I know this question has been asked before, but I tried to prove it myself and I cant finish my prove because im not sure how to write the contradiction in a foraml and correct way.

Let V be a vector space, and $B_1$, $B_2$ an infinite bases. Assume by contradiction that $ ,|B_{1}|\neq|B_{2}| $. So assume that $ |B_{1}|<|B_{2}| $ without loss of generality. So let:

$ |B_{1}|=\aleph_{\alpha}<\aleph_{\beta}=|B_{2}| $

and let:

$ B_{1}=\left\{ u_{j}:j<\aleph_{\alpha}\right\} B_{2}=\left\{ v_{i}:i<\aleph_{\beta}\right\} $

now, for each $v_{i}\in B_{2} $ we will find $ \mathcal{C}_{i}\subseteq\aleph_{\alpha} $ and scalar's $c_j$ such that $ \sum_{j\in C_{i}}c_{j}u_{j}=v_{i} $

and for each $v_i\in B_2 $ define : $ \mathcal{D}_{i}=\left\{ u_{j}:j\in\mathcal{C}_{i}\right\} $

(all the vectors from $B_1$ such that $ \sum_{j\in C_{i}}c_{j}u_{j}=v_{i} $ )

So, it follows that for any $v_i\in B_2 $

$ \mathcal{D}_{i}\in\bigcup_{n\in\mathbb{N}}B_{1}^{n} $

So if I'll define $ \mathcal{D}=\left\{ \mathcal{D}_{i}:i<\aleph_{\beta}\right\} $ we will have:

$ \mathcal{D}\subseteq\bigcup_{n\in\mathbb{N}}B_{1}^{n} $

Also, we know that $ |\bigcup_{n\in\mathbb{N}}B_{1}^{n}|=|B_{1}|=\aleph_{\alpha} $ because all the sequences are finite. Therefore, $ |\mathcal{D}|\leq\aleph_{\alpha} $.

Now, I want to say that for any finite set $ D_i $ there will be an infinite vectors from $ B_2 $ that will share the same $ D_i $ and therefore they will be linear dependent. But I'm not sure how to express it in a correct formal way. If anyone could find a contradiction from the step i have left, it will be very helpful. Thanks in advance.


I think I found a contradiction. So, the are no more then $ \aleph_{\alpha} $ sets in $\mathcal D $ as I stated before. Now, In $ B_2 $ there are $ \aleph_{\beta} $ vectors, so if we will define a function $ f:B_{2}\to\mathcal{D} $ that maps each vector to the appropriate $ D_i $ it will not be injective, so we can define :

$ \mathcal{F}_{k}=\left\{ v\in B_{2}:f\left(v\right)=\mathcal{D}_{k}\right\} $

So it follows that $ B_{2}\subseteq\bigcup_{k<\aleph_{\alpha}}\mathcal{F}_{k} $

Now, notice that $ \bigcup_{k<\aleph_{\alpha}}\mathcal{F}_{k} $ is a union of $ \aleph_{\alpha} $ sets, such that any set has to be finite, because otherwise we'll have infinite vectors that use the same $ \mathcal{D}_{i} $ and therefore they would be linear dependent. So, we can conclude that:

$ |\bigcup_{k<\aleph_{\alpha}}\mathcal{F}_{k}|\leq|\dot{\bigcup_{k<\aleph_{\alpha}}}\mathcal{F}_{k}|\leq\aleph_{\alpha}\times\aleph_{\alpha}=\aleph_{\alpha} $

(because in each set there's finite numbers of vectors, obviously it smaller then $ \aleph_{\alpha} $ )

and therefore $ \aleph_{\beta}=|B_{2}|\leq\aleph_{\alpha} $ In contradiction to our assumption. I will be glad to hear what you think about it. Thanks

  • $\begingroup$ My approach would be to prove that $\aleph_\alpha$ vectors inside a vector space with basis of size $\aleph_\beta$ must fail to span it. $\endgroup$ Jun 19, 2020 at 10:30
  • $\begingroup$ @AnginaSeng: I don't see how you can prove this using cardinal arithmetic style of argument without also assuming that the field itself has size $<\aleph_\beta$. $\endgroup$
    – Asaf Karagila
    Jun 19, 2020 at 10:35
  • $\begingroup$ @AsafKaragila You must be thinking of a different method to mine..... $\endgroup$ Jun 19, 2020 at 10:43

1 Answer 1


Here is a proof, based on the same principles, but somewhat different presentation from what you might see elsewhere: $\DeclareMathOperator{\span}{span}$

We define $F\colon[B_1]^{<\omega}\to[B_2]^{<\omega}$, where $[X]^{<\omega}$ is the set of finite subsets of $X$.

$$F(X)=\min\{Y\mid X\subseteq\span(Y)\}$$

Claim. The function $F$ is well-defined.

Proof. Each $x\in X$ has a unique minimal finite set, $Y_x$, such that $x$ is a non-trivial linear combination of the elements of $Y_x$. So it is enough to look for subsets of $\bigcup_{x\in X}Y_x$. Moreover, if $X$ is a subset of $\span(Y)$ and $\span(Y')$, then $X\subseteq\span(Y)\cap\span(Y')$, but because $Y\cup Y'$ is linearly independent, it has to be that $X\subseteq\span(Y\cap Y')$. So indeed this is well-defined.

Claim. $F$ is finite-to-one.

Proof. If $Y\in[B_2]^{<\omega}$, then $\span(Y)$ is a finite dimensional subspace, and therefore can only contain finite linearly independent subsets, since $B_1$ is linearly independent, that means that only finitely many of its elements can lie in $\span(Y)$, so only finitely many finite subsets are mapped to $Y$.

Claim. $|B_1|=|B_2|$.

Proof. Define the equivalence relation on $B_1$ by $u\sim v\iff F(\{u\})=F(\{v\})$, then by the previous claim, each equivalence class is finite, and therefore $|B_1/{\sim}|=|B_1|$. Taking the union of each equivalence class, which is an element in $[B_1]^{<\omega}$, to its image under $F$, is now injective. Therefore $|B_1|\leq|[B_2]^{<\omega}|=|B_2|$.

Define the same in the other direction, i.e. $F'\colon[B_2]^{<\omega}\to[B_1]^{<\omega}$, etc., and we have that $|B_2|\leq|B_1|$. By Cantor–Bernstein we have equality. (Alternatively, assume that $|B_2|\leq|B_1|$, as you did, and finish one paragraph early.)

  • $\begingroup$ Thank you, I will try to understand your proof. Can you ready my edit in the post and tell me what you think about the contradiction I found (if it is a contradiction at all) $\endgroup$
    – FreeZe
    Jun 19, 2020 at 11:32

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