A question about infinity orders in set theory [duplicate]

I would like to ask this question and i hope it is not obvious or rediculus.

We know about cardinal numbers in set theory and that this ''numbers'' are equivalence classes.

In other words two sets are equivalent(or equipotent) if there exists a bijection from one set to another.This relation is an equivalence relation.

Thus the equivalence classes are the cardinal numbers.

We also know that there are infinitely many cardinal numbers.

For instance $\aleph_0<c<2^c<2^{2^c}<....<$ where $c=2^{\aleph_0}$

My question is that can we form a set that has as its elements all cardinal numbers?

If such set exists then what is its cardinality?(maybe it does not exist because in some way we may derive the existence of the set of all sets..im not sure)

My question is not an exact duplicate of the mentioned question.In the question Cardinality of the set of all cardinals the user assumes that there exist a set of all cardinals and i ask if there exists such set.

marked as duplicate by Noah Schweber, Asaf Karagila♦ cardinals StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 20 '17 at 15:51

• With that definition of cardinal numbers, even a single cardinal is not a set. – user228113 Feb 20 '17 at 15:12
• what is the proper definition of a cardinal other than an equivalence class? – Marios Gretsas Feb 20 '17 at 15:13
• An equivalent definition in ZFC is: $\operatorname{card}X$ is an ordinal $\alpha$ such that (1) there exists a bijection $f:\alpha\to X$ and (2) no proper initial segment of $\alpha$ can be put into bijection with $X$. With this definition, there is no set $Y$ which contains all the cardinals, because its union $\bigcup Y$ would be a set containing all the ordinals. That can't be. – user228113 Feb 20 '17 at 15:17

Using initial ordinals : modulo $AC$, mathematicians only use "initial ordinals" to define cardinals, that is, ordinals that aren't equipotent with any lesser ordinal. With that definition, using $AC$, any set is equipotent with one unique initial ordinal, i.e. cardinal. Then it's a classical theorem to show that there is no set of all cardinals : any set of cardinals is bounded above.
However one can use a trick to smooth it up a bit : using the Von Neumann hierarchy. With the axiom of foundation, one can show that for any $x$, there is an ordinal $\alpha$ such that $x\in V_\alpha$, and then one can say : for any $x$, let $\alpha$ be the least ordinal such that there exists $y\in V_\alpha$ equipotent with $x$, and then you can let the equivalence class of $y$ in $V_\alpha$ for the relation "there exists a bijection between" be denoted by $card(x)$. This makes cardinals into sets, and has the same expression power as your first definition. And then, with this definition, one can answer your question in the same way : no there isn't a set of all cardinals. Indeed, for any distinct initial ordinals $\alpha,\beta$, one would have $card(\alpha) \neq card(\beta)$, and thus if there were a set of all cardinals, there would be a set of all initial ordinals, which is a contradiction.
Notes : - for any ordinal $\alpha$, there is a least ordinal equipotent with $\alpha$ (indeed, there are ordinals equipotent with $\alpha$, and so by the well ordering of ordinals, there is a least one). Ordinals that aren't equipotent with any strictly smaller ordinal are initial ordinals (usually cardinals, but as here there is a risk of confusion, I used the word "initial ordinal"). There isn't a set of all initial ordinals -the Von Neumann hierarchy is defined as follows, by induction : for any ordinal $\alpha$, let $V_\alpha = \displaystyle\bigcup_{\beta <\alpha} \mathcal{P}(V_\beta)$, where $\mathcal{P}(X)$ is the powerset of $X$. Assuming the axiom of foundation, for all $x$, there is $\alpha$ with $x\in V_\alpha$.