Examples for objects which are either finite or uncountable? (like a $\sigma$-algebra)

Question

We know that $\sigma$-algebras are either finite or uncountable (perhaps because their definition allows us to build uncountably many different new sets out of (disjoint) countably many, which we can choose out of an infinite $\sigma$-algebra). another example would be the power set $\mathcal{P}(S)$ for any set $S$.

What are other interesting examples from other areas of math?

Answer 1

The set $Y^X$ of functions from a set $X$ to a non-empty finite set $Y$. If $X$ is finite then $Y^X$ is finite, otherwise $Y^X$ is uncountable.

Answer 2

First, a somewhat serious if ad-hoc example. By Konig's theorem, the continuum has uncountable cofinality (in symbols, $cf(2^\omega)>\omega$). This has the following consequence: letting $\alpha$ be the ordinal such that $2^\omega=\aleph_\alpha$, either $\alpha=\lambda+n$ for some limit $\lambda$ and some finite $n$ or $\alpha$ is a limit cardinal of uncountable cofinality.

We can turn this into an example of the property you're looking at as follows. For an ordinal $\theta$, let $tail(\theta)$ be the least significant term in the Cantor normal form of $\theta$ (e.g. $tail(\omega^2+\omega\cdot 3=\omega\cdot 3$ and $tail(\omega_1)=\omega_1$). Then Konig's lemma implies that $tail(\alpha)$ is either finite or uncountable.

Specifically, we always have that $tail(\theta)$ is either finite or of the form $\omega^\beta\cdot c$ for some nonzero ordinal $\beta$ and some finite nonzero $c$. This means that $cf(\theta)=cf(\omega^\beta)$, and if $\beta$ is countable we have $cf(\omega^\beta)=\omega$. (Note that here ordinal exponentiation is being used.)

• Incidentally, via forcing we can pin down exactly what ZFC can prove about $tail(\alpha)$. This gets a bit technical, though - but it's still nice to know that it can be done.

While related to the powerset example, this is meaningfully more complicated: the proof of Konig's lemma is definitely nontrivial! Of course this example is quite ad hoc, though. Really it's just a rephrasing of two facts: that $cf(2^{\aleph_0})>\omega$ and that successor cardinals are regular (plus the relevant consistency results to rule out triviality).

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Now here's a rather silly class of examples: there are various "generalized finiteness" notions which are equivalent to true finiteness assuming (a weak fragment of) the axiom of choice, but which in ZF alone may have instances which are infinite. The best known example of this is Dedekind-finiteness: a set is Dedekind-finite iff every injection from it to itself is a surjection. It is consistent with ZF that there are infinite Dedekind-finite sets, and even infinite sets which cannot be partitioned into two infinite subsets ("amorphous" sets)!

At this point it's worth clarifying that "finite" here means "in bijection with some natural number."

If we interpret "uncountable" as meaning "having no injection to $\omega$," then these generalized finiteness notions give examples of what you want - e.g. we have:

Every Dedekind-finite set is either finite or uncountable.

Of course, this relies on stretching the definition of "uncountable" arguably to the point of absurdity: infinite Dedekind-finite sets don't admit injections from $\omega$, and amorphous sets don't even admit surjections to $\omega$.

Answer 3

A connected metrizable topological space is either a singleton or has at least has the continuum cardinality.

Answer 4

Let $c:\Bbb R\to X$ be a compactification of $\Bbb R.$ That is, $c$ is a continuous injection, $X$ is compact Hausdorff, $c[\Bbb R]=\{c(r): r\in \Bbb R\}$ is dense in $X,$ and $c:\Bbb R\to c[\Bbb R]$ is a homeomorphism.

If $|X\setminus c[\Bbb R]|>2$ then $|X\setminus c[\Bbb R]|\ge |\Bbb R|.$