## As an F-vector space, an infinite direct sum of F has strictly smaller dimension than an infinite direct power of F over the same index set

Let be a field. Prove that a vector space over having basis (regardless of the cardinality of ) is isomorphic as a vector space to . Prove that is also an -vector space which has strictly larger dimension than that of .

(So a free module on any set is isomorphic to a direct sum. We’ve never gotten around to proving this in the best possible generality, though, so we’ll just prove it for vector spaces here.)

Note that, by the universal property of free modules, the natural injection which sends to the tuple which has 1 in the th component and 0 elsewhere induces a vector space homomorphism . This mapping is clearly surjective, and also clearly injective. So .

Certainly is an -vector space which contains , so that . Suppose these two dimensions are in fact equal. Identify with the usual basis of . By this previous exercise, there is a basis of which contains , and as argued above, . By our hypothesis, in fact and have the same cardinality, and so there exists a bijection . Now induces a vector space isomorphism .

However, note that , while . Since , we have a contradiction. Thus the dimension of is strictly smaller than that of .

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## Comments

How does |F| * |B| < |F|^|B| follow when |B| = |F|, can just use Cantor’s diagonalization argument, since |F| * |B| = max{|F|, |B|}.

That an infinite cartesian product increases the cardinality of a set is intuitive, at least.

My comment above was mangled. I guess the angle brackets were eaten as HTML. What I wrote was…

How does |F| * |B| less than |F|^|B| follow when |B| is smaller than |F|? When |B| is greater than or equal to |F|, we can just use Cantor’s diagonalization argument, since |F| * |B| = max{|F|, |B|}, so

|F| * |B| = |B| is less than 2^|B| is less than or equal to |F|^|B|

That an infinite cartesian product increases the cardinality of a set is intuitive, at least.