Set Theory HT25, Vector spaces

Created: March 08, 2025 | Updated: April 10, 2025 | About these notes | View in context | Study these flashcards

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In ZFC, we consider a vector space $V$ over a field $K$ to be two sets equipped with functions

  • $+, \cdot : K \times K \to K$
  • $+ : V \times V \to V$
  • $\cdot : K \times V \to V$

with the standard definitions of linear independence, spanning sets and bases.

@Prove that every vector space has a basis and any two bases have the same cardinality.

Every vector space has a basis:

Consider the set $\mathcal I \subseteq \mathcal P(V)$ of linearly independent subsets of $V$ as a partial order, ordered by inclusion, $\subseteq$. If $C \subseteq \mathcal I$ is a chain, then its union is also linearly independent, since any finite number of elements of $\bigcup C$ are already elements of some $S \in C$.

Hence by Zorn’s lemma, there exists a maximal element $B \in \mathcal I$.

It remains to show that $B$ is spanning. Suppose not, so that there is some $v \in V \setminus \langle B \rangle$. But then $B \cup \{v\}$ is linearly independent, and hence $v \notin B$, which contradicts the maximality of $B$.

All bases have the same cardinality:

Suppose $B$ and $B’$ are bases. If $B$ or $B’$ is finite, then this is the regular proof in the finite dimensional case. Hence, suppose that $B$ and $B’$ are infinite.

Let

\[P = \mathcal P^{<\omega}(B) := \\{B_0 \subseteq B : |B_0| < \aleph_0\\}\]

be the set of finite subsets of $B$.

Claim: $ \vert P \vert = \vert B \vert $

Since $P = \bigcup _ {n \in \mathbb N} B^{(n)}$ where

\[B^{(n)} := \\{B_0 \subseteq B : |B_0| = n\\}\]

and $B^n \to B^{(n)}$ given by $(b _ 1, \ldots, b _ n) \mapsto \{b _ 1, \ldots, b _ n\}$ is a surjection we have $ \vert B^{(n)} \vert \le \vert B^n \vert = \vert B \vert $. Hence $ \vert P \vert \le \vert B \vert $ by the result that says

If:

  • $\kappa$ is an infinite cardinal
  • $X$ is a set (here $P$)
  • $ \vert X \vert \le \kappa$ (here $ \vert B \vert $)
  • $ \vert a \vert \le \kappa$ for all $a \in X$ (here $ \vert a \vert \le \aleph _ 0 \le \vert B \vert $)

Then:

  • $ \vert \bigcup X \vert \le \kappa$ (so $ \vert P \vert \le \vert B \vert $)

and $ \vert B \vert \le \vert P \vert $ e.g. by the injection sending each element to the singleton.

Claim: $ \vert B’ \vert \le \vert B \vert $

If $B _ 0 \in P$, then $\langle B _ 0 \rangle \cap B’$ is finite by the finite-dimensional case. But then

\[V = \langle B \rangle = \bigcup \\{\langle B_0 \rangle : B_0 \in P\\}\]

so

\[B' = \bigcup \\{\langle B_0 \rangle \cap B' : B_0 \in P\\}\]

is a union of $ \vert P \vert = \vert B \vert $ finite sets, and so by the above result again, $ \vert B’ \vert \le \vert B \vert $.

By symmetry, $ \vert B \vert \le \vert B’ \vert $ also, and hence $ \vert B’ \vert = \vert B \vert $.



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