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06:10
@DanielFischer i get what you saying but, you can't make question for an exam based on UB.. so they just gave simplifying assumptions so we can just work out how the stack would look like .
 
5 hours later…
11:30
@DanielFischer hello
11:57
hola
como estas
esta bien
We have summer ;)
lol nice!
you prefer summer over winter?
Yes. Snow is nice, but guess who is the one who has to shovel it when it comes down.
Also, the snow typically doesn't stay here, it melts soon or is washed away by the rain soon.
haha , that can be annoying
here we dont have much snow, once in 2-3 years , and for a few days
i also prefer the summer because i like the sea.
we defined $\omega$ to be the collection of all natural numbers* where a natural number* is an ordinal $\alpha$ s.t $\alpha = 0$ or $\alpha = \beta +1$ and each $\gamma \lt \alpha$ is also a natural number*
12:19
"An element of the smallest infinite transitive set."
that definition works for you?
i want to show that $\omega$ is an ordinal , and its the first limit ordinal
Well, I more or less know what a natural number is, so it works more or less. Like other definitions.
wait i mixed two ideas of mine haha
But showing $\omega = \bigcup \omega$ would be a good start.
that what i wanted to write!!!!!!
that felt weird to me
so i didn't know how to express it ^^
$\omega$ is a collection of ordinals so $\cup \omega$ is an ordinal
12:27
Yes, that's been shown in an earlier exercise.
So we'd like to show $\bigcup \omega \subset \omega$, which amounts to showing $\alpha \in \omega \implies \alpha \subset \omega$, i.e. the transitivity of $\omega$, and then $\omega \subset \bigcup \omega$.
if $\alpha \in \omega$ then $\alpha \in \alpha+1 \in \omega$ so $\alpha \in \cup \omega$
this shows the reverse inclusion then what you wrote
now if $\alpha \in \omega$
and $\beta \in \alpha$
then $\beta $ is a natural number * so $\beta \in \omega$
so $\alpha \subset \omega$
this shows the inclusion you wrote
@Liad Probably $\alpha \in \omega \implies \alpha + 1 \in \omega$ deserves a remark.
$\alpha +1$ is a successor and each $\beta \lt \alpha +1$ satisfy $\beta \le \alpha$ so it is a natural number* also
if $\alpha$ is some ordinal , it is not true that $\alpha =\cup \alpha$ right?
12:43
Sometimes it is, sometimes it isn't.
i think only if it is a limit ordinal it will hold
or 0
Right. Or $\bigcup \alpha \subsetneqq \alpha \iff \alpha$ is a successor ordinal.
cool.
so $\omega$ is an ordinal. $\omega = \cup \omega$
if i assume it's not a limit ordinal then $\omega = \alpha +1$
Or $\omega = 0$, but we know that's not the case.
yes. im thinking of a way to get a contradiction
12:50
What is $\bigcup (\alpha + 1)$ for an ordinal $\alpha$?
$\alpha$
nice
so $\omega$ is a limit ordinal.
now i need to show it is the first limit ordinal
so if $\beta$ is a limit ordinal i need to show $\omega \le \beta$
or $\omega \subset \beta$
Or that $\alpha < \omega \implies \alpha$ is not a limit ordinal.
if $\alpha \in \omega$ then either $\alpha =0$ or $\alpha = \gamma +1$ for $\gamma $ natural number *
@DanielFischer this is nicer because that's the definition ^^
So $\alpha$ isn't a limit ordinal.
yea we both got to the same result
you used the fact that $<$ is a linear order on the ordinals though ^^
12:58
Modulo the small problem that the ordinals form a proper class, so we should use a different term. But I'm only looking at the set $\{ \alpha : \alpha \leqslant \omega\}$ of ordinals, and there saying "linear order" is totally legit.
13:42
im showing now by induction on $\beta$ that $(\alpha +\beta ,<) $ is iso. to $(\alpha \times \{0\} \cup \{1\} \times \beta , \lt_{lex} )$
and $(\alpha \beta , <)$ iso. to $(\beta \times \alpha , \lt_{lex})$
so $\beta =0$ is fine,assuming it holds for $\beta$ i show the addition is true for $\beta +1$. im not sure how to show that $\alpha (\beta +1) = \alpha \beta + \beta$ is iso. to $(\beta+1)\times \alpha$
in the addition case given $\sigma$ the iso of $\beta$ i defined $\sigma \cup \{<\alpha + \beta , <1,\beta > >\}$
in the multiplication, i cant do this ..
14:00
What is the definition of $\alpha + \beta$ and $\alpha \cdot \beta$ in your course?
i first prove for $\beta +1$
Okay, defined by transfinite induction.
yes
so i prove this exercise by transfinite induction
(it's not a nice proof. too long )
@Liad Yes, that's the natural way.
or the transfinite way :-)
14:12
Good excuse: my transfinite induction is not yet done
i will use it
Have you finished the proof for addition?
yes
29 mins ago, by Liad
in the addition case given $\sigma$ the iso of $\beta$ i defined $\sigma \cup \{<\alpha + \beta , <1,\beta > >\}$
this works
the multipication is weird
There's the case of a limit $\beta$ also to handle.
in the RHS only one element is added
nvm its from $\alpha \beta + \alpha$ to $(\beta +1 )\times \alpha$
Do use parentheses. It's $(\beta + 1) \times \alpha$.
so its $\beta \times \alpha \cup \{\beta\} \times \alpha$
Right. And an iso $\{\beta\} \times \alpha \to \alpha$ (or $\{1\}\times \alpha$) is easy to find.
i know!
yea
i had to write it down
wait its not to alpha
$\alpha \beta + \alpha \to \beta \times \alpha \cup \{\beta\} \times \alpha$
i have $\alpha \beta \to \beta \times \alpha$
14:24
and you extend that
yes that is what i want
but how an element of $\alpha\beta +\alpha$ that is not in $\alpha\beta$ looks like
i want to define the map
@Liad recall the iso for ordinal addition
nice!
so first i say it iso to what i already showed
do you have pets?
Only pet peeves.
14:40
you dont like pets ?
Depends on the pets. I like cats in general, and some other animals. I wouldn't like having a pet, however.
dogs?
Some kinds of dogs, but most are just loud and ugly.
Ah.
in the addition
when i prove for $\beta$ which is limit
i defined $\sigma = \cup \sigma_{\gamma}$
where $\gamma \lt \beta$ , and $\sigma_{\gamma} $ is the iso. i assume exists
Yeah, you need to show that this is actually well-defined, and an iso. That's not hard, but a wee bit tedious.
14:51
i need to show that if $\gamma_1 \le \gamma_2 \lt \beta$
then $\sigma_{\gamma_1 } $ and $\sigma_{\gamma_2}$ settle?
settle? Do you mean $\sigma_{\gamma_1}$ is the restriction of $\sigma_{\gamma_2}$?
yes
i meant $\sigma_{\gamma_1} \cup \sigma_{\gamma_2}$ is a function
we called it "settle"
Yes, you need to show that for the well-definedness. Then verify that it's an order-preserving bijection.
how can i show that?
@DanielFischer that
What can you use? The most elementary way is by induction.
15:01
maybe $\alpha +\gamma_1 \subset \alpha +\gamma_2$ so the restriction must be $\sigma_{\gamma_1}$
You need some more. $\sigma_{\gamma_1}$ is an order-isomorphism between an initial segment of $\alpha + \gamma_2$ and an initial segment of $\{0\}\times \alpha \cup \{1\} \times \gamma_2$. If you know enough about order-isomorphisms between well-oredered sets, that does it. Otherwise, you have to prove it.
yes this finish it.
nice..
15:27
in the multipication if $\beta$ is a limit then the same $\sigma$ as in the addition would work
Yes, but again one needs an argument that it is well-defined and does what it should do.
(essentially the same as for addition)
i just wrote that it is identical to the addition case
i mean i wrote that the same arguments hold
 
