A soft question on the concept of cardinality and infinity As a layman, I think I have a notion of aleph-zero—that it is countable. I think I have a notion of aleph-one in the sense that it is uncountable (for example, I cannot tell what comes after a given real number).
To my extremely limited understanding, the real number are "more" than natural numbers, because whatever strategy you come up with to make a one-to-one correspondence, there will be infinitely many real numbers left out.
Without getting too technical, could someone please explain, how does one show that we cannot make a one-to-one correspondence between elements of a set of aleph-two with real numbers?
(For example, I remember reading in a book by Gamov, where he said that the cardinality of the set of all possible curves on a plane is higher than the cardinality of the set of real numbers. Is that possible to demonstrate without getting technical, like the diagonalization example when comparing natural numbers and real numbers?)
 A: Though the way you asked for it is confused, I believe the argument you need is this one:
The key result is:

If $A$ is a set and $\scr A$ is a collection of subsets of $A$ such that there is a one-to-one correspondence between $A$ and $\scr A$, then there is some subset $B \subseteq A$ with $B\notin \scr A$.

If $a \in A$, then let $S(a)$ be the corresponding element of $\scr A$. Define $B$ by $$B = \{a \in A\mid a \notin S(a)\}$$
Suppose $C \in \scr A$. There must be some $c \in A$ with $S(c) = C$. There are two possibilities:

*

*if $c \in C$, then $c \in S(c)$ and therefore $c \notin B$ by definition. Therefore $C \ne B$ since $C$ contains an element that $B$ does not contain.

*if $c \notin C$, then $c \notin S(c)$ and therefore $c \in B$ by definition. Once again, $C \ne B$, since $B$ contains an element that $C$ does not contain.

Thus $B$ is not equal to any element of $\scr A$. I.e., $B \notin \scr A$. QED.

This is actually a generalization of Cantor's diagonal argument. $\big(c,S(c)\big)$ represents the diagonal of $A \times \scr A$, and we've chosen $B$ to disagree with each entry along the diagonal. You may also note I made no use of the one-to-one-ness of the correspondence. It actually works for any correspondence between $A$ and $\scr A$ where each element of $A$ corresponds to only one set in $\scr A$, and every element of $\scr A$ corresponds to at least one element of $A$ (a "surjection from $A$ to $\scr A$").
The set $\mathscr P(A) = \{C \mid C \subseteq A\}$ of all subsets of $A$ is called the "power set of $A$". A corollary of the above result is

There can be no one-to-one correspondence between $A$ and $\mathscr P(A)$.

If there were, there would be some $B \subseteq A$ with $B \notin \mathscr P(A)$. But $\mathscr P(A)$ contains every subset of $A$, so that cannot be. But $A$ is obviously in one-to-one correspondence with the set $\big\{ \{a\}\mid a \in A\big\} \subset \mathscr P(A)$. Thus we can say

For any set $A$, $|A| < |\mathscr P(A)|$

where $|A|$ indicates the cardinality of $A$. But $\mathscr P(A)$ is also a set, so we can say $$|\mathscr P(A)| < |\mathscr P(\mathscr P(A))|.$$
In particular,
$$|\Bbb N| < |\mathscr P(\Bbb N)| < |\mathscr P(\mathscr P(\Bbb N))| < |\mathscr P(\mathscr P(\mathscr P(\Bbb N)))| < \dots$$
Thus there must be infinitely many infinite cardinals.

Now $\aleph_0 = |\Bbb N|$ by definition. But also by definition $\aleph_1$ is the smallest cardinal $> \aleph_0$ (any non-empty collection of cardinals will always have a smallest element). Similarly $\aleph_2$ is defined to be the smallest cardinal $> \aleph_1$, and so on.
So as Asaf Karagila pointed out, the strict answer to your question is $\aleph_2$ cannot be put into one-to-one correspondence with $\aleph_1$ because by definition it is a larger cardinal that $\aleph_1$. What the argument above shows is that there actually is a cardinal $\aleph_1 > \aleph_0$ and actually is a cardinal $\aleph_2 > \aleph_1$, etc.
We know that $|\Bbb R| = |\mathscr P(\Bbb N)|$. What we do not know (and what makes your question confused) is whether $\aleph_1 = |\mathscr P(\Bbb N)|$. It must be that $\aleph_1 \le |\mathscr P(\Bbb N)|$, since it is the smallest cardinal above $\aleph_0 = |\Bbb N|$. But whether or not there are any cardinals between $\aleph_0$ and $|\mathscr P(\Bbb N)| = |\Bbb R|$ is actually an undecidable question. You can assume it to be true or assume it to be false, and never run into a contradiction either way.
