Hi, $f: R \ ^ k \to R$ defined by $f(t) = x_0 + tv$ for fixed $x_0,v \in R \ ^ k$. i have a book im reading and its written in it that $f'(0) = v$. i don't see why, can someone explain?

sorry, $f:R \to R \ ^ k$

@Liad At any given time t, the curve is moving in the direction of $v$.

So, $f'(t)$ is the constant vector $v$, meaning $f'(0) = v$ automatically.

@DHMO depends what you mean by interesting, Lebesgue and Hausdorff are thr common ones

@AlessandroCodenotti apart from them?

we discussed $f(A) = |A \cap \{x_0\}|$

and $f(A) = (A \ne \varnothing)$

@Fargle thanks , i just wrote it by definition and saw that it is actually easy.

@Liad Yep! The derivative is always pretty easy for linear functions because it always provides a linear approximation to the function at each point, and if the function is already linear, you don't really have any work to do.

How many continuous real functions are there?

@DHMO Cardinality or measure?

I know that a continuous function is uniquely determined by its values on rational inputs

@Fargle cardinality

but that doesn't mean every function in Q->R can be extended to a continuous real function

@DHMO It is at least $|\Bbb R|$ because of every constant function.

@Fargle agreed

oh!

thanks, then

what is the theorem called? $|k| \ge |\Bbb N| \implies |k \times k| = |k|$

and does it have an easy proof?

I've seen its proof using ordinals...

@DHMO Yeah, what you stated is enough to ensure that the cardinality is precisely that of the continuum

@Fargle agreed

@DHMO It's called axiom of choice

(Not really, but it's equivalent)

@AlessandroCodenotti do you have an easy proof?

At least, can I prove that it works for all the beta-numbers?

@DHMO there are $mathfrak{c}$ continuous functions $\Bbb R\to\Bbb R$, because of the thing about rationals you said

@AlessandroCodenotti ya, I got that thanks to Fargle

Can I prove that $|k \times k| = |k| \implies |2^k \times 2^k| = |2^k|$?

@DHMO It should work for all infinite sets, apparently. Zermelo proved it at some point

@Fargle yes, I'm looking for an easy proof

@DHMO You can probably find a few links here: Overview of basic results on cardinal arithmetic. The links given there for this result are: About a paper of Zermelo and For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice.

@DHMO The fact that this does not hold in ZF and you need to invoke AC at some step suggests that there is probably not a proof which would be very very easy.

Zermelo proved that for well orderable sets $|S\times S|=|S|$, AC says that all sets are well orderable

@MartinSleziak I know, so I restricted my problem to the beta numbers

@DHMO why are they not $2^{\mathfrak{c}}$?

@DHMO This fact is true even without AC.

@AlessandroCodenotti why should they be?

@DHMO You have $2^k\cdot2^k = 2^{k+k} = 2^k$, if you can prove $k+k=k$.

@Fargle How should I prove that?

@DHMO Not sure.

@Fargle then how do you know?

@MartinSleziak I'm having a hard time translating, you know, $2^k \times 2^k$ to $2^{k+k}$

@DHMO .there's $2^{\mathfrak{c}}$ functions $\Bbb R\to\Bbb R$

@AlessandroCodenotti I said they're uniquely determined by the values they take on the rationals

and $|\Bbb Q \to \Bbb R| = |\Bbb R|^{|\Bbb Q|} = (2^{|\Bbb Q|})^{|\Bbb Q|} = 2^{|\Bbb Q \times \Bbb Q|} = 2^{|\Bbb N|} = |\Bbb R|$

@AlessandroCodenotti $2^{\mathfrak{c}}$ functions, but only $\mathfrak{c}$ continuous.

@DHMO Read on MSE...

I know

@DHMO In general you have $a^b\cdot a^c = a^{b+c}$ for cardinals. The proof is constructing a bijection between two set of functions. It does not involve AC and can probably found in several posts on this site.

@Fargle alright

@MartinSleziak ok, I'll try it myself, thanks

Again from the Overview of basic results on cardinal arithmetic: Let $A,B,C$ be sets, and $B \cap C=\emptyset$. Show $|A^{B \cup C}|=|A^B \times A^C|$ and Notation on proving injectivity of a function $f:A^{B\;\cup\; C}\to A^B\times A^C$.

The proof of $a^b\cdot a^c = a^{b+c}$ is not terribly difficult, so I am sure you will figure it out by yourself.

$|\mathbb R\times \mathbb R| = |\mathbb R|$ is the same as $2^{\aleph_0}\cdot 2^{\aleph_0} = 2^{\aleph_0}$. I do not see need for AC in there.

Then the bijection is done lol

Of course, AC is easy to miss, but I do not think it is really necessary in any part of the proof of equalities needed above. Namely proofs of $\aleph_0+\aleph_0=\aleph_0$, proof of $|\mathbb R|=2^{\aleph_0}$ and proof about $a^{b+c}$.

