# Proof that vector spaces $k^M$ and $k^{(M)}$ are not isomorphic

Assume $$k$$ is a division ring and $$M$$ an infinite set. There are two vector spaces over $$k$$ to consider: $$k^M$$ and $$k^{(M)}$$ (the latter is the subset of $$k^M$$ containing all infinite sequence with only a finite number of nonzero components).

Obviously $$|k^{(M)}|\leq |k^M|$$. I know that $$k^{(M)}$$ has a canonical basis $$(e_\mu)_{\mu\in M}$$. This means that $$(e_\mu)_{\mu\in M}$$ is a family of linearly independent elements of $$k^M$$. How can I show that the $$e_\mu$$'s don't generate $$k^M$$? I have to find an element of $$k^M$$ that isn't a linear combination of the $$e_\mu$$. What is a good choice for this supposed element? (Or is a proof by contradiction more appropriate?)

• Hint: take a nonzero constant sequence. Jan 15, 2021 at 18:33
• One way to do is to note that your first observation is sharp: $|k^{(M})|<|k^M|$. Another way is to show that $k^{(M)}$ doesn't have the universal property of being the product of $|M|$ many $k$'s. Jan 15, 2021 at 18:34
• @Berci The inequality need not be strict, for example if $k$ has the cardinality of the continuum and $M$ is countably infinite. Jan 15, 2021 at 18:38
• @AndreasBlass: Indeed, that's right. Jan 15, 2021 at 18:40

Finding such an element would not be enough, as it would simply show that the inclusion $$k^{(M)}\to k^M$$ isn't an isomorphism, but that doesn't mean that there is no isomorphism.

For the record, the constant family with value $$1$$ is not in the span of the $$e_\mu$$.

The usual way to prove that they are not isomorphic is to prove that any $$M$$-indexed family does not span $$k^M$$ - in fact that would be equivalent because one can extract a basis from any such family, and it must then have cardinality $$\geq |M|$$ and $$\leq |M|$$ so $$=|M|$$. But for some reason I always forget the standard argument for this - it's a sort of diagonal argument where you build a new element not in the span of any $$M$$-indexed family of elements.

So another way to proceed is to prove that $$k^{(M)}$$ has too many linearly independent linear forms. Indeed, $$(k^{(M)})^* \cong k^M$$ (canonically), and so these linearly independent linear forms contradict the existence of an isomorphism.

For this I will use the following well-known set-theoretic lemma:

If $$M$$ is an infinite set, there is a bijection between the set of (nonempty) finite subsets of $$M$$, and $$M$$.

The proof then goes as follows (hidden, to leave some suspense):

It follows from this lemma - actually, a variation on that lemma that I'll let you figure out- that $$k^{(M)}$$ is isomorphic, as a $$k$$-vector space, to $$k[(X_\mu)_{\mu\in M}]$$, the polynomial algebra in $$M$$-many indeterminates. Now for any $$z\in k^M$$, $$ev_z$$,"evaluation at $$z$$", defined by sending a polynomial $$P$$ to $$P(z)$$, defines a linear form on $$k[(X_{\mu})_{\mu\in M}]$$.

The claim is that these are linearly independent. But this is classical polynomial algebra: if you have a finite family $$(z_i)_i \in k^M$$, then I can find a finite subset $$J\subset M$$ such that the restrictions of these families to $$J$$ are distinct - and now in a finite number of variables, I can of course design a polynomial that takes any value I like on a finite list of points, in particular I can make the values leave any hyperplane. This is left as an exercise.

The coup-de-grâce is in claiming that these are too many. But Cantor's theorem ensures that $$|k^M|> |M|$$, as $$k$$ is a field and so $$|k|\geq 2$$, so we are done.