The order in (1) is known as the lexicographic order on $\Bbb N\times\Bbb N$ and is actually a total order. To show that it’s antisymmetric, suppose that $\langle m_1,n_1\rangle\le\langle m_2,n_2\rangle$ and $\langle m_2,n_2\rangle\le\langle m_1,n_1\rangle$.
- Since $\langle m_1,n_1\rangle\le\langle m_2,n_2\rangle$, we know that either $m_1<m_2$, or $m_1=m_2$ and $n_1\le n_2$.
- Since $\langle m_2,n_2\rangle\le\langle m_1,n_1\rangle$, we know that either $m_2<m_1$, or $m_2=m_1$ and $n_2\le n_1$.
We cannot have $m_1<m_2$: that would imply that $m_2\not<m_1$ and $m_2\ne m_1$, contradicting (2). Thus, $m_1=m_2$, and $n_1\le n_2$. And from (2) we can then see that $n_2\le n_1$ as well, since $m_2=m_1$. Finally, the usual order on $\Bbb N$ is antisymmetric, so $n_1\le n_2$ and $n_2\le n_1$ imply that $n_1=n_2$. We’ve now shown that $m_1=m_2$ and $n_1=n_2$, i.e., that $\langle m_1,n_1\rangle=\langle m_2,n_2\rangle$, as desired.
In the second problem you’re given a transitive relation $<$ on some set $X$, and you’re further told that $<$ has the trichotomy property: for any $x,y\in X$, exactly one of the statements $x<y$, $x=y$, and $y<x$ is true. Now you’re to define a new relation, $\le$, on $X$ by setting $x\le y$ if and only if either $x<y$ or $x=y$. Finally, you’re to prove that this new relation $\le$ is a total order.
To do this you must show that $\le$ is reflexive, antisymmetric, transitive, and total. Reflexivity and transitivity are pretty straightforward, and I’ll leave them to you. For antisymmetry, suppose that $x\le y$ and $y\le x$. Since $x\le y$, we know that either $x<y$ or $x=y$. Suppose that $x\ne y$. Then $x<y$, and by the trichotomy property we know that it is not the case that $y<x$. But then we have $y\not<x$ and $y\ne x$, so be definition it’s not the case that $y\le x$. This contradiction shows that we cannot have $x\ne y$ and hence that $x=y$, as desired. Finally, to show that $\le$ is total, use trichotomy again: for any $x,y\in X$, either $x<y$, in which case $x\le y$; or $x=y$, in which case $x\le y$; or $y<x$, in which case $y\le x$.
This exercise shows how you can turn any so-called strict linear (or total) order $<$ on a set $X$ into an ordinary total order $\le$ in such a way that the $\le$ symbol has the obvious, expected meaning.
This is just the opposite of the previous problem: starting with a total order $\le$ on $X$, turn it into a strict linear order by ‘throwing away the equals part of it’. That is, given the total order $\le$ on $X$, define a new relation $<$ on $X$ by $x<y$ if and only if $x\le y$ and $x\ne y$. Now you’re to show that this new relation is transitive and has the trichotomy property: for any $x,y\in X$, exactly one of $x<y$, $x=y$, and $y<x$ holds. Proving transitivity is very straightforward, and I’ll leave it to you. Trichotomy isn’t much harder, but you do have to check two things: you must show that for any $x,y\in X$ at least one of $x<y$, $x=y$, and $y<x$ holds, and you must also show that it’s never the case that more than one holds. To see, for example, that it’s not possible to have both $x<y$ and $y<x$, note that $x<y$ implies that $x\le y$, and $y<x$ implies that $y\le x$. The relation $\le$ is antisymmetric, so $x=y$. But then by definition $x\not<y$, and we have a contradiction. I’ll leave the rest to you, but feel free to ask for help if you get stuck.
Questions 1 and 2 do highlight slight inconsistencies in the approach, but these can be easily fixed. Question 3 follows immediately from the definition of $\omega_\alpha$ when $\alpha$ is a limit.
One easy fix is to view this as a proof by contradiction: we begin by supposing that $\omega_\gamma$ is uncountable, singular, and so large that no singular $\alpha\ge\omega_\gamma$ has $\omega_\alpha=\alpha$. From this we then know that the sequence $(\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_\gamma}},...)$ is increasing (specifically this uses the fact that if $\kappa$ is singular then $\omega_\kappa$ is singular), and hence we can define its limit $\alpha$ and proceed unworried. Since $\alpha$ has countable cofinality (and is clearly uncountable) by construction, we'll be done if we can show that $\omega_\alpha=\alpha$.
