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I'm working through some old Harvard Math 55 problem sets, and one problem in particular has got me stumped:

If $F$ is field, then prove that the vector space $F^\infty$ over $F$ (the space of infinite tuples or sequences) has no countable spanning set.

In the case that $F$ is countable or finite, we can show that the space spanned by a countable set is necessarily itself countable and therefore cannot span $F^\infty$, which is uncountable. However, that argument fails when the field is uncountable, since the span of even a single vector is an uncountable set. Any hints on how to proceed?

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CJ Dowd
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    One idea ... given a countable set $C$, make a sequence $s$ that is not in the span of the countable set, which is to say is not a linear combination of any finite subset of the set. For set $T$ of $k$ vectors in the set $C$, if we pick a collection of $k+1$ new coordinates of $s$, then the $k$ vectors in $R^{k+1}$ we get from restricting $T$ to those $k+1$ coordinates can't span $R^{k+1}$, so we can choose values for those coordinates in $s$ to guarantee $s$ is not in the span of $T$. I seem to remember there is a more high-powered argument, but I can't recall it. – Carl Mummert Sep 27 '16 at 22:27
  • There is a high-powered argument for Banach spaces, which says that any basis for a Banach space is either finite or uncountable. – Stanley Sep 27 '16 at 22:45
  • @Stanley: yes, that rings a bell - thanks. I found a write up of it here, after your hint: http://math.stackexchange.com/questions/217516/let-x-be-an-infinite-dimensional-banach-space-prove-that-every-hamel-basis-of Of course, in this case we just have a vector space. – Carl Mummert Sep 28 '16 at 01:43

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Here is one idea/hint for a proof by diagonalization. Given a countable set of sequences $C$, we want to make a sequence $s$ that is not in the span of $C$, which is to say $s$ is not a linear combination of any finite subset of $C$.

To do this, it is sufficient to make an $s$ such that, for each finite subset $T$ of $C$, say of size $k_T = k$, there are some $k+1$ coordinates of $s$ such that the vectors we get in $\mathbb{R}^{k+1}$ by restricting $T$ to those coordinates do not span the vector in $\mathbb{R}^{k+1}$ obtained by restricting $s$ to those coordinates. This guarantees that $s$ is not in the span of $T$.

No set of $k$ vectors in $\mathbb{R}^{k+1}$ can span $\mathbb{R}^{k+1}$, so if we choose a set of new coordinates for each finite subset $T$ of $C$, we can choose values of $s$ on those coordinates appropriately to ensure the previous condition holds. This means we just need to set up the appropriate construction to make $s$.

Separately, there is a more well known argument that shows that no infinite dimensional Banach space can have a countable basis; see the question Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable. In this case, we only have a vector space, so the techniques used there don't seem to apply.

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