# A simpler proof that every basis of a vector space has the same cardinality

I'm trying to prove this well-known result. My proof is very simple that I suspect I made some subtle mistakes. Could you have a check on it?

Let $$V$$ be a linear space. A set $$B \subseteq V$$ is called a basis of $$V$$ if each $$x\in V$$ is a finite linear combination of elements in $$B$$ and if each finite subset of $$B$$ is linearly independent. By axiom of choice, such $$B$$ always exists. Then $$\dim V := \operatorname{card} B$$ is independent of the choice of $$B$$.

My attempt: Let $$C$$ be another basis of $$V$$. We now define a map $$T:B \to C$$.

• For each $$b \in B$$, $$\mathbb Rb$$ is a vector subspace of $$V$$, then there is $$c_b\in C$$ such that $$\mathbb Rc_b=\mathbb Rb$$. Notice that $$c_b$$ is linearly independent of each element in $$C\setminus\{c_b\}$$. Hence such $$c_b$$ is unique. We define $$T(b):=c_b$$.

• Assume that $$T(b_1)=T(b_2)$$. Then $$\mathbb Rb_1 =\mathbb Rb_2$$. Notice that $$b_1$$ is linearly independent of each element in $$B\setminus\{b_1\}$$, so $$b_1=b_2$$ and thus $$T$$ is injective. It follows that $$\operatorname{card} B \le \operatorname{card} C$$.

• By symmetry, we have $$\operatorname{card} C \le \operatorname{card} B$$. This completes the proof.

Update: As mentioned in the comment, above proof is not correct. I fix it as follows.

WLOG, we assume $$B$$ and $$C$$ are infinite. Let $$F_B, F_C$$ denote the collections of all finite subsets of $$B$$ and $$C$$ respectively. By axiom of choice, there is a map $$T:B \to F_C$$ such that $$b$$ is a linear combination of vectors in $$T(b)$$. We have $$V=\operatorname{span} B \subseteq \operatorname{span} \left ( \bigcup_{b\in B} T(b) \right )\subseteq V.$$

Because $$C$$ is linearly independent, it is the only subset of itself that spans $$V$$. So $$\bigcup_{b\in B} T(b) = C.$$

We have $$|C| = \left | \bigcup_{b\in B} T(b) \right | \le |B| \cdot \aleph_0 = |B|.$$ By symmetry, $$|B| \le |C|$$. This completes the proof.

• why is there a $c_b \in C$ such that $c_bV = bV$? Jan 24, 2022 at 11:07
• No, $\Bbb Rb$ being a subspace does not imply there exists $c\in C$ with $\Bb Rb=\Bbb Rc$. You can take $c$ to be a linear combination of elements of $C$ (for example $c=b$). Jan 24, 2022 at 11:08
• Why is the downvote??? Jan 24, 2022 at 11:13
• Your proof fails already for $\Bbb R^2$. When a claim is supposed to be true for every linear space, it's a good litmus test to try it for some trivial examples. Jan 24, 2022 at 11:16
• The downvote probably has to do with the proof being totally wrong... Jan 24, 2022 at 11:42

The assertion “there is a $$c_b\in C$$ such that $$\Bbb Rc_b=\Bbb Rb$$” is false. Take $$V=\Bbb R^2$$, $$B=\{(1,0),(0,1)\}$$, and $$C=\{(1,1),(1,-1)\}$$. There is no $$c\in C$$ such that $$\Bbb Rc=\Bbb R(1,0)$$.
• Same problem. Consider $V$, $B$, and $C$ as in my answer and take $M=\{(1,0)\}\in F_B$. Then $\operatorname{span}(M)=\Bbb R(1,0)$ and there is no subset $N$ of $C$ such that $\operatorname{span}(N)=\Bbb R(1,0)$. Jan 24, 2022 at 12:13