Over a countable field, any two bases of an infinite dimensional vector space have the same cardinality

Let F be a field whose cardinality is countable (that is, finite or countably infinite). Let V be an infinite dimensional vector space over F having basis B. prove that V and B have the same cardinality.


Note that V = \oplus_B F. Certainly we have \mathsf{card}\ B \leq \mathsf{card}\ V, since B \subseteq V.

Let X be an infinite set. Given a natural number n, denote by \mathcal{P}_n(X) the set of all subsets of X having cardinality n. Note that \mathsf{card}\ X \leq \mathsf{card}\ \mathcal{P}_n(X) \leq \mathsf{card} X^n = \mathsf{card}\ X, since X is infinite. Denote by \mathcal{P}_F(X) = \bigcup_\mathbb{N} \mathcal{P}_n(X) the set of all finite subsets of X; we also have \mathsf{card}\ X \leq \mathsf{card}\ \mathcal{P}_F(X) = \mathsf{card}\ \bigcup_\mathbb{N} \mathcal{P}_n(X) = \sum_\mathbb{N} \mathsf{card}\ X = \mathsf{card}\ X. (This sum denotes a cardinal sum.) Thus the set of all finite subsets of X has the same cardinality as X.

Now every element of V is uniquely an F-linear combination of finitely many elements of B. Thus we have \mathsf{card}\ V \leq \mathsf{card}\ \bigcup_{T \in \mathcal{P}_F(B)} \prod_T F \leq \sum_{\mathsf{card}\ B} \mathsf{card}\ F = \mathsf{card}\ B.

Thus we have \mathsf{card}\ V = \mathsf{card}\ B.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: