Bijection between infinite bases of vector spaces

[andytoh]

I am reading "The linear algebra a beginning graduate student ought to know" by Golan, and I encountered a puzzling statement:

Let V be a vector space (not necessarily finitely generated) over a field F. Prove that there exists a bijective function between any two bases of V. Hint: Use transfinite induction.

If V is generated by a finite set (with n elements), then I know how to prove that any basis has at most n elements, and thus all bases will have the same number of elements. But for infinite-dimensional vector spaces, I'm confused. How do I use transfinite induction to prove that there is a bijective correspondence between two bases of V if V is infinite-dimensional?

Sorry: I moved this to the algebra forum.

 

[andytoh]

I think I have a solution now. Here it is. Opinions are welcomed.

 

[tacman]

The statement that there exists a bijective function between any two bases of a vector space V is known as the Steinitz Exchange Lemma. This lemma is a fundamental result in linear algebra and plays a crucial role in understanding the structure of vector spaces.

To prove this lemma, we can use transfinite induction. Transfinite induction is a generalization of mathematical induction that allows us to prove statements for infinite sets. In this case, we want to prove that for any two bases B and B' of V, there exists a bijective function f: B → B'.

First, we consider the case when V is generated by a finite set (with n elements). In this case, we can use the proof you mentioned to show that any basis has at most n elements. This means that any two bases of V will have the same number of elements, and hence there exists a bijective function between them.

Now, let's consider the case when V is infinite-dimensional. We will use transfinite induction to show that there exists a bijective function between any two bases of V.

Step 1: Base case
Consider the base case when V is generated by a set with one element. In this case, any basis of V will also have one element. Hence, there exists a bijective function between any two bases of V.

Step 2: Inductive hypothesis
Assume that for any vector space W generated by a set with k elements, there exists a bijective function between any two bases of W.

Step 3: Inductive step
Now, let V be a vector space generated by a set with k+1 elements. By the definition of a basis, any basis of V will have at most k+1 elements. This means that we can remove one element from the basis without losing the property of being a basis. Let's call this element v. Then, the remaining k elements form a basis of V. By our inductive hypothesis, there exists a bijective function between any two bases of V.

Step 4: Conclusion
Since we can always remove one element from a basis of V without losing the property of being a basis, we can use this bijective function to map the basis with k+1 elements to the basis with k elements. This shows that there exists a bijective function between any two bases of V.

In conclusion, by using transfinite induction, we have shown that for any vector space V, there exists a