Relationship between $\aleph$ and the symbol $|A|$ or $\operatorname{card}(A)$ Aleph ($\aleph$) is the symbol used in mathematics to indicate the cardinality of the numerable. We know that an infinite set has cardinality $\aleph_0$ if there exists a bijection that relates it biunivocally to the set $ \mathbb {N}$ to $ \mathbb {N}$ of the natural numbers.
When I give the definition of cardinality of a set $A$, I use, for the students of an high school (14 years-old) the symbol $|A|$ or $\operatorname{card}(A)$.

Just a curiosity: Is there a correlation between the cardinality of $A$ and the symbol $\aleph$ (or $\aleph_0$)  (i.e. is it the same notation and the same significance)?

PS: I prefer a possible answer. Thank you very much.
 A: Aleph numbers are used to denote infinite cardinalities.
Before we go any further, we need to discuss some basic facts about cardinalities.
Given sets $A$, $B$, we say $A \approx B$ exactly when there is a bijection $A \to B$. For every set $A$, there is a cardinal $|A|$. Every cardinal can be expressed as $|A|$ for some $A$. And for any sets $A, B$, we have $|A| = |B|$ if and only if $A \approx B$.
Given two cardinals $|A|$ and $|B|$, we say $|A| \leq |B|$ if and only if there exists some injection $A \to B$. We can verify that $\leq$ is transitive and reflexive using the fact that the identity map $1_A : A \to A$ is injective and using the fact that the composition of injective maps is injective. The Schroder-Bernstein theorem further states that $\leq$ is a partial ordering on cardinals. In other words, if $|A| \leq |B|$ and $|B| \leq |A|$ then $|A| = |B|$.
When $\kappa$ and $\lambda$ are cardinals, we say that $\kappa < \lambda$ if and only if both $\kappa \leq \lambda$ and $\kappa \neq \lambda$.
Using the axiom of choice, we can further prove that $\leq$ is a total ordering. That is, given any cardinals $\lambda$ and $\kappa$, either $\lambda \leq \kappa$ or $\kappa \leq \lambda$.
In fact, the axiom of choice allows us to prove that $\leq$ is a well-ordering. This means that given any nonempty collection of cardinals $F$, there is a smallest element of $F$.
There are plenty of obvious cardinals that we know how to deal with. These obvious cardinalities are the cardinalities of finite sets. It's easy for us to check which of two finite sets is larger, and many operations of "cardinal arithmetic" turn out to be ordinary math when we are working with finite cardinals.
But what about the infinite cardinals? Are there any cardinals which are not finite?
This is where we require the axiom of infinity, which allows us to prove that there is set $\mathbb{N}$ of all natural numbers. $\mathbb{N}$ is infinite, so $|\mathbb{N}|$ is larger than any finite cardinal.
We can also prove, using the axiom of countable choice, that if $B$ is any set which is not finite, then $|\mathbb{N}| \leq |B|$. This means that $|\mathbb{N}|$ is the smallest infinite cardinal.
Since $|\mathbb{N}|$ is an infinite cardinal and there are $0$ infinite cardinals smaller than $|\mathbb{N}|$, we define $\aleph_0 = |\mathbb{N}|$.
The next question is: are there any other infinite cardinalities? In order to answer this question, we require the axiom of power sets, which state that power sets exist.
We can use Cantor's Diagonalisation Argument to prove that if $S$ is any set, then $|S| < |P(S)|$. This proves in particular that there is no largest cardinality.
In particular, we can expand Cantor's argument to prove that given any set $S$ of cardinals, there is a cardinal $\kappa$ which is bigger than all the cardinals in $S$. This argument is somewhat technical.
Given any cardinal $\lambda$, we can consider the class $\mathcal{F}$ of all cardinals $\kappa$ such that $\kappa > \lambda$. We know that there is at least one $\kappa \in \mathcal{F}$, since there is no biggest cardinal. Therefore, because cardinals are well-ordered, there is a smallest element of $\mathcal{F}$. Call this element $\lambda^+$. Then $\lambda^+$ is the smallest cardinal $> \lambda$.
