Set theory or something

[waxen sedge]

So in class our professor told us that the reason |(0,1)| equals |(0,2)| is because we can connect them through the epimorphic+monomorphic function x-->2x, or something similar. Can someone correct this statement? as I don't remember how the prof told it. Also, why is this (corrected version) statement true?

[noble sequoia]

$(0, ,1)\overset{x\to 2x}{\cong}(0, ,2)$

but for your statement you just need bijectivity of the function

[barren oracle]

filippo

[waxen sedge]

is that what I wrote? I'm not sure what the equal symbol is in the middle

[waxen sedge]

does that mean that the function is only 1-1?

I usually know what things mean in english, but I've heard most of these terms for the first time and i dont know how to translate them

[noble sequoia]

if there is a 1-1 function between two sets, they have the same cardinality

[noble sequoia]

you wrote the conditions for certain types of isomorphicity

so i wrote that (0,1) and (0,2) are isomorphic as regions of metric spaces

[waxen sedge]

I know of three kinds of functions. is the one you referring to, the one where you can fill one set with the other set's members, but repetition is allowed?

[noble sequoia]

1-1, repetition isn't allowed

[waxen sedge]

So if we use a 1-1 function we must prove that the right side fits perfectly in the left and other way around, right?

[noble sequoia]

yes

but we do the opposite, we use 1-1 functions to prove that LHS fits into RHS and viceversa

[waxen sedge]

yes true. so, how do we prove this?

i know its x-->2x but Im not sure what it means

[noble sequoia]

if you can invert the function s.t. the domain of the inverse is the LHS, then the function is 1-1

[waxen sedge]

I dont know how to do that. the picture of the function x-->2x is quite hazy to me

[short brook]

Not just one to one. A "bijection" is the term you are looking for

[waxen sedge]

What does bijective mean?

[noble sequoia]

1-1

[short brook]

No

An injection is one to one

A bijection is both an injection and a surjection

[noble sequoia]

1-1 is referred to the range, not the codomain

[waxen sedge]

so its 1-1 with repetition allowed, right?

[noble sequoia]

no

[short brook]

That's a contradictory statement

[waxen sedge]

lol so 1-1 without repetition?

[short brook]

But that's vague. That's why it's better to call it a bijection

[noble sequoia]

i guess it depends on the national conventions

[short brook]

That's redundant. That's what 1-1 means

[waxen sedge]

what do we have left then?

[short brook]

Go to Wikipedia and search for injective function and surjective function

A bijection is both

[noble sequoia]

surjectivity, but it depends on conventions

[waxen sedge]

y'all are talking different languages

but idk either of them so I'll educate myself to the basics before asking again

[noble sequoia]

it's just that 1-1 isn't formally used, so it's ambiguously used

[waxen sedge]

ohhhhhh bijection means it is 1-1 and fits fully, right?

so how do we prove that x-->2x is bijective?

[noble sequoia]

.

[short brook]

You need to specify a domain and codomain first. They should be your two sets in question

[noble sequoia]

note that my 1-1 is your bijective

[waxen sedge]

I am not educated enough to understand what you said there lol

[waxen sedge]

How do we do that?

i got confused

maybe (0,1) and (0,2)?

respectively

[noble sequoia]

yes

[waxen sedge]

okay and now what do we do?

[noble sequoia]

can you invert x->2x ?

[waxen sedge]

I have no idea what the criteria for it would be. Probably that they both can span R, but not sure

wait that wouldnt make sense

[noble sequoia]

you know what the compositional inverse of a function is?

[waxen sedge]

Maybe I saw it today but I forgot

[noble sequoia]

you know what function composition is?

[waxen sedge]

The high school definition, yes

[noble sequoia]

$f\colon X\to Y$ is bijective $\iff\ \ \exists f^{-1}\colon Y\to X$ s.t. $\ f\circ f^{-1}=\mathrm{id}{X}\ f^{-1}\circ f=\mathrm{id}{Y}$

[waxen sedge]

I understand what this means

[barren oracle]

filippo

[waxen sedge]

not anymore

[noble sequoia]

f(f^{-1}(x))=f^{-1}(f(x))=x
(without domain reference)

[waxen sedge]

sure

that means that X has a one by one correspondence with Y and vice versa in other words

[noble sequoia]

yes

now, what's the inverse of x->2x (hint, try to solve x=2y for y)

[waxen sedge]

wait thats what this means?

[noble sequoia]

f(x)=2x, f^{-1}(x)=?

[waxen sedge]

f^(-1)(y) = x/2

and we can change the thingies

[noble sequoia]

f^{-1}(y)=x

x=y/2

[waxen sedge]

we can just do that?

[noble sequoia]

yes

so we get f^{-1}(x)=x/2

[waxen sedge]

yes

[noble sequoia]

now, the domain of that function contains (0,2)

[waxen sedge]

I don't see how

[noble sequoia]

x->x/2 is defined on all R

perhaps i used a slight abuse of notions

[waxen sedge]

why cant we say that about x-->2x?

