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I am reading the following question:

Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

And I am wondering why

$$X=\bigcup_{n\in \mathbb N}X_n$$

Since the right hand side is just an union of some sets and the left hand side is the complete space. Shouldn't it be:

$$X=lin(\bigcup_{n\in \mathbb N}X_n)$$ ?

Community
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Duke
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  • Note that the $X_n$'s form an increasing chain of subsets, so at each stage $\bigcup_{n\le N}X_n=X_N$. That's why we don't need $\operatorname{span}(\bigcup_{n\in\Bbb N}X_n)$, since any vector is spanned by finitely many basis vectors, which is in some $X_n$. – Mario Carneiro Feb 05 '15 at 11:53
  • But I still don't understand it. $\cup_{n\in \mathbb N}X_n$ is a collection of independent vectors. Not the whole space. :/ – Duke Feb 05 '15 at 11:56
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    Oh, now I see the issue. $X_n=\operatorname{span}(v_1,\dots,v_n)$, not $X_n={v_1,\dots,v_n}$. The author appears to be using square brackets to denote this. – Mario Carneiro Feb 05 '15 at 11:59
  • Thanks a lot. That was stupid from me. I only know the notation <...>. – Duke Feb 05 '15 at 12:46

1 Answers1

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The notation $[v_1,\ldots,v_n]$ refers to the vector space spanned by $B=\{v_1,\ldots,v_n\}$ and not $B$ itself. If $x\in X$ then by definition of a Hamel basis, $x=\sum_{j=1}^ka_jv_j$ for some $k\ge1$, $a_j\in\mathbb{K}$. This implies $x\in X_k\subseteq\bigcup_{n\in\mathbb{N}}X_n$.

Jason
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