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.

  • $\begingroup$ why is there a $c_b \in C$ such that $c_bV = bV$? $\endgroup$
    – C Bagshaw
    Jan 24, 2022 at 11:07
  • $\begingroup$ 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 $c=b$). $\endgroup$ Jan 24, 2022 at 11:08
  • $\begingroup$ Why is the downvote??? $\endgroup$
    – Akira
    Jan 24, 2022 at 11:13
  • 1
    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Jan 24, 2022 at 11:16
  • $\begingroup$ The downvote probably has to do with the proof being totally wrong... $\endgroup$ Jan 24, 2022 at 11:42

1 Answer 1


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 $\Bbb Rc=\Bbb R(1,0)$.

  • $\begingroup$ I have just fixed the proof. Could you have a check on my fix? $\endgroup$
    – Akira
    Jan 24, 2022 at 12:09
  • 1
    $\begingroup$ 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 $\operatorname{span}(N)=\Bbb R(1,0)$. $\endgroup$ Jan 24, 2022 at 12:13
  • $\begingroup$ I have fixed it again. Could you please have a re-look? $\endgroup$
    – Akira
    Jan 24, 2022 at 21:31
  • 1
    $\begingroup$ It looks fine now, although there is no need to mention the axiom of choice here. $\endgroup$ Jan 24, 2022 at 21:38
  • $\begingroup$ Thank you so much for your dedicated help! $\endgroup$
    – Akira
    Jan 24, 2022 at 21:38

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