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Let $(X, \| \cdot \|)$ be a normed vector space with algebraic basis $e_1, e_2, ...$ (dim $X = \infty$ and for all $x \in X, x = \sum_{i=1}^{\infty} a_i e_i$ for unique $a_i$.). Then $X$ is incomplete.

I have written a proof of this theorem, but it is quite long and I have a feeling there should be an easier way of doing it. Please let me know if my proof is correct, and if possible, how to write a better proof.

My proof:

Proof by contradiction: assume $X$ is complete, then $X$ is a Banach space. Then, as finite dimensional subspaces of a normed vector space, $C_n = \text{span} \{e_1, ..., e_n\}$ is closed for all $n \in \mathbb{N}$. Furthermore, since $X = \cup_n C_n$, by Baire's theorem there is some $n \in \mathbb{N}$ such that int$(C_n) \neq \emptyset$. Hence, let $x = \sum_{i=1}^n x_i e_i \in$ int$(C_n)$.

Note $C_{n+1}$ is itself a finite dimensional normed vector space, and as such all norms on $C_{n+1}$ are equivalent and $C_{n+1}$ is complete with respect to every norm.

Hence, define the norm $\| \sum_{i=1}^{n+1}a_ie_i\|_{\infty} = \sup_{i =1, ..., n}|a_i|$ on $C_{n+1}$ (it is easily verified that this is a norm). Consider the sequence $y_i = \sum_{i=1}^nx_ie_i + \frac{1}{i}e_{n+1} \in C_{n+1}$.

Clearly, $\lim_i\|y_i - x\|_{\infty} = \lim_i \frac{1}{i}=0$, which means that $y_i \to x$ in $(C_n,\| \cdot \|_{\infty})$. Since $\| \cdot \|_{\infty}$ is equivalent to $\| \cdot \|$ on $C_{n+1}$, we therefore have $y_i \to x$ in $\|\cdot\|$. But, $x \in \text{int}(C_n)$ so $y_i \to x$ implies that there is some $M \in \mathbb{N}$ such that for all $m \geq M, y_m \in C_n$, a contradiction since $y_m = \sum_{i=1}^n x_i e_i + \frac{1}{m} e_{n+1} \notin C_n$.

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  • $\begingroup$ The statement in your title is wrong. Countable dimensional normed vector spaces are incomplete. $\endgroup$ – Cave Johnson Jun 19 '18 at 3:41
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Now that you know BCT and the fact that all finite dimensional normed vector spaces are Euclidean, the proof is very straight-forward. As you argued, there is some $n\in\mathbb N$ such that $\operatorname{int}C_n\ne\emptyset$. In particular, $\operatorname{int}C_n$ is an non-empty open set in the subspace topology on $C_{n+1}$. However, $C_{n+1}$ is linearly homeomorphic to $\mathbb R^{n+1}$, whose $n$-dimensional subspace contains no non-empty open sets.

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  • $\begingroup$ Thanks! Does that mean my proof is correct? I don't already have a theorem saying $C_{n+1}$ is linearly isomorphic to $\mathbb{R}^{n+1}$, or that $\mathbb{R}^{n+1}$ contains now non-empty open sets (although I can see that these things are true) $\endgroup$ – user2139 Jun 20 '18 at 13:51
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    $\begingroup$ @user2139 Yes, your proof is correct :-) $\endgroup$ – Cave Johnson Jun 21 '18 at 3:14

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