4 hours later…
19:17
@DanielFischer is it still summer?
@Liad Yes, about half an hour until sunset.
doesn't it like 21:22 in Germany?
Yes, but that doesn't last long.
wow i just checked.. you have sunlight untill 22!! ? ?!!
amazing
and sunrise in 4:45
wow..
That's nothing. Go to Hammerfest, they have sunlight the whole day this time of year.
19:24
here , in the summer the sunset is in like 19:30 and sunrise 5:40
at the winter its the worst.. you have sunset in 5:30 .. that's bad
that is another reason why i prefer the summer
Yeah, but in winter you still have ten or eleven hours of daylight, while here it's less than eight.
right. and i guess its less cold here
generally, yes
Though the winters used to be much colder in the seventies than they are now.
We used to have ice enough for skating on the creeks and lakes for months back then, nowadays it's at most a week or two. Poor today's kids.
19:42
Today's kids skating on the iPhone..
I thought they were surfing on the iPhone.
that too
Did you did a phd in mathematics?
so how is that you know mathematics in that level?
It's interesting.
19:57
i agree
but you know math in depth
like you investigate it
have you ever built an app. ?
for iphone or android
20:28
Apps aren't my thing. Nor are phones having an OS. I like my phones dumb. (Actually, I dislike them too, but I acknowledge that sometimes a telephone is useful, so it's something I'm prepared to put up with. To some extent.)
20:40
So i guess you dont have whatsapp ^^
I haven't.
cool
it was a long day today.. gonna go to sleep
good night
21:00
'night

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