Suppose we have a set $M = (0,1) \subset R$ of reals well-ordered as the first uncountable ordinal. Let $M(a) = \lbrace x \in M : x < a \rbrace$. For every $a \in M$ set $M(a)$ is countable. That's why every increasing sequence is bounded: $$(*) ~~~~~~~~~~ \forall \lbrace a_1,...,a_n,...\rbrac...

I've been pondering the following question. How can we measure the amount of "space" above an element $p$ in a partially ordered set $P$? One way would be to try to cram the elements of increasingly large ordinals into the space above $p$, respecting order. The least ordinal whose elements ...

Prove that no uncountable family of subsets of $\mathbb{N}$ is well-ordered by relation of inclusion. I had two ideas how to do it. First was to show that if such family is uncountable, then it is not true that all proper beginning segment (is it correct english name for that?) is of the form $O...

I'm trying to understand the following. I'll denote by $X$ the set of all ordinals that are at most $\omega_1$, the first uncountable ordinal. To avoid confusion, they can be equal to $\omega_1$ as well. Equip $X$ with the order topology. Then how come $\omega_1$ is an accumulation point of $X\...

While I agree that @Austin Mohr ‘s answer is a good picture of $\omega_1$, it is technically problematic since it is circular. You cannot define an uncountable ordinal in terms of uncountable union since you have not yet defined uncountable or uncountable union. Also, the notation may give the i...

$\omega_1$ is the first uncountable ordinal, or, equivalently, the set of all countable ordinals. The countable ordinals in turn can be constructed by the following rules: 0 is a countable ordinal If $\alpha$ is a countable ordinal, then so is $\alpha + 1$. If for each $i \in \mathbb{N}$ $\alph...

While I agree that @Austin Mohr ‘s answer is a good picture of $\omega_1$, it is technically problematic since it is circular. You cannot define an uncountable ordinal in terms of uncountable union since you have not yet defined uncountable or uncountable union. Also, the notation may give the i...

What you want is the usual argument that the bounding number $\mathfrak{b}$ is uncountable. Given a countable family $\mathscr{F}=\{f_n:n\in\omega\}\subseteq{}^\omega\omega$, define $$f:\omega\to\omega:n\mapsto 1+\max_{k\le n}f_k(n)\;;$$ then for each $n\in\omega$ we have $f(k)>f_n(k)$ for all ...

Consider the following trigonometric identity, valid for any set of angles $u,v,t$: $$\cos t⋅\cos u⋅\cos v =\frac14\left[\cos(t + u + v)+\cos(t + u - v)+\cos(u+v-t)+\cos(v + t - u)\right]$$ This identity and its derivation have previously appeared as a question on this site (though the version...

The underlying rationale will be as follows. The Tree(n) question can be translated into a problem of the length of effectively given bad sequences. I think this goes directly back to Friedman. Such effectively given bad sequences can be mapped into effectively descending sequences of ordinals us...