Talk:Finite Dimensional Subspace of Normed Vector Space is Closed
A gap in the proof
The last step of the proof uses that "sequence of vectors converges iff each component converges" but the norm on $V$ is not the product norm, so the convergence in the norm on $V$ doesn't imply convergence in the product norm, so the sequence of components may not converge to component of the limit.
Take the $\mathbb{Q}$-vector space $\mathbb{Q}[\sqrt{2}]$ as an example, there exists a sequence $q_n\in\mathbb{Q}$ that converges to $\sqrt{2}$. Take $\{1,\sqrt{2}\}$ as a basis, then $q_n$ is $(q_n,0)$ and $\sqrt{2}$ is $(0,1)$, but $(q_n,0)$ doesn't converge to $(0,1)$ in the product norm, and the second component of $(q_n,0)$ is always 0 which doesn't converge to the second component $1$ of $\sqrt{2}$. --Hbghlyj (talk) 17:03, 13 March 2024 (UTC)
- Always good to have a careful eye on these things, thanks. Would I be correct in saying we are implicitly using that all norms on $\R^k$ (resp $\C^k$) are equivalent (I am guessing this is not true of $\Q$ because of the use of topological properties of $\R$), so we get the result for the norm on $V$ restricted to that subspace from equivalence to the product norm? Caliburn (talk) 18:24, 13 March 2024 (UTC)
- Yes, I think it is true for $K=\R$. I added a reference to Norms on Finite-Dimensional Real Vector Space are Equivalent. --Hbghlyj (talk) 12:02, 14 March 2024 (UTC)
Do we need to add a condition "$V$ is complete"?
Proof 2 depends on Subspace of Complete Metric Space is Closed iff Complete which requires $V$ to be complete. --Hbghlyj (talk) 12:23, 14 March 2024 (UTC)
- You wrote it. If you're not certain what you are proving, and you realise you're trying to prove something that the initial statement does not say, then you do not change the original statement. Instead, you write a new page dedicated to the statement you are trying to prove. --prime mover (talk) 12:45, 14 March 2024 (UTC)
- As above you generally retain hypotheses as is unless they are insufficiently general. We do not yet have a good mechanism for proofs that only work under restricted hypotheses. You could split up into "Proof for Complete Normed Vector Spaces" for example. I would guess this fact is equivalent to the ground field $K$ being complete? If every finite dimensional subspace is closed, take a unit vector $x$ so that the span of $x$ is closed. Then for each Cauchy sequence $\sequence {\alpha_n}_{n \in \N}$ in $K$, $\sequence {\alpha_n x}_{n \in \N}$ is a Cauchy sequence in the span of $x$, which converges to some limit $\alpha x$, at which point you can easily see $\alpha_n \to \alpha$ in $K$. I don't have time to write this up currently - but I think proof 2 might be missing some detail as well. Caliburn (talk) 14:00, 14 March 2024 (UTC)
- Please take on board the fact that we emphatically do not support the practice of linking to proofs on a different website, for obvious reasons. Hence I have removed the link to the planetmath proof.
- If you want to use that proof, add a page on $\mathsf{Pr} \infty \mathsf{fWiki}$ containing the material on that external website, then adding the link to that website in the approved form using the appropriate template. --prime mover (talk) 14:21, 14 March 2024 (UTC)
- I suggest separating the necessity (closed implies complete) and sufficiency (complete implies closed) parts of the proof of Subspace of Complete Metric Space is Closed iff Complete and point out the sufficiency part doesn't require the larger space to be complete, and link to the sufficiency part of the proof only. But I know how to separate a proof. Could you help me with that? --Hbghlyj (talk) 14:52, 14 March 2024 (UTC)