Is $\aleph_0$ "bigger", or "smaller" than $\omega$? What sets are of cardinality $\omega$ or $\aleph_0$? I am new to set theory and need a little clarification.
Cardinals are generalization of natural number, so they can tell us how big a set is. After we run out of natural numbers 1,2,3,... 10000000,...... we continue with $\aleph_0$, $\aleph_1$, $\aleph_3$,...¨
The $\aleph_0$ is called the first transfinite cardinal number.
Then there is $\omega$ - the lowest transfinite ordinal number, according to Wikipedia.
My questions are:

*

*Which of $\aleph_0$ and $\omega$ is "bigger" in terms of cardinality?


*$\aleph_0$ is cardnality of natural numbers. What are some examples of sets that have cardinality $\omega$, or also cardinality $\aleph_0$?


*$\mathbb{R}$ has cardinality $\mathfrak{c}$ - "infinite cardinal number". I know $\mathfrak{c}$ > $\aleph_0$, but what about $\omega$ and $\aleph_1$, $\aleph_2$ etc.?
Thank you in advance. If I get something in this concept completely wrong, please correct me so I can edit the question accrodingly.
 A: *

*$\omega$ considered as an ordinal is not "bigger in terms of cardinality" since $\omega$ is an ordinal, $\aleph_0$ is a cardinal - the two represent different things, and cannot be compared like this. Note: see Rob Arthan’s comment about how $\omega$ has the cardinal $\aleph_0$ associated with it, and in a certain context it is valid to say $\omega=\aleph_0$, but always bear in mind a difference between ordinal and cardinal - a difference in usage, since I have learnt now that formally we take $\omega$ and $\aleph_0$ to be the very same object.

*No sets have cardinality $\omega$, since $\omega$ is not a cardinal. Examples of "countable" sets (i.e. have cardinality lesser or equal to $\aleph_0$) are the evens, the odds, the integers, the rationals, the algebraic numbers,... Note: same note as in $1)$ - see the comments.

*$\frak{c}$ is an infinite cardinal, but so is $\aleph_0$. The idea that $\frak{c}$ is the second smallest infinite cardinal is called the continuum hypothesis, which I believe is not provable in ZF. $\aleph_1=\frak{c}$ is the continuum hypothesis.

A set which has $\omega$ as its greatest ordinal has cardinality $\aleph_0$. However, $\omega+1\gt\omega$, but the set associated with this does not naively have cardinality $\aleph_0+1\gt\aleph_0$. See Hilbert's Hotel for a classic description of this. Two sets have the same cardinality if and only if there exists a bijection between them. Two sets have the same ordinals (I think) if each element can be uniquely attached a different ordinal. In this way, strangely, $1+\omega=\omega$ but $\omega+1\gt\omega$. See ordinal arithmetic.
I should also add for the sake of your understanding that all sets described by the maximum ordinal of $\omega,\omega+1,\omega+2,\omega+3,\omega+4,\omega+5,\cdots,2\omega,3\omega,\cdots,\omega^2,\omega^3,\cdots,\omega^\omega,\cdots,\omega^{\omega^\omega},\cdots$ have cardinality $\aleph_0$ - they are countable ordinals. Moreover, $\omega^{\omega^{\omega^{\omega^\cdots}}}=\varepsilon_0$ is also a countable ordinal. It is only when a set has $\Omega=\omega_1$ as its largest ordinal that the set has a cardinality greater than $\aleph_0$. This set is the set of all countable ordinals, and an example of this (if you take the continuum hypothesis to be true) is the real numbers.
Even more unintuitively (for some), the set of reals has the same cardinality as the interval $(0,1)$. In fact, $\mathbb{R}^2,\mathbb{R}^3,\mathbb{R}^4,\cdots$ all have the same cardinality, the cardinality of the continuum $\frak{c}$. We know this because there are ways to construct bijections between all of these.
P.S. any expert reading this who knows a proof / a link to a proof that $\mathbb{R}^n$ has a bijection with $\mathbb{R}$, it would be much appreciated if you could comment a link!
A: $\omega$ and $\aleph_0$ are the same set (the natural numbers $\{0,1,2,3,\ldots\}$) by different names. $\omega$ stresses the fact that we see it as an ordinal number (the smallest one that is infinite) while $\aleph_0$ is a name that says it is the first (i.e. of index $0$) in the transfinite sequence of cardinal numbers. All cardinals are special ordinal numbers, which in turn are transitive sets well-ordered by $\in$. All this is standard modern set theory (assuming ZFC standard axioms, including well-foundedness). As to the continuum $\mathfrak{c}$ we can say that $\mathfrak{c} \ge \aleph_1$ (it's uncountable) but not much more. Also, there is a theorem that $\text{cf}(\mathfrak{c})$ (the cofinality) cannot be $\omega$, so that $\mathfrak{c}=\aleph_{\omega}$ is not possible; but other than that restriction it can be any uncountable aleph (consistently).
So it's more correct to say sets of cardinality $\aleph_0$ (not cardinality $\omega$, though one does see it occasionally). Ordinals are for "order things", cardinals for "sizes" of sets.