That $\alpha=\lim_{n\rightarrow\omega}\alpha_n\implies \omega_\alpha=\lim_{n\rightarrow\omega}\omega_{\alpha_n}$ then follows immediately from the definition of the ordinal $\omega_\theta$: recall that by definition, if $\theta$ is a limit then $\omega_\theta=\sup_{\beta<\theta}\omega_\beta$. Now since $\alpha$ is a limit ordinal (the limit of any increasing sequence is a limit ordinal) we have $\omega_\alpha=\sup_{\beta<\alpha}\omega_\beta$, but since the set $\{\alpha_n: n\in\omega\}$ is cofinal in $\alpha$ this implies $$\omega_\alpha=\sup_{\beta<\alpha}\omega_\beta=\sup_{\beta\in\{\alpha_n:n\in\omega\}}\omega_\beta,$$ and this is just $\sup_{n<\omega}\omega_{\alpha_n}.$
We could also modify the construction of the sequence: let $\alpha_0=\omega_\gamma$ and let $(\alpha_{i+1}=\omega_{\alpha_i})^+$. Then the sequence $(\alpha_i)_{i\in\omega}$ is clearly increasing, and its limit is uncountable and has cofinality $\omega$, hence is singular.
Not entirely related, but worth mentioning: we can also modify the definition of "limit" to apply to a broader class of sequences. Specifically, whenever $(\alpha_\eta)_{\eta<\lambda}$ is a sequence of ordinals which is nondecreasing - that is, which satisfies $\eta<\delta<\lambda\implies\alpha_\eta\le\alpha_\delta$ - then $\sup_{\eta<\lambda}\alpha_\eta$ exists, and we can call this the limit of that sequence (note that this matches up with the notion of limit of a $\lambda$-indexed sequence coming from topology).
With this modification, everything works nicely: it's clear that the sequence $\omega_\gamma,\omega_{\omega_\gamma},\omega_{\omega_{\omega_{\gamma}}}$,... is nondecreasing, and so we can define its limit.
Ignoring singularity for a moment, it's a good exercise to show that even with this modified definition, the following remains true:
If $(\alpha_\eta)_{\eta<\lambda}$ is a nondecreasing sequence of ordinals with limit $\alpha$, then the sequence $(\omega_{\alpha_\eta})_{\eta<\lambda}$ is also nondecreasing and has limit $\omega_\alpha$.
HINT: if the sequence $(\alpha_\eta)_{\eta<\lambda}$ is not eventually constant, WLOG assume it is increasing (if necessary, pass to a subsequence) and use the definition of $\omega_\theta$ for limit $\theta$ as above; otherwise, things are fairly trivial ...
So this gives that there are arbitrarily large fixed points of the map $\alpha\mapsto\omega_\alpha$ (note that any such fixed point must be an uncountable cardinal). It doesn't give singularity, of course, but it's still worth understanding. In fact, here's an important principle you should really prove and understand:
Suppose $A$ is any set of ordinals. Then $$\omega_{\sup A}=\sup\{\omega_\alpha:\alpha\in A\}.$$
So this really isn't about sequences at all, just how the map $\alpha\mapsto\omega_\alpha$ behaves with respect to suprema.
Best Answer
As far as I know, the standard definition of $|A|\lt|B|$ is "there is an injection from $A$ to $B$, but there is no injection from $B$ to $A$". (Unfortunately, your book defines $|A|\lt|B|$ as "there is an injection from $A$ to $B$ but there is no bijection between them." Fortunately, in view of the Cantor–Bernstein theorem, that is equivalent to the natural definition, which is the one I gave.)
If $|A|\lt|B|$ and $|B|\lt|A|$ hold simultaneously, then we have: $$\text{there is an injection from }A\text{ to }B;\tag1$$ $$\text{there is no injection from }B\text{ to }A;\tag2$$ $$\text{there is an injection from }B\text{ to }A;\tag3$$ $$\text{there is no injection from }A\text{ to }B;\tag4$$
All you need is basic logic to get a contradiction from $(1)$ and $(4)$, or from $(2)$ and $(3)$.
From $|A|\lt|B|$ and $|B|\le|C|$, we have injections $A\to B\to C$; composing these we get an injection $A\to C$. We also have to show that there is no injection $C\to A$. If we had an injection $C\to A$, composing this with the given injection $B\to C$ would provide an injection $B\to A$, contradicting the assumption that $|A|\lt|B|$.