# Why is the $\aleph$ hierarchy a discrete hierarchy? [duplicate]

The $\aleph$ hierarchy, I think, is a sequence of infinite cardinals. $\aleph_0$ is $|\mathbb{N}|$, possibly by definition. $\aleph_1$ is the next highest cardinal.

A lot has been written about whether $\aleph_1 = |\mathbb{R}|$ or, equivalently $\aleph_1 = \beth_1$ and how the continuum hypothesis is independent of ZFC.

But there definitely isn't a cardinal $\kappa$ such that $\aleph_0 < \kappa < \aleph_1$.

Why do we know that there's a successor cardinal to $\aleph_0$ / least cardinal greater than $\aleph_0$?

What would break if we insist, by fiat, on the existence of a hierarchy of cardinals indexed by the non-negative reals or non-negative rationals (in the way that $\aleph_n$ is indexed by $n \in \mathbb{N_0}$)? If cardinals aren't primitive things and the existence of a successor falls right out of how they're defined, this question might not make sense.

## 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(); } ); }); }); Sep 19 '18 at 6:54

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• cardinals are ordinals; ordinals are well-ordered – Lord Shark the Unknown Sep 19 '18 at 6:19
• @LordSharktheUnknown true only if assuming choice – Holo Sep 19 '18 at 6:24
• @LordSharktheUnknown When I think of cardinality, I think of equivalence classes on sets based on the existence of bijections. I think of a "cardinal" or "cardinality" as a name for one of the classes. That's probably naive. Are cardinals defined as particular ordinals in the ordinal hierarchy? Is the fact that bijections exist between same-sized sets just another result and not fundamentally related to how cardinality is defined? – Gregory Nisbet Sep 19 '18 at 6:28
• @GregoryNisbet do you assume axiom of choice? If not than your definition is the correct way to think about cardinals. But shark is correct with a little change in either case. I'll post an answer about this – Holo Sep 19 '18 at 6:31
• @Holo - Well, if the question is about $\kappa$ such that $\aleph_0 < \kappa < \aleph_1$, then Choice isn't the issue. There is just no such cardinal, since $\kappa < \aleph_1$ implies that $\kappa$ is an $\aleph$ number. What the failure of Choice might do is a bit stranger. – Malice Vidrine Sep 19 '18 at 6:39

## 1 Answer

First: assuming the axiom of choice we have that the class of cardinals is equal to the union of all limit ordinals hence equal to $On$(or $Ord$) thus it is well ordered.

If we do not assume choice we need to go back to the definition of $\aleph_a$, we have $\aleph_0=\omega$, and $\aleph_{a+1}=\aleph_a^+$ and if $a$ is limit ordinal then $\aleph_a=\bigcup\{ \aleph_b\mid b< a\}$, now because ordinals are well founded there exists unique successor cardinal for every aleph number, thus aleph one is well defined

• Your first paragraph confuses me. – Asaf Karagila Sep 19 '18 at 7:14