2

Let $V$ be a vector space with bases $B_1$, $B_2$. For all $b\in B_1$ there exists $U_b\subset B_2$ such that $U_b$ is finite and $b\in span(U_b)$. Hence, $V=span(B_1)=span(\cup_{b\in B_1}U_b)$. Since $B_2$ is a minimal spanning set this implies that $B_2=\cup_{b\in B_1}U_b$. At this point, I use an intuition-istic argument to show that $|B_1|\leq |B_2|$. i.e $$|B_2|=|\cup_{b\in B_1}U_b|\leq_{(1)} \sum_{b\in B_1}|U_b|\leq_{(2)} \sum_{b\in B_1}|\aleph_0|=_{(3)}|B_1||\aleph_0|=_{(4)}|B_1|$$

Using the same argument we can show that $|B_2|\leq |B_1|$. Hence by Schröder-Bernstein, $|B_1|=|B_2|$.

I did a course on set theory once and would have come across the operations involved in cardinal arithmetic. However, that was a long time ago. Can someone care to state or elucidate the actual cardinal operations or theorems involved in the argument above? Thanks.

Asaf Karagila
  • 393,674
Jhon Doe
  • 2,902

1 Answers1

2

  1. $|\bigcup_{i\in I X_i}|\leq\sum_{i\in I}|X_i|$, this is because the sum is defined as the cardinality of a disjoint union, whereas the union may have overlapping components.

  2. Each $U_b$ is finite, so its cardinality is less than $\aleph_0$, and since cardinal arithmetic is monotonic, increasing the cardinals in the sum is not going to decrease the cardinality of sum itself.

  3. Cardinal arithmetic satisfies the rule that repeated summation is the same as multiplication. In other words, the disjoint union of $\{b\}\times\Bbb N$, for $b\in B$, is exactly the set $B\times\Bbb N$. So we can replace the sum by a product.

  4. Infinite cardinals are idemmultiple, i.e. $|X|^2=|X|$. And consequently, $|X|\cdot|Y|=\max\{|X|,|Y|\}$. If $X$ is infinite and $Y=\aleph_0$, then $|Y|\leq|X|$, so $|X|\cdot|Y|=|X|$.

Asaf Karagila
  • 393,674
  • Hello, just wanted to clarify that the same argument would also work for free modules with infinite bases right? – Jhon Doe Jun 15 '20 at 02:14
  • I think you need the free module to be over a PID or something, but to be honest, my algebra is very rusty at this point. – Asaf Karagila Jun 15 '20 at 06:42
  • I was referring to this link where cardinality arguments are used. https://math.stackexchange.com/questions/2005708/let-m-be-a-free-r-module-and-let-it-have-an-infinite-basis-then-all-bases. I just wanted to make sure. But thanks for the answer. – Jhon Doe Jun 15 '20 at 08:23
  • 1
    Well, the argument is the same, then the argument is the same. – Asaf Karagila Jun 15 '20 at 08:51