Continuum Hypothesis says that every infinite cardinal smaller than $2^{\aleph_0}$ is equal to $\aleph_0$. In the other hand I've heard that Martin Axiom says that these cardinals are "similar" in some sense (and that's a quote from Polish Wikipedia). What exactly does it mean? I couldn't find anything about it.
In what sense does Martin’s axiom say that cardinals between $\aleph_0$ and $2^{\aleph_0}$ are “similar” to $\aleph_0$
cardinalsset-theory
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There is a general argument without choice: Suppose ${\mathfrak m}+{\mathfrak m}={\mathfrak m}$, and ${\mathfrak m}+{\mathfrak n}=2^{\mathfrak m}$. Then ${\mathfrak n}=2^{\mathfrak m}.\,$ This gives the result.
The argument is part of a nice result of Specker showing that if CH holds for both a cardinal ${\mathfrak m}$ and its power set $2^{\mathfrak m}$, then $2^{\mathfrak m}$ is well-orderable. This shows that GCH implies choice, and that the proof is "local". It is still open whether CH for ${\mathfrak m}$ implies that ${\mathfrak m}$ is well-orderable.
Anyway, here is the proof of the statement above: Note first that $2^{\mathfrak m}\cdot 2^{\mathfrak m}=2^{{\mathfrak m}+{\mathfrak m}}=2^{\mathfrak m}={\mathfrak m}+{\mathfrak n}$.
Let $X$ and $Y$ be disjoint sets with $|X|={\mathfrak m}$, $|Y|={\mathfrak n}$, and fix a bijection $f:{\mathcal P}(X)\times{\mathcal P}(X)\to X\cup Y$.
Note that there must be an $A\subseteq X$ such that the preimage $f^{-1}(X)$ misses the fiber $\{A\}\times {\mathcal P}(X)$. Otherwise, the map that to $a\in X$ assigns the unique $A\subseteq X$ such that $f^{-1}(a)$ is in $\{A\}\times {\mathcal P}(X)$ is onto, against Cantor's theorem.
But then, for any such $A$, letting $g(B)=f(A,B)$ gives us an injection of ${\mathcal P}(X)$ into $Y$, i.e., $2^{\mathfrak m}\le {\mathfrak n}$. Since the reverse inclusion also holds, we are done by Schroeder-Bernstein.
(Note the similarity to Apostolos's and Joriki's answers.)
The original reference for Specker's result is Ernst Specker, "Verallgemeinerte Kontinuumshypothese und Auswahlaxiom", Archiv der Mathematik 5 (1954), 332–337. A modern presentation is in Akihiro Kanamori, David Pincus, "Does GCH imply AC locally?", in "Paul Erdős and his mathematics, II (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.
Note that assuming that ${\mathfrak m}$ is infinite is not enough for the result. For example, it is consistent that there are infinite Dedekind finite sets $X$ such that ${\mathcal P}(X)$ is also Dedekind finite. To be Dedekind finite means that any proper subset is strictly smaller. But if $2^{\mathfrak m}$ is Dedekind finite and $2^{\mathfrak m}={\mathfrak n}+{\mathfrak l}$ for nonzero cardinals ${\mathfrak n},{\mathfrak l}$, then we must have ${\mathfrak n},{\mathfrak l}<2^{\mathfrak m}$.
To prove the existence of $\aleph_1$ we use the concept of Hartogs number. The question asks, really, why are there uncountable ordinals, since $\aleph_1$ is by definition the least ordinal which is not countable.
Take a set of cardinality $\aleph_0$, say $\omega$. Now consider all the orders on $\omega$ which are well-orders, and consider the order isomorphism as an equivalence relation. The collection of all equivalence classes is a set.
Fact: If $(A,R)$ is a well-ordered set, then there exists a unique ordinal $\alpha$ such that $(A,R)\cong(\alpha,\in)$.
Map every equivalence class to the unique ordinal which is order isomorphic to the members of the class. We now have a set and all its members are ordinals which correspond to possible well-ordering of $\omega$.
Fact: The union of a set of ordinals is an ordinal, it is in fact the supremum of the elements in the union.
Let $\alpha$ be the union of the set defined above. We have that $\alpha$ is an ordinal, and that every ordinal below $\alpha$ is a possible well-ordering of $\omega$ (and therefore countable).
Suppose $\alpha$ was countable too, then $\alpha+1$ was also countable (since $\alpha+1=\alpha\cup\{\alpha\}$), and therefore a possible well ordering of $\omega$. This would contradict the above fact that $\alpha$ is greater or equal to all the ordinals which correspond to well-orderings of $\omega$, since $\alpha<\alpha+1$.
This means that $\alpha$ is uncountable, and that it is the first uncountable ordinal, since if $\beta<\alpha$ then $\beta$ can be injected into $\omega$, and so it is countable. Therefore we have that $\alpha=\omega_1=\aleph_1$.
Note that the above does not require the axiom of choice and holds in $\sf ZF$. The collection of all well-orders is a set by power set and replacement, so is the set of equivalence classes, from this we have that the collection of ordinals defined is also a set (replacement again), and finally $\alpha$ exists by the axiom of union. There was also no use of the axiom of choice because the only choice we had to do was of "a unique ordinal" which is a definable map (we can say when two orders are isomorphic, and when a set is an ordinal - without the axiom of choice).
