Cardinality: Why is there no "ℵ½"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:09:55Z http://mathoverflow.net/feeds/question/25132 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25132/cardinality-why-is-there-no Cardinality: Why is there no "ℵ½"? NevilleDNZ 2010-05-18T14:47:52Z 2010-05-18T16:04:54Z <p>A wikipedia page/paragraph on <a href="http://en.wikipedia.org/wiki/Aleph-null#Aleph-one" rel="nofollow">ℵ₁</a> states: </p> <ol> <li>"The definition of ℵ₁ implies (in ZF, <a href="http://en.wikipedia.org/wiki/Zermelo-Fraenkel_set_theory" rel="nofollow">Zermelo-Fraenkel set theory</a> without the axiom of choice) that no cardinal number is between ℵ₀ and ℵ₁."</li> <li>"If the <a href="http://en.wikipedia.org/wiki/Axiom_of_choice" rel="nofollow">axiom of choice</a> (AC) is used, it can be further proved that the class of cardinal numbers is totally ordered, and thus ℵ₁ is the second-smallest infinite cardinal number."</li> </ol> <p>Can someone point me at a lay explanation of this please?</p> <p>Is it simply saying that ℵ½'s existence is up to definition, or choice? And this has been shown by the <a href="http://en.wikipedia.org/wiki/Axiom_of_choice" rel="nofollow">axiom of choice</a>?</p> http://mathoverflow.net/questions/25132/cardinality-why-is-there-no/25135#25135 Answer by Andrej Bauer for Cardinality: Why is there no "ℵ½"? Andrej Bauer 2010-05-18T15:01:17Z 2010-05-18T15:01:17Z <p>It says the following: consider all cardinal numbers greater than $\aleph_0$. Among them there is a smallest one, let us call it $\aleph_1$.</p> <p>If you are confused by the notation, change $\aleph_0$ to $X$ and $\aleph_1$ to $Y$. Then the text reads:</p> <blockquote> <p>Let $X$ be the least infinite cardinal number, i.e., the cardinality of the set of natural numbers. Then there is a cardinal number $Y$ which is greater than $X$ and such that there is not cardinal number between $X$ and $Y$.</p> </blockquote> http://mathoverflow.net/questions/25132/cardinality-why-is-there-no/25136#25136 Answer by Tom Boardman for Cardinality: Why is there no "ℵ½"? Tom Boardman 2010-05-18T15:01:32Z 2010-05-18T15:01:32Z <p>One can prove using ZF the existence and uniqueness of a first uncountable ordinal $\omega_1$, this is by definition $\aleph_1$. See <a href="http://en.wikipedia.org/wiki/Hartogs_number" rel="nofollow">http://en.wikipedia.org/wiki/Hartogs_number</a>.</p> http://mathoverflow.net/questions/25132/cardinality-why-is-there-no/25152#25152 Answer by Joel David Hamkins for Cardinality: Why is there no "ℵ½"? Joel David Hamkins 2010-05-18T15:54:06Z 2010-05-18T16:04:54Z <p>The point is that without the Axiom of Choice, cardinalities are not linearly ordered, and it is possible under $\neg AC$ that there are additional cardinalities to the side of the $\aleph$'s. Thus, the issues is not additional cardinalities between $\aleph_0$ and $\aleph_1$, but rather additional cardinalities to the side, incomparable with these cardinalities. </p> <p>Let me explain. We say that two sets $A$ and $B$ are <em>equinumerous</em> or <em>have the same cardinality</em> if there is a bijection $f:A\to B$. We say that $A$ has smaller-or-equal cardinality than $B$ if there is an injection $f:A\to B$. It is provable (without AC) that $A$ and $B$ have the same cardinality if and only if each is smaller-or-equal to the other (this is the Cantor-Shroeder-Bernstein theorem). </p> <p>Under AC, every set is bijective with an ordinal, and so we may use these ordinals to select canonical representatives from the equinumerosity classes. Thus, under AC, the $\aleph_\alpha$'s form all of the possible infinite cardinalities. </p> <p>But when AC fails, the cardinalities are not linearly ordered (the linearity of cardinalities is equivalent to AC). Let me mention a few examples:</p> <ul> <li><p>It is a consequence of the Axiom of Determinacy that there is no $\omega_1$ sequence of distinct reals. Thus, in any model of AD, the cardinality of the reals is uncountable, but incomparable to $\aleph_1$. Thus, in such a model, it is no longer correct to say that $\aleph_1$ is the smallest uncountable cardinal. One should say instead that $\aleph_1$ is the smallest uncountable well-orderable cardinal.</p></li> <li><p>A more extreme example is provided by the Dedekind finite infinite sets. These sets are not finite, but also not bijective with any proper subset. It follows that they can have no countably infinite subsets. In particular, they are uncountable sets, but their cardinality is incomparable with $\omega$. Thus, in a model of $\neg AC$ having a Dedekind finite infinite set, it is no longer correct to say that $\aleph_0$ is the smallest infinite cardinal. </p></li> </ul> <p>Thus, the issue isn't whether there is something between $\aleph_0$ and $\aleph_1$, but rather, whether there are additional cardinalities to the side of these cardinalities.</p>