# Prove that any 2 bases of a vector space has the same cardinality

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.

Edit:

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

• 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. Jun 19, 2020 at 10:30
• @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 $<\aleph_\beta$. Jun 19, 2020 at 10:35
• @AsafKaragila You must be thinking of a different method to mine..... Jun 19, 2020 at 10:43

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.)

• 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) Jun 19, 2020 at 11:32