# What is the Cardinality of the euclidean space $\mathbf{R}^N$?

We know that the set of real numbers is not countable by Cantor's proposition and hence higher dimensional Euclidean space is not countable too. However I couldn't find any result about the cardinality of $\mathbf{R}^N$? Is it $\aleph_{n}$ or it has the same cardinality of $\mathbf{R}$ which is $\aleph_1$?

• The cardinality of $\mathbb{R}$ is not $\aleph_0$, which is the cardinality of $\mathbb{N}$, but $\mathfrak{c}$, which may or may not be $\aleph_1$. The cardinality of $\mathbb{R}^N$ is also $\mathfrak{c}$. – Daniel Fischer Oct 5 '13 at 15:01

The cardinality of $\mathbb N$ is $\aleph_0$. The cardinality of $\mathbb R$ is definitely not $\aleph_0$. It is usually denoted $\mathfrak c$ (continuum) and coincides with the cardinality $2^{\aleph_0}$ of the Cantor set. Since $\aleph_0\times n=\aleph_0\times\aleph_0=\aleph_0$ for any $n$ with $0<n<\aleph_0$, we see that the cardinality of $\mathbb R^n$ and even the cardinality of $\mathbb R^{\mathbb N}$ is $\mathfrak c$ again.

The point is that $(2^{\aleph_0})^n=2^{\aleph_0\times n}=2^{\aleph_0}$, and similarly for $(2^{\aleph_0})^{\aleph_0}$.

By the way, that the Euclidean spaces all have size continuum is due to Cantor himself. His proof consisted of interleaving the decimal expansions of two reals to produce another real, thus giving an injection of $(0,1)\times(0,1)$ into $(0,1)$. This, and the circumstances surrounding this result, is discussed in detail in this nice paper by Fernando Q. Gouvêa.

The technique of forcing can be used to show that the usual axioms of set theory do not suffice to decide whether $2^{\aleph_0}$ is $\aleph_1$, $\aleph_2$, some other $\aleph_n$, or something else. (Cantor's famous Continuum Hypothesis is the claim that $2^{\aleph_0}=\aleph_1$. Again, both that this is true and that this is false are statements unprovable from the standard axioms.)

You can prove that the cardinality of $\Bbb{R}$ and $\Bbb{R}^n$ are same. Also, you can prove that the cardinality of $\Bbb{R}$ is $2^{\aleph_0}$. But it is done. You can't decide what $n$ satisfies $2^{\aleph_0}=\aleph_n$. In fact, it is known that $\mathsf{ZFC}+(2^{\aleph_0}=\aleph_n)$ is consistent if $n\ge 1$ (If $\mathsf{ZFC}$ is consistent.) That means, you can additionally assume that $2^{\aleph_0}=\aleph_n$ and it makes no contradiction (unless $\mathsf{ZFC}$ derives a contradiction.)

• sorry for my mistake , i have updated my question by replace $\aleph_0$ with $\aleph_1$. – behrad mahboobi Oct 5 '13 at 15:06
• can you bring up the injective mapping between $\Bbb{R}$ and $\Bbb{R}^N$ that acknowledge your statement ? – behrad mahboobi Oct 5 '13 at 15:09
• I'm no set theorist, but I believe it is provable that $2^{\aleph_0} \not= \aleph_\omega$, so you may want to be precise about what $n$ is ranging over here. – Ben Millwood Oct 5 '13 at 16:06
• – Ben Millwood Oct 5 '13 at 16:07