If $[R:Q]$ is cardinality equal to $R$?

Question

We know real number set $\mathbb R$ as a linear space over rational numbers $\mathbb Q$ is infinite dimensional, if $[\mathbb R:\mathbb Q]$ is cardinality equal to $\mathbb R$?

We can do field extension for $\mathbb Q$ to get $\mathbb R$, I think do countable extension canont get $\mathbb R$ ,so the cardinal of $[\mathbb R:\mathbb Q]$ is strictly greater than the cardinality of $\mathbb Q$.

I have another guess: maybe the problem is like "Continuum hypothesis" is undecidable?

Answer 1

The answer depends on whether or not we accept the axiom of choice.

With choice, Ned’s comment is the solution. Take $B$ to be a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Every point in $\mathbb{R}$ can be written as a $\mathbb{Q}$-linear combinations of points of $B$. The number of such combinations is given by $\sum_{n \in \mathbb{N}}|(\mathbb{Q} \times B)^n|$. Since we have choice, for all $n \in \mathbb{N}$ we have that $| (\mathbb{Q} \times B)^n| = |\mathbb{Q} \times B| = \max\{|\mathbb{Q}|, |B|\}$. It follows that $|\mathbb{R}| = \sum_{n \in \mathbb{N}} \max\{|\mathbb{Q}|, |B|\} = \max\{|\mathbb{Q}|, |B|\}$. Thus we may conclude $|\mathbb{R}| = |B|$, which is equivalent to saying that $[\mathbb{R}:\mathbb{Q}] = |\mathbb{R}|$. This cardinal arithmetic argument relies on choice though.

The rules we used to make this argument are:

1. For infinite sets $A,B$, we have $|A \times B| = |A + B| = \max\{A \times B\}$, where $+$ is disjoint union.
2. $\sum_{n \in \mathbb{N}} |A| = |\mathbb{N}| \times |A|$

Without choice, there is no guarantee that such a basis $B$ exists. Even if we assume there is a basis, I think there are models of set theory in which the cardinality of $B$ cannot be pinned down as $|\mathbb{R}|$.