What does it mean for $\omega_1$ to contains initial segment $A$ such that $A \cong \mathbb{N}_{discrete}$ and $\omega+1$

Question

I read this as a comment and I don't quite know what this means. In particular I am not sure what "initial segment" means. If this is something widely known, please inform!

Def 1:

Let $\omega+1$ be defined as:

$$\omega+1 = \mathbb{N} \cup \{\omega\}$$

Where $\omega \geq n, \forall n \in \mathbb{N}$, where $\geq$ is the usual ordering relationship

Def 2: Definition adapted from http://www.math.fsu.edu/~bellenot/class/su08/found/other/omega-one.pdf

Let $\omega_1$ be defined as: $\omega_1 = \{\alpha \in W| \{x \in W| x < \alpha\} \text{ is countable }, W \text{ is an uncountable well order }\}$

or

$\omega_1 = \{\alpha| \text{pred}(\alpha) \text{is countable}\}$, where $\text{pred}(\alpha)$ is the predecessor set

Proposition: $\omega_1$ to contains initial segment $A$ such that $A \cong \mathbb{N}_{discrete}$ and $A \cong \omega+1$

1. Showing $A \cong \mathbb{N}_{discrete}$:

Since $\omega_1$ is a well order, there exists $m_1 := \min(\omega_1)$. Similarly, we can define $m_2 := \min(\omega_1\backslash \{m_1\})$, then $m_2$ is the immediate successor of $m_1$. Continue this way, we have $m_n := \min(\omega_1\backslash \{m_k|k \leq n\})$ $\implies m_1 < m_2 < m_3 < \ldots$. Let $A = \{m_k|k \in \mathbb{N}\}$ be the "initial segment"

Then $f(m_k) = k$ is a continuous bijection into $\mathbb{N}$ with a continuous inverse, hence $f$ is a homeomorphism $\implies$ $A \cong \mathbb{N}_{discrete}$

1. Showing $A \cong \omega+1$:

This means we need to cap $\mathbb{N}$. So take $\alpha = \min(\omega_1\backslash \{m_k\}_k)$, where $\{m_k\}$ is as constructed in previous. Then modify the above function $f(m_k) = k, f(\alpha) = \omega$. Then $f$ is again a homeomorphism and $A \cong \omega+1$

Can someone check that my interpretation of this question and the proofs are all correct. Thanks!

Answer 1

If $\langle X,\le\rangle$ is a linear order, a set $A\subseteq X$ is an initial segment of $X$ if and only if it is ‘downward closed’, meaning that if $a\in A$ and $x\le a$, then $x\in A$. Equivalently, $\{x\in X:x\le a\}\subseteq A$ whenever $a\in A$. In particular,

$$\operatorname{pred}(x)=\{y\in X:y<x\}$$

is always an initial segment of $X$. Yes, this is a standard notion.

Your adaptation of Bellenot’s definition of $\omega_1$ isn’t quite correct. To use his definition, you need to start by assuming that $\langle W,\le\rangle$ is some specific uncountable well order that has at least one element with uncountably many predecessors. Then you can define

$$\omega_1=\big\{\alpha\in W:\{x\in W:x<\alpha\}\text{ is countable}\big\}\;.$$

There are a notational problem and what is probably an oversight in your first argument. Setting $m_1=\min\omega_1$ is fine, if for you $\Bbb N$ denotes the set of positive integers. (For me it’s the set of non-negative integers, so I’d set $m_0=\min\omega_1$.) Then you should have $m_2=\min\big(\omega_1\setminus\{m_1\}\big)$, not $m_2=\min(\omega_1\setminus m_1)$, and in general you want

$$m_{n+1}=\min\big(\omega_1\setminus\{m_k:k\le n\}\big)\;:$$

you want to remove all of the previously determined $m_k$ from consideration, not just the last one previously determined. Then $\{m_k:k\in\Bbb N\}$ is indeed the desired initial segment, and the map $k\mapsto m_k$ is an order-isomorphism and hence a homeomorphism between the sets with their respective order topologies.