[noble sequoia]

f(x)=x/2 is not the function, the function is f

[noble sequoia]

you can, but then (0,2) wouldn't be the range of f

[waxen sedge]

so we are converting (0,2) into (0,1)? If yes, I dont see how we achieved that

[noble sequoia]

f(x)=x/2 is just the underlying analytical expression of f, together with a domain and a codomain you get a function

[waxen sedge]

so the inverse of the function x-->2x is x-->x/2, right?

[noble sequoia]

yes, we achieved that because as the definition of the inverse function says, the codomain of f^{-1} (also the range because of bijectivity) is X

[waxen sedge]

I can see that actually. the function is (0,1)-->(0,2) and the inverse is (0,2)-->(0,1)

[noble sequoia]

and if the analytical expression can be inverted on all Y, then the domain is Y, so the expression together with the domain and codomain is indeed the inverse

i'll give you an exercise, prove |[0,1)|=|(0,2)|

[waxen sedge]

I see. I had no idea that we were talking about our everyday functions so thats really cool

[waxen sedge]

Isnt that what we just did

[noble sequoia]

look at the brackets

[waxen sedge]

ohhhh

damn okay

so now we have what we had earlier, plus one term

maybe we do f: x+1-->2x ?

it makes sense in my brain

actually, choosing the domain wont be so sensical

honestly my brain has frozen

im sure its some trick

ughhhh I give up

Its my first time thinking of functions in this way

we want a function with domain [0,1) and range R

[noble sequoia]

codomain R

the range should be (0,2)

[waxen sedge]

True

I see why

maybe then x-->x+1 with domain [0,1)?

but im rly not sure

that would mean that the range is [1,2) so no

I cannot think of one sorry

the idea of these functions is too fresh for me

[noble sequoia]

hint, you don't need to construct the analytical expression, you just need a consistent pairing

[waxen sedge]

well i think thats whats confusing me. the analytical expression really helped

so we want [0,1)-->(0,2)

we wanna prove that there is one more place for one member exactly so sit

one more place than in the case of (0,1)

I mean we have shown that (0,1) and (0,2) are essentially the same

so how about we find a function [0,1)-->(0,1)

[noble sequoia]

there's a really clever proof, but first you need to prove a lemma

[waxen sedge]

is it the diagonal thing?

[noble sequoia]

no

$\lvert\mathbb{N}\rvert=\lvert\mathbb{N}\cup{0.5}\rvert$

[barren oracle]

filippo

[noble sequoia]

prove it

[waxen sedge]

n-->n+a?

I remember something similar

Actually idk if this is provable

nvm it is

but idk how

i suck at this

We can say that we added a discrete amount of numbers in N

[foggy cape]

bro do you remember all the llemmas

[waxen sedge]

so the result is still numerable

[foggy cape]

cos i cant remember anything lol

i just understand it

[waxen sedge]

are you tryna destroy my confidence lol

im struggling over here

[foggy cape]

no as in i understand the llemmas then move on (the llemmas i learn are pretty easy basic fundamentals)

but i couldn't imagine myself remembering them

as well

[noble sequoia]

i'm making them up

[foggy cape]

💀

bro too smart

it takes me ages to understand some of these llemmas

[noble sequoia]

n>0 -> n-1
0 -> 0,5

a piecewise function

[waxen sedge]

I understand it logically, not mathematically

you basically added a place for 0.5 to sit in

[noble sequoia]

yes

you can do it because there is a thing such that the next real number

you can't do it for reals

[waxen sedge]

I can see why

but there is a different way to do it there

[noble sequoia]

exactly

$x\in (0,1)\to 2x\ x\in{n\in\mathbb{N}\mid n>2}\to x+1\ 0\to 3$

[barren oracle]

filippo

[noble sequoia]

it's a bijection that makes $\lvert[0,1)\cup{n\in\mathbb{N}\mid n>2}\rvert=\lvert(0,2)\cup{n\in\mathbb{N}\mid n>2}\rvert$

[waxen sedge]

This feels really weird

[barren oracle]

filippo

[noble sequoia]

why?

[waxen sedge]

first of all are there "<=>"s implied between the lines?

or at least =>

[noble sequoia]

it's the piecewise definition of a function

[noble sequoia]

as i did here

[waxen sedge]

yeah it just confuses me to the point where im not sure what to ask

[noble sequoia]

to understand better a function you plug in values

what is 0.7 mapped to?

[waxen sedge]

1.4?

[noble sequoia]

yes

where is 0 mapped?

[waxen sedge]

1

no

3

Is it a function with three different thingies?

I see why its called piecewise

we start with 0, then we have all the real numbers between (0,1) with range (0,2) and then we take the rest of the naturals mapping to n+1

maybe I said it wrong but something like that

[noble sequoia]

do you understand this?

[waxen sedge]

no

You dont have to fully make me understand if you dont wanna. weve been here a long time

[noble sequoia]

mb here it should be >1

and 0 to 2

[noble sequoia]

and here too

[waxen sedge]

$x\in (0,2)\to 2x\ x\in{n\in\mathbb{N}\mid n>1}\to x+1\ 0\to 3$

[barren oracle]

fijokazż

[waxen sedge]

this?

