# Are cardinal numbers well defined? Or could we have something like $\aleph_{1/2}$? [duplicate]

From what I understood, cardinal numbers are defined as:

$\aleph_0$ = the cardinality of $\mathbb{N}$
$\aleph_{n+1}$ = is the least cardinal number greater than $\aleph_n$

The continuum hypothesis states that the cardinality of $\mathbb{R}$ is $\aleph_1$. I more or less understand this cannot be proven (does not follow from ZFC).

However, can it be proven dat there in fact is a 'least cardinal number' greater than $\aleph_0$?

Is it possible that there are sets with cardinality $\aleph_{1/2}$, $\aleph_{1/3}$, et cetera, so to speak?

Or expressed differently: is it possible that for every set A with cardinality CA > $\aleph_0$, there exists another set B with cardinality CB where CA > CB > $\aleph_0$?

## marked as duplicate by 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(); } ); }); }); Nov 15 '15 at 12:03

• I suspect it might be duplicated several times. – Hanul Jeon Nov 15 '15 at 11:37
• Have you heard about ordinal numbers? – Hanul Jeon Nov 15 '15 at 11:38
• I think if you could answer the question in the negative then you'd have resolved the CH, which we know cannot be resolved. So I'm going to say the answer is no this cannot be proven. – Gregory Grant Nov 15 '15 at 11:39
• @Gregory: No, the answer is always negative. – Asaf Karagila Nov 15 '15 at 12:03

Cardinal numbers are special ordinal numbers, and the class of ordinals is well-ordered. So it does make sense to talk about the first ordinal that is not in a bijective correspondence with $\omega$, this is a unique and well-defined ordinal number and is defined to be $\aleph_1$. By definition all smaller ordinals are countable, so there is no intermediate cardinality between $\aleph_0$ and $\aleph_1$. The same goes for $\aleph_\alpha$ and $\aleph_{\alpha+1}$, mutatis mutandis.
More precisely, $\aleph_0$ is defined as the least infinite ordinal, called $\omega$, and is same as the set of all natural numbers set-theoretically. $\aleph_1$ is defined as the least uncountable ordinal (in other words, least ordinal whose cardinailty is not less than $\omega$) $\omega_1$, and $\aleph_2$ is defined as the least ordinal whose cardinality is not less than $\omega_1$ and so on. $\aleph_1$ is the least cardinal greater than $\aleph_0$, and there is no cardinal between them.
I haven't heard about such as $\aleph_{1/2}$, and I strongly suppose such objects are not treated in set theory. However as far as I know, the surreal number contains 'fractional ordinal' such as $\omega_{1/2}$. If you want, it might be okay to say that $\aleph_{1/2}$. Of course, they are not real cardinals and no set has such 'cardinality'.