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This is an assignment question:
Prove that an infinite dimensional banach space has an uncountable (hamel) basis.
I understand that there are many proofs here in MSE but all of them uses Baire Category theorem. When I sat down to prove it, I couldn't come up with the idea of using Baire's theorem to write the proof. It didn't seem natural to me and everyone in my class used Baire only. I tried a different approach but I couldn't get anywhere.
Here is what I tried :
Let us assume to the contrary that the Banach space $X$ has a countable (hamel) basis $\{v_1, v_2, v_3, \ldots \}$. Then for all $n \in \mathbb{N}$ let $M_n = \text{span}(\{v_1,v_2,\ldots,v_n\})$. By Riesz lemma, for all $n$ there exist $x_n\in X $ such that $d(x_n, M_n) \geq 1 - 1/n $ and $||x_n|| = 1$. The consider the sequence $\left( x_n \right) _{n\geq 1}$ I got stuck here to get a contradiction. If only I could show that this sequence has a limit point say $x$ and then if I could show that $d(x_n,M_n) \to d(x, M)$ then I will be done. Since the former distance tends to 1 but the latter distance is zero.
Is this approach correct or am I totally wrong to assume that a proof without using Baire is not at all possible??
Thank you.

1 Answers1

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I don't think it is realistic to expect to be able to show your bounded sequence $\{x_n\}$ has a limit point in $X$. Thinking this should be possible, given how little structure you have to work with in $X$, seems to be based on your experience in $\mathbf R^n$, which is locally compact. And it turns out that Banach spaces that are locally compact in their norm-topology must be finite-dimensional. (The Banach-Alaoglu theorem is a compactness theorem for unit balls in spaces that might not be finite-dimensional, which doesn't contradict what I just said since the topology in the Banach-Alaoglu theorem is not the norm topology.)

KCd
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