction. –  Oct 09 '14 at 05:43
  • To make sure we're talking about the same thing: By $|X|$ I mean cardinality, not dimension (that is why I wrote something different from $\dim$). It is assumed that $X$ is separable. You can show that every separable metric space has cardinality at most $2^{\aleph_0}$. The other direction is harder; I don't understand what you've written, but you can find a solution at http://math.stackexchange.com/q/141535/ – Jonas Meyer Oct 09 '14 at 05:45

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Let $Q$ be a countable dense subset of your Banach space $X$. In particular, every element of $X$ may b written as the limit of a sequence of points in $Q$, so there is an onto map $Q^{\mathbb{N}} \to X$, hence $$|X| \leq |Q^{\mathbb{N}}| = \mathfrak{c}.$$ On the other hand, $$|X| = |\mathbb{K}| \cdot \dim (X)$$ where $\mathbb{K}= \mathbb{R}$ or $\mathbb{C}$ is your field. Therefore, $$\mathfrak{c} \geq |X|= \max( \mathfrak{c}, \dim(X))$$ hence $\dim(X) \leq \mathfrak{c}$.

Seirios
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  • but isn't your proof assumes that $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ or in general $\dim\mathbb{K}=\mathfrak{c}$? –  Oct 09 '14 at 07:05
  • Yes. Do you work is an arbitrary field? In that case, you should precise your definition of normed vector space: the field has to be at least valued. – Seirios Oct 09 '14 at 07:14
  • What do you mean by "valued"? e.g I can also talk about $\mathbb{H}^{\aleph_0}$. –  Oct 09 '14 at 10:13
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    I mean that the field has to have an absolute value. See for example http://mathworld.wolfram.com/Valuation.html. What does $\mathbb{H}^{\aleph_0}$ denote? – Seirios Oct 09 '14 at 11:23
  • Banach spaces which are neither real nor complex are silly. – Tomasz Kania Oct 11 '14 at 13:33