@MartinSleziak I just proved it

Always when I come here, people are laughing out loud at me.

I'd say that the idea of the general proof of $a^b\cdot a^c=a^{b+c}$ is - to some extent - similar.

agreed, although I have a hard time figuring out what $\Bbb N + \Bbb N$ means

@DHMO Of course, to make the proof complete, you should also check that the map $(f,g)\mapsto h$ is indeed bijection $2^{\mathbb N}\times 2^{\mathbb N} \to 2^{\mathbb N}$. (I'm not saying that it is difficult, I am just saying that this should be pointed out explicitly.)

And of course, if you are interested in whether or not you have used AC somewhere, you should probably also check that you did not use it in $|\mathbb R|=2^{\aleph_0}$.

I don't think I used it

I should probably go somewhere where I a not object of ridicule. :-)

how should I present $\Bbb N + \Bbb N$?

$\mathbb N\times\{0\}\cup\mathbb N\times\{1\}$ seems pretty standard way to go

oh, nice!

but I'm interested in what it means

like formally

I know that $\times$ is the Cartesian product

but I never learnt about $+$

It's cardinal addition

@AlessandroCodenotti I figured that out lol

$\kappa+\lambda$ is defined as $|A\cup B|$ where $|A|=\kappa$, $|B|=\lambda$ and $A$ is disjoint from $B$

it is an addition that acts on cardinal numbers

@AlessandroCodenotti $\Bbb N \cup \Bbb N$ doesn't look well to me

it's just itself

wait what lol it must be itself

Well, $\mathbb N$ and $\mathbb N$ are not disjoint.

That's the $\times\{0\}$ part

oh, sorry

I think I have seen something like $A \sqcup B$ to be used as notation for disjoint union.

But from the point of cardinal arithmetic, it does not mater what sets with $|A|=a$ and $|B|=b$ you take to get $a+b=|A\cup B|$, as long as they are disjoint.

In other words, additions of cardinals defined in the way Alessandro Codenotti wrote is well-defined.

@MartinSleziak Why can't I define it as just $|A \cup B|$?

Then you would get $4=2+2=2$.

You want it to work for finite cardinalities, too.

oh, ok

For infinite cardinalities, you would probably have no problems, if you work in ZFC.

I am not sure whether some problems occur without AC.

alright

But perhaps it is easier to work with simpler definition which also covers finite case.

does $|k+k|=|k|$ need AoC?

I am not sure, but I guess so.

what about on beta numbers?

This must be well-known - and probably discussed before here or on MO.

somehow using beta numbers eliminates the need for AoC

You mean $2^k \le 2^k+2^k \le 2^k\cdot 2^k = 2^{k+k} = 2^k$?

eh, why does $2^k+2^k \le 2^k \times 2^k$?

If you prefer, you can write there $2^k+2^k = 2^{k+1} = 2^k$.

But my original inequality follows from $2^k+2^k= 2^k \cdot 2 \le 2^k \cdot 2^k$.

oh, ok

@DHMO This might be relevant: The relationship of ${\frak m+m=m}$ to AC

wtf is that letter

m in fraktur

$\mathfrak{mmmmmmm}$ delicious

Fraktur is quite commonly used for cardinals. For example, cardinality of continuum is usually denoted by $\mathfrak c$.

I have only seen $\frak c$

Let's only work with beth numbers from now on

which for some reason I have been calling beta numbers and nobody has corrected me

@DHMO We knew what you meant I guess

I did not notice. Which only show my lack of knowledge in set theory.

so here goes my long question:

among the results listed here which still needs AoC when restricted to the beth numbers?

Anyway, it seems that $a+a=a$ is not provable in ZF alone.

Oh, my own post is used against me. To be honest, I did not work with beth numbers that much.

@MartinSleziak I mean, when restricted to beth numbers!

@DHMO I guess most of them do not need AC for arbitrary cardinals.

Some which need AC for arbitrary cardinals:

$a^2=a$

@MartinSleziak that isn't what I asked

$b+c = \max\{b,c\}$

Any two cardinals are comparable: $a\le b$ or $b\le a$.

@DHMO I am just trying to help you clarify your question. You said that it is a long question - probably because you linked to long list of cardinal identities.

@MartinSleziak alright, sorry

But you are only asking about those of them which involve AC. So perhaps checking which of them you are interested in can help clarify what you are asking.

sorry, it's my fault, please continue

In fact comparability is not even listed there. Or did I miss it?

And you have already mentioned $a+a=a$.

Am I missing something or are these all of the things from that list which are not provable in ZF?

@MartinSleziak a < 2^a

@DHMO I do not think Cantor's theorem needs Axiom of Choice.

@MartinSleziak "Elegant and remarkably simple." Yeah, no kidding. That's pretty sweet.