[noble sequoia]

maybe it is better to get intuition first

[noble sequoia]

0 to 2

[waxen sedge]

I agree. I am not supposed to be in that class thats why im struggling lol

[noble sequoia]

search cardinal numbers on google

[waxen sedge]

ohhh okeoke

$x\in (0,1)\to 2x\ x\in{n\in\mathbb{N}\mid n>1}\to x+1\ 0\to 2$

[barren oracle]

fijokazż

[noble sequoia]

yes

cardinal numbers are numbers that obey algebraic rules induced by the set-operations composed with the map that sends a set to its cardinality

[waxen sedge]

A lot of formalities but it kinda makes sense

[noble sequoia]

cardinal numbers are much more intuitive to work with

[waxen sedge]

the reason its called piecewise is because it picks up where the previous part left it at , right?

[noble sequoia]

it's because the function has a set of overlined analytical expressions that has more than one element

[waxen sedge]

that confused me further lol

[waxen sedge]

im tryna construct the inverse of this but im not sure how

[noble sequoia]

but it's more of a meta-mathematical concept, it can be formalized in more than a way

[noble sequoia]

try to invert it piecewisely

[waxen sedge]

so it would look kinda like
xE(0,2)-->x/2
xE{xEN | x>2}-->x-1
2-->0

domains go to ranges and vice versa

and how does this help?

we proved that the function has an inverse

which prolly means it is bijective

[noble sequoia]

you wrote the domain of the inverse and it matched the codomain/range of the function

[waxen sedge]

yes

[noble sequoia]

so this is true

with >1

[waxen sedge]

well piecewisely we proved that
|(0,1)| = |(0,2)|
|{xEN | x>1}| = |{xEN | x>2}
and that 0 maps to 2 and 2 maps to 0

right?

[noble sequoia]

yes

can you see how this argument generalizes?

[waxen sedge]

I'm thinking about it

I can see how the first two are a union, because we have that if xE(0,1) we can match it with exactly one x' belonging in (0,2), and if xE{N|x>1} then blah blah

so we have that
|(0,1)U{x>1 xEN}| = |(0,2)U{x>2 xEN}|

wrote the natural things a little short

but I understand this far

Now I dont get how we can integrate the 0 and 2 thingy in here

[noble sequoia]

$|A\cup B|=|A|+|B|$ if $|A|$ isn't finite

[noble sequoia]

.

[noble sequoia]

can you see how you can algebraically manipulate this now?

[waxen sedge]

what does A+B even mean? if A and B are sets then the addition operator is pretty weird

[barren oracle]

filippo

[noble sequoia]

i made a typo

[waxen sedge]

I guess that we can add any term to our initial thingy and itd still be the same

[waxen sedge]

i see its okay

[noble sequoia]

use the identity to separate unions

[waxen sedge]

because its a union we can choose to add one extra term from the naturals on the LHS but 0 extra on the RHS and itd result in [0,1) having the same numerability as (0,2), right?

[waxen sedge]

ohhh we gotta use that

[noble sequoia]

then prove |N| leq |[0,1)| and |N| leq |(0,2)|

[waxen sedge]

|(0,1)U{x>1 xEN}| = |(0,2)U{x>2 xEN}| <=> |(0,1)| + |{xEN, x>1}|= |(0,2)| + |{xEN, x>2}|

[noble sequoia]

yes

[waxen sedge]

okay wait

what if

we take x=2 out of the first one

lol

and we add it in (0,1)

as 0

like embed it

actually nvm

[waxen sedge]

idk how to do this step

[noble sequoia]

prove that there is an injective function from N to the sets

btw there are obv simpler formal proofs but they involve the concept of limits

[waxen sedge]

I dont know

[noble sequoia]

1 -> 0.1
2 -> 0.11
3 -> 0.111
.
.
.

[waxen sedge]

thats enough to prove it?

I mean it makes sense

[noble sequoia]

now, if |B| leq |A|, |B\cup A| leq 2|A|

[waxen sedge]

If A is non finite

[noble sequoia]

if it's finite you subtract a |A intersected B| so it's still true, but we are assuming |A| non finite

2|A|=|A cup A|=|A|

so we get |[0,1)| leq |(0,2)|, but |[0,1)|=|(0,1)|+1 =|(0,2)| +1 geq |[0,1)|

so |[0,1)|=|(0,2)|

[waxen sedge]

lemme think abt it for a moment

I think I lost you again, sorry

[noble sequoia]

we could have done it for B={0} but i wanted to give you more than an example to |A cup B|=|A|+|B|-|A intersected B|

there are other intuitive proofs but they don't use cardinal numbers, which i now think you would prefer

[waxen sedge]

I dont get our example because we have an equality already, which means we will use "=>", if that makes sense. and it just confuses me idk

[waxen sedge]

It's okay, youve helped me with a lot more than what I came here for

[noble sequoia]

good luck with your course

[waxen sedge]

I will learn how to understand this stuff eventually, i try not to force things like that in my mind because it gets messed up

[waxen sedge]

thank you

.solved