@DanielFischer hello

hi

are you feeling sufficiently good today?

tired, but otherwise yes

Ah. me too. tomorrow is my 5th exam in a week and two days.

i had only today to learn to it. it is an exam in set theory

You've probably learnt some things during the semester.

of course.. i just dont think i have enough time to learn cardinals. it was the last topic and i just dont have time to get into it unfortunately. i hope it wont appear in the exam

i got a small question:

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assuming that $\{A_n\}$ are countables. i want to show using AC that there is a sequence $<g_n:n\in N>$ of bijections between $N$ and $A_n$

so i defined $X_n = \{f:N\to A_n : f$ is a bijection $\}$

if i let $X =\{X_n\}$ and taking a choise function on $X$ say $f$ then $f(X_n)$ is the sequence

one thing im not sure about - how to justify that $X$ is a set

i know.. $X = G(N)$ where $G$ takes n to $X_n$

Yes.

Of course that uses that each $X_n$ is a set.

$X_n \subset P(N\times A_n)$

:)

right

@DanielFischer if i say $(A,\lt)$ is well order, does $\lt$ needs to be parital order or sharp order?

by sharp i mean $\lt$ and by partial $\le$

The customary terms are strict order and weak order. Typically $<$ is used for strict orders and $\leq$ for weak orders. Whether "order" without qualifier is supposed to mean weak order or strict order varies. Look up which one was used in your course.

@DanielFischer trying an exercise : if $A$ is a set of ordinals s.t $otp(A) =\delta $ is a limit ordinal, then $sup(A)$ is also a limit ordinal

if $sup(A)=0$ that means that $A = \emptyset $ or $A=\{\emptyset\}$ ?

@Liad What is "$otp(A)$"?

order type

every well order set has a unique ordinal that iso. to that set

this ordinal is its order type

Okay. So suppose $\sup A$ were a successor, and show that the order type of $A$ can in that case not be a limit ordinal.

first i want to show it can not be $0$

@Liad The natural interpretation of $\sup \varnothing$ in this context is $0$, so $\sup \varnothing = \sup \{\varnothing\} = 0$.

alright

now if $sup(A) = \alpha +1$

then $\alpha +1 \in A$

Yes. Why?

because otherwise $\alpha$ would be the sup

Not necessarily the supremum, but an upper bound in any case.

not the sup but a smaller element that in that set

now what ^^

$otp(A)= \alpha +1$ ?

So $\sup A \in A$, in other words, $\sup A = \max A$, in other words …

@Liad No, the order type can be smaller than that. Consider $A = \{ \omega_5 + k : k \in \omega + 1\}$.

dont know then.

$A$ has a maximum..

Do limit ordinals have a maximum?

no

so?

A is a set of ordinals

with a maximum

hence its order type …

$A$'s order type is an ordinal that isomorphic to $A$

what am i missing O_o

if $A$ has a maximal element, then?

i dont know

If you have an order isomorphism $A \to B$, what does it do with $\max A$?

it sends it to maxB

In particular, it follows that $B$ has a maximum.

So what about the order type of $A$?

if it has a maximum then so does $otp(A)$

yes

what does it mean for an ordinal to have a maximum element?

i think that what i was missing earlier

it means it's a successor

why?

If $\alpha = \max \beta$, then $\beta = \alpha + 1$. And conversely.

i cant see why

$\alpha = max \beta$ means that there is no $\gamma \in \beta$ s.t $\gamma\gt \alpha$

$$\alpha = \max \beta \iff \beta = \{ \gamma : \gamma \leqslant \alpha\} = \{\gamma : \gamma < \alpha\} \cup \{\alpha\} = \alpha \cup \{\alpha\} = \alpha + 1$$

got it

your explanation is nicer though. ^^