With the axiom of choice this can be even easier:
From the axiom of choice we know that the continuum is bijectible with some ordinal. Let this order type be $\alpha$, now since the ordinals are well-ordered there exists some $\beta\le\alpha$ which is the least ordinal which cannot be injected into $\omega$ (that is there is no function whose domain is $\beta$, its range is $\omega$ and this function is injective).
From here the same argument as before, since $\gamma<\beta$ implies $\gamma$ is countable, $\beta$ is the first uncountable ordinal, that is $\omega_1$.
As to why there is no cardinals strictly between $\aleph_0$ and $\aleph_1$ (and between any two consecutive $\aleph$-numbers) also stems from this definition.
- $\aleph_0 = |\omega|$, the cardinality of the natural numbers,
- $\aleph_{\alpha+1} = |\omega_{\alpha+1}|$, the cardinality of the least ordinal number which cannot bijected with $\omega_\alpha$,
- $\aleph_{\beta} = \bigcup_{\alpha<\beta}\aleph_\alpha$, at limit points just take the supremum.
This is a function from the ordinals to the cardinals, and this function is strictly increasing and continuous. Its result is well-ordered, i.e. linearly ordered, and every subset has a minimal element.
This implies that $\aleph_1$ is the first $\aleph$ cardinal above $\aleph_0$, i.e. there are no others between them.
Without the axiom of choice, however, there are cardinals which are not $\aleph$-numbers, and it is consistent with $\sf ZF$ that $2^{\aleph_0}$ is not an $\aleph$ number at all, and yet there are not cardinals strictly between $\aleph_0$ and $2^{\aleph_0}$ - that is $\aleph_0$ has two distinct immediate successor cardinals.
For the second question, there is no actual limit. Within the confines of a specific model, the continuum is a constant, however using forcing we can blow up the continuum to be as big as we want.
This is the work of Paul Cohen. He showed that you can add $\omega_2$ many subsets of $\omega$ (that is $\aleph_2\le 2^{\aleph_0}$), and the proof is very simple to generalize to any higher cardinal.
In fact Easton's theorem shows that if $F$ is a function defined on regular cardinals, which has a very limited set of constraints, then there is a forcing extension where $F(\kappa) = 2^\kappa$, so we do not only violate $\sf CH$ but we violate $\sf GCH$ ($2^{\aleph_\alpha}=\aleph_{\alpha+1}$) in a very acute manner.
Best Answer
Roughly speaking, Martin's axiom says that many statements that can be proven about countable sets by simple diagonalization arguments also hold for sets of all cardinalities below $2^{\aleph_0}$. Here is a typical example.
Two subsets of $\omega$ are called almost disjoint if their intersection is finite. Consider the following simple theorem.
Now you might ask, what if $S$ is not countable? Well, clearly the theorem cannot remain true without some additional constraint on $S$, since you could take $S$ to be a maximal infinitely family of pairwise almost disjoint subsets of $\omega$ (which exists by Zorn's lemma, say). In particular, the theorem cannot be true for all $S$ of cardinality $2^{\aleph_0}$. But what if $S$ is uncountable, but $|S|<2^{\aleph_0}$? The proof of the theorem certainly wouldn't apply to $S$; trying to continue the construction of $(n_k)$ transfinitely runs into trouble already at $k=\omega$, since the $n_j$ you have chosen already could exhaust the entire complement of $\bigcup_{j<k} A_j$.
However, Martin's axiom implies that the theorem does apply to such $S$. In this sense, Martin's axiom says that sets of size less than $2^{\aleph_0}$ behave "as though they were countable" for these purposes.
Of course, this raises the question of exactly which "purposes" Martin's axiom applies to. The exact statement is quite technical (and probably won't mean much to you if you aren't familiar with forcing). It says that if $P$ is a poset satisfying the countable chain condition (given any uncountable subset $A\subseteq P$, there exist distinct $a,b\in A$ and $c\in P$ such that $c\leq a$ and $c\leq b$) and $S$ is a collection of fewer than $2^{\aleph_0}$ dense subsets of $P$ (a subset $D$ of $P$ is called dense if for each $a\in P$, there exists $b\in D$ such that $b\leq a$), then there exists a filter on $P$ which intersects every element of $S$. In the case that $S$ is countable, it is easy to construct such a filter by a diagonalization argument (pick a descending sequence of elements of each set in $S$ one by one, and take the filter they generate). So, this says that in a wide variety of situations (anything that can be encoded by finding a filter in a ccc poset meeting dense sets), sets of fewer than $2^{\aleph_0}$ conditions can be simultaneously satisfied in the same way that countable sets of conditions can be simultaneously satisfied by building an object to meet the conditions one by one.
Here's another example that may be more familiar. By the Baire category theorem, any intersection of countably many open dense subsets of $\mathbb{R}$ is dense. A consequence of Martin's axiom is that this is also true for intersections of fewer than $2^{\aleph_0}$ open dense subsets of $\mathbb{R}$.