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I suddenly thought of it while looking at one problem. That question was a basic question, but I wondered what would happen if I expanded it a little more.

First of all, I started thinking about Stone-Weierstrass Theorem. for every $f\in C([0,1])$, there exists a sequence of polynomial function ${\{p_n\}}_{n\in\mathbb{N}}$ that converges to $f$.

Then, I let the space of polynomial function as $\mathbb{P}([0,1])$. Then there exists a subspace of ${\mathbb{P}([0,1])}^\omega$ named $P$ such if ${\{p_n\}}\in P$, then ${\{p_n\}}$ converges.

It is obvious that the dimension of $P$ has smaller dimension than ${\mathbb{P}([0,1])}^\omega$, and there exists surjective $\mathbb{C}$-linear map from $P$ to $C([0, 1])$.

Then $\dim_{\mathbb{C}}{C([0, 1])} \le \dim_{\mathbb{C}}P \le \dim_{\mathbb{C}}{\mathbb{P}([0,1])}^\omega$.

Dimension of ${\mathbb{P}([0,1])}^\omega$ is countable, then $C([0,1])$ has countable dimension.

Did I make any mistakes in the proof process?

MH.Lee
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    Yes: the dimension of $\mathbb{P}([0,1])^{\mathbb{N}}$ is not countable. The dimension of any complete normed space (a Banach) cannot be countable by the Baire Category Theorem. – Aphelli Oct 18 '21 at 17:50
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    I don't understand the surjectivity claim. How do I write $e^x$ as a finite sum of basis vectors? – Randall Oct 18 '21 at 17:50
  • "Converges to" is, unfortunately, not enough for what is called a "Hamel basis" (a basis in the usual vector space meaning). It suffices for a Hilbert basis (which only requires that you be able to arbitrarily approximate). – Arturo Magidin Oct 18 '21 at 17:50
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    I'm not sure what you mean by "dimension" here, but I know that $C[0, 1]$ has no countable Hamel basis (which is linearly independent and spanning in the sense that every point is a unique finite linear combination of points in the basis). That means the Hamel dimension is uncountable. Indeed, no Banach space can have countably infinite Hamel dimension, which is a lovely application of Baire Category Theorem. – Theo Bendit Oct 18 '21 at 17:52
  • @Mindlack Ohh... I thought $\mathbb{P}([0,1])$ has countable basis, so countable product of countable dimension has countable basis. Where I missed in this process? – MH.Lee Oct 18 '21 at 17:54
  • @Randall A sequence of partial sum of Maclaurin series of $e^x$...? – MH.Lee Oct 18 '21 at 17:56
  • And the linear combination is..... – Randall Oct 18 '21 at 17:57
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    @Nightflight A countable product of countable sets is not necessary countable. For example, even $2^{\omega}$ is an uncountable set. – Mark Oct 18 '21 at 17:57
  • @Mark I know that, but I thought that dimension do not follows that. In finite cases, $\dim V^n = n(\dim V)$, and I thought it will be applied at infinite cases. – MH.Lee Oct 18 '21 at 18:00
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    @Nightflight Note that $V^n$ is the direct sum of $n$ copies of $V$. As for infinite direct sums, the direct sum of spaces $\sum_{\alpha\in I} V_{\alpha}$ is the set of formal sums $\sum_{\alpha\in I} v_{\alpha}$ where $v_{\alpha}\in V_{\alpha}$ and all but finitely many of the vectors are zeros. So in your case, the direct sum of $\omega$ copies of $P([0,1])$ is the space of all sequences which have finitely many nonzero elements. For this space, the formula you wrote would indeed hold. But the space of all sequences (which is what you defined) is a much larger space. – Mark Oct 18 '21 at 18:04
  • I will just point out this older question: Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of $X$ is uncountable. (You can find here the proof based on Baire Category Theorem that was mentioned in some comments.) Some of the questions linked there might be of interest, too. – Martin Sleziak Oct 19 '21 at 14:08

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