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I am trying to show:

Algebraic dimension of infinite-dimensional Banach Space is uncountable.

By algebraic dimension it is meant that the cardinality of the Hamel Basis of the space.
Suppose we defined $V$ to be an infinitely dimensional normed linear space. So far I found out the followings:

  • Interior of any proper subspace of $V$ is empty
  • Every proper closed subspace of $V$ is nowhere dense
  • According to Baire's Theorem, $V$ can not be written as a union of countable union of nowhere dense closed sets.
  • By the way of contradiction, I assume there is a countable Hamel Basis for $V$.
  • Using the Hamel Basis, I need to construct closed sets such that their union gives me the whole space $V$.
  • Suppose $\{ x_n \}_{n \in \mathbb{N} }$ is the Hamel Basis and suppose $(x_1,x_2,...,x_n)$ be the subspace generated linearly by $x_1,x_2,...,x_n$.

Trouble Now all i need to show is the subspace $(x_1,x_2,...,x_n)$ is closed. I pick an element from the closure of that subspace and argue it can be written as an linear combination of $x_1,x_2,...,x_n$. I use bunch of triangle inequalities, cauchy sequences but i feel i am lost. Would please help me to complete this proof?

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On a finite-dimensional ($\mathbb{R}$ or $\mathbb{C}$) vector space, all norms are equivalent.

The finite-dimensional subspace $\operatorname{span} \{ x_1,\dotsc, x_n\}$ is complete if we endow it with the norm induced by the Euclidean norm on $\mathbb{R}^n$ (or $\mathbb{C}^n$) via the isomorphism $(\alpha_1,\dotsc,\alpha_n) \mapsto \sum\limits_{k=1}^n \alpha_k\cdot x_k$. Hence it is complete in the norm induced from $V$.

A complete subset of a metric space is closed.

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  • $\begingroup$ So are you saying that $span \{ x_1,...,x_n \}$ is complete because $\mathbb{R}^n$ is complete? $\endgroup$ – iamvegan Jan 12 '14 at 17:46
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    $\begingroup$ Sort of. Because $\mathbb{R}^n$ is complete, and the norms induced by the ambient space and the one induced via an isomorphism to $\mathbb{R}^n$ are equivalent. Both facts together yield the completeness of $\operatorname{span} \{x_1,\dotsc,x_n\}$. $\endgroup$ – Daniel Fischer Jan 12 '14 at 18:27
  • $\begingroup$ Okay i just looked up the definition of equivalent norms. I saw that convergence is preserved in equivalent norms. Now i understand what you mean. $\endgroup$ – iamvegan Jan 12 '14 at 18:48

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