Anyway, still at least the question whether beth numbers are comparable (without AC) was still not answered.

I should probably get back to work.

@Adeek No it's not the torus. Identify one of the inner circles with the outer and you get a torus with a hole. Now identify the hole with one of the meridians - that's your space

It's a cell complex, not a manifold

So apparently I don't show up in the room when I switch broser at the same time as exiting it on another broser.

@user400188 welcome back

Hi again

I think I have a solution to the problem yesterday

nice!

We had X={{a},{b,c}}

I know what we had

now what I did is I re-wrote the Union axiom with the $\forall X$ removed. I thought then what could Y be?

I decided that Y could be {a} OR {b,c}

so I removed the $\exists Y$ part and looked at u

no it couldn't

ah darn

the rests hinges of that bit so if its wrong it wont be correct

we need 4 to be true for Y={a}

oh I should meantion by OR I should have written XOR

we need 3 to be true for all u we choose

@user400188 doesn't matter

we need the truth values of 1 and 2 to be the same for all u we choose

@user400188 can you demonstrate ^ for Y={a}?

If Y was {a} then u must be a

this means that z must contain a

again, u has no domain!

u can be {{{{{a,b},c},d},e}} for all intends and purposes

the fact that u comes after Y merely means that we define u after we define Y

not that we define u based on Y

I have a question: if u is in Y and Y is {a}; how could u be anything but a?

there is only one thing to choose from in Y

you can't pick only the values of u such that 1 is true

you also need to check the values of u such that 1 is false

I see

so U could be something larger; it just has to be false in the case that it is

and when 1 is false, 2 also have to be false

or the rest of the axiom needs to be false

Hi @Steamy @Balarka

so lets let u={a,b} and let y={a}

in this case (1) is false

this means that z must have a u in it; so z may be {{a,b},.....}

however {a,b} is not in x; nor is {{a,b}...}

so (2) fails

@user400188 yes

u will not be an element of Y in this case becuase 2 fails

which means (3) fails

@user400188 (3) holds

because (1) and (2) fail at the same time!

this means that (4) fails too. Becuase for all u allows us to choose a silly u like {{a,b}}

so in conclusion 1,2,3 and 4 will fail if Y is chosen to be {a}

@user400188 are you choosing to ignore me lol

sorry: i was scrolling up and down a lot so I didnt see the edits

I see that (1) ad (2) fail at the same time now. Which means the Iff holds so (3) holds

which means 4 holds for the case we jst choose. Howver at this moment I don't see how we could test whether all cases work

@user400188 try to find cases in which (2) is true and (1) is false

without re-defining Y?

we're trying to prove that Y cannot be {a}

so assume that Y is {a} and find a contradiction

Hi @Alessandro

@user400188 so can you?

$a\leftrigtharrow b$ doesn't make sense

let u={b,c} now 1 is false. let z ={a} {b,c}. Now z is in X. So 2 is true when 1 is false

@DHMO I was making a and b the same thing to see what would happen

what is z?

{a},{b,c} I think this is ok to write

I'm not certain though

doesn't make sense to me

do you mean {{a},{b,c}}

becuase it has no brackets around it?

@user400188 ya

If I meant that then 2 would not be true

so I guess that example does not work

I find it strange that {a} {b,c} is wrong but "a" is fine on its own

{a} {b,c} simply isn't defined

you can't put two objects together

a is fine because it is an object

{a} is still an object

it is the set containing only a

well ok lets let u be an arbitary letter C. where c may be any other letter or string of letters except for a

oh sorry I missed that last comment

In the context of your last comment: is there any difference between an object and a question about the object?

what is a question about the object?

if you don't like the fact that we use a,b,c

for instance a might have a truth value of T, but is there a difference between that and the question: "does a have a truth value of T"?

do you find {{{}},{{{},{}},{{},{},{}}}} more comprehensible?

unfortunately not

@user400188 well one is an object and one is a question

@user400188 so let's continue to use a,b,c

ok. By the way I was just asking that out of curiocity not beucase it was making the work harder

I honestly cant see a difference though appart from the name of each. They have the same truth value and properties in every case

back to the oringinal question though

let u=c now u is not in y

brilliant

Now z must contain c

and if z it will be in x so long as it doesnt get too big

so 1 fails while 2 holds

nice

so I think we have shown that for will not always hold for Y={a}

yes

similarly; this will not hold for y={b,c}

yes

I have to ask here is it ok to write things like b,c ?

or does it have to be just b or just c?

it has to be just b or just c

b,c means nothing

in that case; Y must be X

why?

@user400188 u in z and z in X doesn't mean u in X

we have some fundamental misconceptions going on here

u in z will mean z will have something like {u} where u is whatever we choose it to be

and z in X will be {{u}}

yes

but u is not in {{u}}!

was about to ask that

so does that mean Y must be {a,b,c}?

bingo!

didn't I aks that yesterday near the start?