I'm currently trying to understand the basic notions concerning infinity. I think I understand that $\aleph_0\cdot\aleph_0=\aleph_0$ but how about $\aleph_1$? Is $\aleph_1\cdot\aleph_1=\aleph_1$ i.e. is there a bijection between a line and the plane?
I did some research, but I couldn't find anything. I can't imagine such a bijection, but I couldn't find a prove for the contrary. Any help will be greatly appreciated.
$\aleph_1$ is by definition the cardinality of the set of all countable ordinals.
The cardinality of the line is $2^{\aleph_0}$. Whether $\aleph_1 = 2^{\aleph_0}$ is not an easy question to answer. Cantor conjectured that it is. In Zermelo--Fraenkel set theory without the axiom of choice, it is at least consistent that $\aleph_1$ and $2^{\aleph_0}$ are not comparable, i.e. they are not equal and neither is less than the other. With the axiom of choice, one can prove that $\aleph_1\le 2^{\aleph_0}$.
Cantor proved, however, that $(2^{\aleph_0})^2 = 2^{\aleph_0}$, so there is a bijection between the line and the plane.
Among continuous surjections from the line to the plane are Peano curves. Google that term. Getting a bijection from that can be done in a number of ways, but no bijection can be a continuous function from the line to the plane.
As was pointed out $\aleph_1$ is the second smallest cardinal, which is not necessarily the cardinality of the line i.e. it is consistant with $\textsf{ZF}$ and $\textsf{ZFC}$ that $\aleph_1 \neq 2^{\aleph_0}$.
It is worth noting that $\aleph_\alpha \cdot \aleph_\alpha = \aleph_\alpha$ for all ordinal $\alpha$. (see theorem 2.1 of Hrbacek and Jech "Introduction to Set Theory). If one assumes the axiom of choice than $\mathfrak{c}:= 2^{\aleph_0}$ is an $\aleph$ therefore $\mathfrak{c} = \mathfrak{c}^2$, i.e. there exists a bijection between the line and the plane.
However it should be stressed that the existance of a bijection between the line and the plance can also be obtained without $\textsf{AC}$ using a the following different approach. In fact Cantor proved a far stronger result in $\textsf{ZF}$: for any positive integer $n$, there exists a 1-to-1 correspondence between the points on the line and all of the points in an $n$-dimensional space.
$$ \mathfrak{c} = 2^{\aleph_0} = 2^{\aleph_0 + \cdots +\aleph_0} = 2^{\aleph_0}\cdots 2^{\aleph_0} = \mathfrak{c}^n$$
About this discovery Cantor wrote to Dedekind: "I see it, but I don't believe it!"
In fact, given any aleph $\aleph_\alpha,$ we have $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha.$
We can prove this by transfinite induction. Suppose $\kappa$ is the least aleph for which the claim has not yet been determined to hold. Since $\kappa\precsim\kappa\cdot\kappa$, (here, $\precsim$ indicates "is injectable into") then it suffices to show that $\kappa\cdot\kappa\precsim\kappa$.
For each ordinal $\beta<\kappa$, then we may define the $\beta$th section of $\kappa\times\kappa$ recursively by \begin{equation*} S_\beta = (\beta+1)\times(\beta+1) \smallsetminus \left(\bigcup_{\xi<\beta}S_\xi\right). \end{equation*} Since each $S_\beta$ is a subset of $(\beta+1)\times(\beta+1)$, then we can well-order each in lexicographic order, so that $|S_\beta|$ is a well-orderable cardinal for all $\beta<\kappa$. Moreover, since the $S_\beta$ are pairwise-disjoint, and indexed by the well-ordered set $\kappa$, then the union of any set of $S_\beta$ sections is also well-orderable. In particular, $\kappa\times\kappa$ is the union of all the $S_\beta$, and so can be well-ordered, say by $\sqsubset$.
Take any $\langle\gamma,\delta\rangle\in\kappa\times\kappa$, and set $\zeta=\max\{\gamma,\delta\}+1$. Since $\max\{\gamma,\delta\}<\kappa$ and $\kappa$ is a limit ordinal, then $|\zeta|\leq\zeta<\kappa$. Note that $\langle\gamma,\delta\rangle\in\zeta\times\zeta$, so that $\langle\gamma,\delta\rangle$ has no more than $|\zeta\times\zeta|=|\zeta|\cdot|\zeta|$ $\sqsubset$-predecessors in $\kappa\times\kappa$. If $\zeta$ is finite, then this is only finitely-many $\sqsubset$-predecessors. If $\zeta$ is infinite, then $|\zeta|$ is an infinite cardinal less than $\kappa$, so by inductive hypothesis, there are no more than $|\zeta|\cdot|\zeta| = |\zeta| \prec \kappa$ $\sqsubset$-predecessors of $\langle\gamma,\delta\rangle$ in $\kappa\times\kappa$. Since this holds for all $\langle\gamma,\delta\rangle\in\kappa\times\kappa$, then the order type of $\kappa\times\kappa$ under $\sqsubset$ is an ordinal $\alpha\le\kappa$. Hence, $\kappa\cdot\kappa=|\kappa\times\kappa|=|\alpha|\le\alpha\le\kappa,$ and so $\kappa\cdot\kappa\precsim\kappa,$ as desired.
Added: If one assumes the Continuum Hypothesis, then one finds that the line and the plane have the same cardinality as a results of the work above, as the line has cardinality $\aleph_1$. Alternatively, if one assumes the Axiom of Choice, then the cardinality of the line is an aleph (though not necessarily $\aleph_1$), and again we find that the line and the plane have the same cardinality by the work above. Unfortunately, it may be that the line is not well-orderable otherwise! In that case, we must proceed more directly, by (for example) demonstrating that $A\times A\precsim A$ for some set $A$ having the same cardinality as the line. (off the top of my head, it seems like if we take $A$ to be the Cantor ternary set, then an interleaving map using ternary expansions should do the trick, but I haven't sat down to verify that.)
Let $\mathfrak c$ be the cardinal of the Euclidean line $\mathbb R$. Then $\mathfrak c^2 = \mathfrak c \cdot \mathfrak c$ is the cardinal of the Euclidean plane $\mathbb R^2$. Using some basic cardinal arithmetic, we can compute $$ \mathfrak c = 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0}\cdot 2^{\aleph_0} = \mathfrak c \cdot \mathfrak c $$
note
This does not require the Axiom of Choice. (But AC is required to show every cardinal is an aleph. Which, in turn, yields another proof of
$\mathfrak c = \mathfrak c^2$, as you find in another answer.)
