The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measurable ...

Feb 22, 2025 · $\begingroup$ @DaveLRenfro The “obvious” statement I was point out was that $\exists\lambda.(\kappa=\lambda^{\aleph_0}) \;\Leftrightarrow\; \kappa = \kappa^{\aleph_0}$: the $\Leftarrow$ direction is trivial, but the $\Rightarrow$ direction is ever-so-slightly less obvious, as it consists of saying that $(\lambda^{\aleph_0})^{\aleph_0} = \lambda^{\aleph_0}$, something which indeed everyone ...

The answer is yes: there are $2^{\aleph_0}$ countable Boolean algebras up to isomorphism, or equivalently $2^{\aleph_0}$ homeomorphism class of metrizable totally disconnected compact Hausdorff spaces. This is the main result of: Reichbach, M.

Jul 22, 2024 · In your case, you have $\kappa=2^{\aleph_0}$, and so $\kappa^\omega=(2^{\aleph_0})^{\aleph_0}$, which is the same as $2^{\aleph_0}$. Although it is true that GCH simplifies cardinal arithmetic, and it is useful to consider a complex calculation first under the case of GCH as a rough estimation, it is the usual practice not to assume GCH in an ...

Sep 12, 2022 · It's also worth noting that the situation you're interested in is provable from $\mathsf{ZFC+MA+2^{\aleph_0}>\aleph_{\omega}}$, the point being that $\mathsf{MA}$ gives $2^\kappa=2^{\aleph_0}$ for every $\kappa<2^{\aleph_0}$. Of course, it takes effort (= iterated forcing) to show that that theory is consistent in the first place.

The category of Banach spaces and contractions (over the reals or any other complete normed field, I think) is an example of an $\aleph_{1}$-presentable category which is not $\aleph_{0}$-presentable.

Jul 19, 2020 · You are confusing several notions of homogeneity. Saturated structures, and therefore also $\aleph_0$-categorical structures, are homogeneous, but not necessarily ultrahomogeneous. This means that every finite partial elementary mapping …

Sep 23, 2024 · Note that the question trivialises when we drop "unstable", since the trivial reduct is $\aleph_0$-categorical. Note also that it is enough to consider reducts with only one relational symbol, since that's all we need to witness unstability.

Aug 23, 2023 · There exists an $\aleph_0$-universal partially ordered system. Moreover, if the Generalized Continuum Hypothesis holds, and if $\alpha$ is any positive ordinal, then there exists an $\aleph_\alpha$-universal partially ordered system.

Jul 15, 2020 · This is a theorem of Rosenstein from the paper $\aleph_0$-categoricity of groups. But note that he uses the terminology bounded order rather than finite exponent. I say that a group is $\aleph_0$-categorical if its complete theory in the language of groups is $\aleph_0$-categorical (i.e., has a unique countable model).