Now in particular, let's define $\aleph_1 = \aleph_0^+$. The only infinite cardinal $\kappa$ such that $\kappa < \aleph_1$ is $\kappa = \aleph_0$, since if we had some other $\kappa$, then we would have $\aleph_0 < \kappa < \aleph_1$ which contradicts the very definition of $\aleph_1 = \aleph_0^+$. That is, there is only 1 infinite cardinal smaller than $\aleph_1$.
Similarly, we can define $\aleph_2 = \aleph_1^+$, $\aleph_3 = \aleph_2^+$, and, in general, $\aleph_{n + 1} = \aleph_n^+$ for all $n \in \mathbb{N}$. We can show that for all $n$, there are exactly $n$ infinite cardinals smaller than $\aleph_n$.
So we've defined an infinite collection of infinite cardinals, $\aleph_n$ for all natural numbers $n$. The next question is: is there a cardinal which is even larger than all the $\aleph_n$s we've defined so far?
It turns out that we need the axiom of replacement to prove that the answer is "yes". Using the axiom of replacement, we can form the infinite set $S = \{\aleph_n \mid n \in \mathbb{N}\}$. But recall that given any set of cardinals $S$, we can form a cardinal $\kappa$ which is bigger than any element of $S$. So there is some cardinal which is even bigger than all the $\aleph_n$s where $n \in \mathbb{N}$. Call the smallest such cardinal $\aleph_\omega$. Note that there are exactly $|\omega| = |\mathbb{N}|$ cardinals smaller than $\aleph_\omega$.
Once we have $\aleph_\omega$, it's only natural to define $\aleph_{\omega + 1} = \aleph_\omega^+$. Note that there are exactly $|\omega + 1| = |\omega|$ infinite cardinals smaller than $\aleph_\omega$.
So once we get to infinite $\alpha$s, we can't just define $\aleph_\alpha$ as the infinite cardinal such that there are exactly $|\alpha|$ smaller infinite cardinals, since there are exactly $|\omega|$ infinite cardinals smaller than $\aleph_\omega$, and there are also $|\omega|$ infinite cardinals smaller than $\aleph_{\omega + 1}$.
Instead, we consider the fact that given an infinite cardinal $\kappa$, the set $S_\kappa := \{\lambda \mid \lambda $ an infinite cardinal, $\lambda < \kappa\}$ is well-ordered by $<$. What matters is not $|S_\kappa|$ - what actually matters is the order-type of $(S_\kappa, <)$.
To illustrate this, consider that $S_{\aleph_\omega} = \{\aleph_n \mid n \in \mathbb{N}\}$ has a very simple order $<$. We have $\alpha_a < \aleph_b$ if and only if $a < b$ as natural numbers. In other words, the order type of $(S_{\aleph_\omega}, <)$ is exactly the order type of the natural numbers, which is written as $\omega$.
By contrast, consider that $S_{\omega + 1} = S_\omega \cup \{\aleph_\omega\}$. This set's order type is $\omega + 1$, since the set can be broken down into $S_\omega$, which has order type $\omega$, and an element $\aleph_\omega$ which is bigger than all the elements of $S_\omega$.
Given every well-ordered set $S$, there is another well-order $type(S)$ such that $type(S)$ and $S$ are equivalent well-orders. Furthermore, $type(A) = type(B)$ if and only if $A$ and $B$ are well-orders. We say that the "ordinals" are the well-orders of the form $type(S)$.
So in general, we have the following definition:

Given an ordinal $\alpha$, the cardinal $\aleph_\alpha$, if it exists, is defined to be the infinite cardinal such that $(S_{\aleph_\alpha}, <)$ has order type $\alpha$.

Note that we can prove that if there are two cardinals $\kappa, \lambda$ such that $(S_\kappa, <)$ has the same order type as $(S_\lambda, <)$, then we can prove by induction on that order type that $\kappa = \lambda$. Thus, we know that there is at most one cardinal satisfying the definition of $\aleph_\alpha$ for all ordinals $\alpha$.
From this definition, it follows immediately that every infinite cardinal is an aleph. In particular, given an infinite cardinal $\kappa$, let $\alpha$ be $type(S_\kappa, <)$. Then it's immediate that $\kappa = \aleph_\alpha$.
We can prove using the axiom of replacement that for every ordinal $\alpha$, $\aleph_\alpha$ exists. The proof uses a technique called "ordinal induction".
We have thus discovered that the ordinals and infinite cardinals can be put into correspondence using the $\aleph$ function. This is a profound and philosophically satisfying result.
