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The following is from Gowers's https://www.dpmms.cam.ac.uk/~wtg10/tensors3.html:

In fact, one can even do away with the condition that $X$ should be finitedimensional, as follows. If $f : V \times W \longrightarrow X$ is a bilinear map such that $$a_1 f (v_1 , w_1 ) + ... + a_n f (v_n , w_n ) = x$$ for some non-zero vector x, then let g be a linear map from $X$ to $\mathbb R$ such that $g(x)$ is not zero. The existence of this map can be proved as follows. Using the axiom of choice, one can show that the vector $x$ can be extended to a basis of $X$. Let $g(x) = 1$, let $g(y) = 0$ and extend linearly. Once we have $g$, we have a bilinear map $gf : V \times W \longrightarrow \mathbb R$ such that $$a_1 gf (v_1 , w_1 ) + ... + a_n gf (v_n , w_n )$$ is non-zero.

I understand the other parts of this article. But this paragraph is not really clear. All of vector spaces have infinite dimensions? How exactly use the axiom of choice? What is the choice function?

lbarczikay
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    Duplicate of https://math.stackexchange.com/questions/461436/vector-space-bases-without-axiom-of-choice and https://math.stackexchange.com/questions/1650069 and https://math.stackexchange.com/questions/707732 and https://math.stackexchange.com/questions/2352392 – Martin Brandenburg Jan 14 '25 at 17:45
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    I don't think this is a dupe. The question is not asking "why does choice imply every vector space has a basis", it is asking "how is choice used in this specific remark". It is quite possible that the OP does not see the connection between "$x$ can be extended to a basis of $X$" and "$X$ has a basis". As someone who knows that choice implies $X$ has a basis, I did not immediately make the connection myself. – preferred_anon Jan 15 '25 at 08:11

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You need the axiom of choice to prove that every vector space has a basis. It follows from this that given a non-zero vector $v$ from a vector space $V$, there is some basis of $V$ to which $v$ belongs. And Gowers uses this fact.

José Carlos Santos
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