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Joel David Hamkins
Joel David Hamkins
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The point is that without the Axiom of Choice, the cardinality concept iscardinalities are not so tidy. With AC, every set is bijective with a unique smallest ordinallinearly ordered, and we may takeit is possible under $\neg AC$ that ordinal asthere are additional cardinalities to the cardinalityside of the set. The cardinal numbers $\aleph_\alpha$ arise in this way as all$\aleph$'s. Thus, the possibleissues is not additional cardinalities ofbetween (well-orderable) infinite sets$\aleph_0$ and $\aleph_1$, but rather additional cardinalities to the side, incomparable with these cardinalities.

Without AC, however, there can be uncountable sets that are not well-orderable, and hence not bijective with any $\aleph_\alpha$Let me explain. In the non-AC context, weWe say that two sets $A$ and $B$ have the same cardinalityare equinumerous or have the same cardinality if there is a bijection $f:A\to B$. This is equivalent to the assertionWe say that there is an injection $f:A\to B$ and another injection $g:B\to A$. The associated partial pre$A$ has smaller-order on cardinalities is thator-equal cardinality than $B$ is at least as large as $A$ if there is an injection $f:A\to B$.

Now, the point is that without AC, this order It is not linear.provable (The assertionwithout AC) that any two sets are comparable in$A$ and $B$ have the same cardinality if and only if each is actually equivalentsmaller-or-equal to AC.the other (this is the Cantor-Shroeder-Bernstein theorem).

To give some examples, it is a consequence of the Axiom of Determinacy that there is no $\omega_1$ sequence of distinct reals. Thus, in any model of ADUnder AC, the cardinality of the realsevery set is uncountablebijective with an ordinal, but incomparableand so we may use these ordinals to $\aleph_1$select canonical representatives from the equinumerosity classes. Thus, in such a modelunder AC, it is no longer correct to say that $\aleph_1$ is the smallest uncountable cardinal. One should say instead that $\aleph_1$ is$\aleph_\alpha$'s form all of the smallest uncountable well-orderable cardinalpossible infinite cardinalities.

A more extreme example is provided byBut when AC fails, the Dedekind finite infinite sets. These setscardinalities 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 withlinearly ordered $\omega$. Thus, in a model(the linearity of $\neg AC$ having a Dedekind finite infinite set, itcardinalities is no longer correctequivalent to say that $\aleph_0$ is the smallest infinite cardinalAC). Let me mention a few examples:

  • 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.

  • 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.

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.

The point is that without the Axiom of Choice, the cardinality concept is not so tidy. With AC, every set is bijective with a unique smallest ordinal, and we may take that ordinal as the cardinality of the set. The cardinal numbers $\aleph_\alpha$ arise in this way as all the possible cardinalities of (well-orderable) infinite sets.

Without AC, however, there can be uncountable sets that are not well-orderable, and hence not bijective with any $\aleph_\alpha$. In the non-AC context, we say that two sets $A$ and $B$ have the same cardinality if there is a bijection $f:A\to B$. This is equivalent to the assertion that there is an injection $f:A\to B$ and another injection $g:B\to A$. The associated partial pre-order on cardinalities is that $B$ is at least as large as $A$ if there is an injection $f:A\to B$.

Now, the point is that without AC, this order is not linear. (The assertion that any two sets are comparable in cardinality is actually equivalent to AC.)

To give some examples, 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.

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.

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.

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.

Let me explain. We say that two sets $A$ and $B$ are equinumerous or have the same cardinality 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).

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.

But when AC fails, the cardinalities are not linearly ordered (the linearity of cardinalities is equivalent to AC). Let me mention a few examples:

  • 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.

  • 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.

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.

Source Link
Joel David Hamkins
Joel David Hamkins
  • 236.5k
  • 44
  • 777
  • 1.4k

The point is that without the Axiom of Choice, the cardinality concept is not so tidy. With AC, every set is bijective with a unique smallest ordinal, and we may take that ordinal as the cardinality of the set. The cardinal numbers $\aleph_\alpha$ arise in this way as all the possible cardinalities of (well-orderable) infinite sets.

Without AC, however, there can be uncountable sets that are not well-orderable, and hence not bijective with any $\aleph_\alpha$. In the non-AC context, we say that two sets $A$ and $B$ have the same cardinality if there is a bijection $f:A\to B$. This is equivalent to the assertion that there is an injection $f:A\to B$ and another injection $g:B\to A$. The associated partial pre-order on cardinalities is that $B$ is at least as large as $A$ if there is an injection $f:A\to B$.

Now, the point is that without AC, this order is not linear. (The assertion that any two sets are comparable in cardinality is actually equivalent to AC.)

To give some examples, 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.

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.

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.