/sci/ - Science & Math

>>12204650
Maybe.

>>12204650
Yes

>>12204656
It seems to contradict the definition of a subset. How can it be true?

Dane here, what are you doing with our Ø?

>>12204650
No, that's not true, I think what you meant was 'is a subset of the power set of every set'.

>>12204660
[math]A\subset B \iff \forall a \in A, a \in B[/math]
Is there an element of [math]/emptyset[/math] which isn't in A?

>>12204730
The answer is no.
Someone prove it for me. I'm a brainlet.

>>12204730
Which one are you defining to be the empty set?

>>12204721
It's a statement I pulled from a book verbatim. But to continue what you brought up. Why would the empty set also be a subset of the power set given the power set only includes sets which contain elements of the original set.

>>12204750
Assume there is. But that means the empty set has an element, which is a contradiction. QED.

(Counterpositive)

Yes by definition of subset for ever x that is in a set it must exist in what it is subsetting. So for an empty set it doesn’t not violate this defintion

>>12204650
Yes.

E.g. there are 4 subsets of the 2-element set {a,b}:
{{}, {a}, {a}, {a,b}}

>>12204828
>exist
Be in i mean

>>12204828
Except if a set contains an element then it is not empty, and emptiness, or null, is therefore not contained in the set. Another way to put it: Is emptiness contained in what is filled? Or, is there empty space (void) in a plenum?

By allowing the empty set be a subset of a non-empty set you are essentially appending the empty or null element in the set. E.g. Inherently, A = {1}, B = {_}, B is a subset of A, so A = {1, _ }

>>12204830
Follows by definition of a power set. But I am asking whether or not it is true.

>>12204774
The empty set contains no elements, said 'no elements' are in say set1 whose power set contains the empty set. Taking the contrapositive of that statement just means that the contents of the empty set are not in said set1.

>>12204875
Nothing is not an element in a set

>>12204875
No because then you are implying having nothing in a set is an element. An empty set has no elements thus can’t have an element that the other set doesn’t have this is a subset

>>12204898
>this
Thus I mean

>>12204774
If the empty set contains no elements, then said 'no elements' are in say set1 whose power set contains the empty set. Taking the contrapositive of that statement just means that if the contents of the empty set are not in said set1 then the empty set does contains some elements, which is a contradiction. This applies to all elements of a set.

>>12204886
If I understand you correctly.

A = {}
P(A) = 2^0 = 1
So the set that contains no elements has a power set which contains 1 elements. That is, P(A) ={{}}
There should immediately be an issue with this if we can take the power set of P(A).

>>12204887
>>12204898
This was the point I was trying to make when it is said that the empty set is a subset of all sets. There is no element of the empty set to be a subset of a non-empty set.

>>12204921
See my update, here>>12204916

>>12204921
>There should immediately be an issue with this if we can take the power set of P(A).
what?

>>12204875
>Follows by definition of a power set. But I am asking whether or not it is true.
I don't understand what you're saying.

The empty set is always in the power set. The powerset has all the subsets. The empty set is always a subset.
Or do you mean true in an ontological sense beyond formal math?

>>12204921
By definition you don’t need an element that is in the set you are subsetting. Only requirement is that all elements you are in the set you are subsetting

>>12204774
Empty set is both a subset of the power set and an element of the power set.

>>12204931
P(A) is the power set of A. So what is the power set of the power set of A? The power set is also a set, so we must be able to take the power set of it.

>>12204934
I wouldn't consider what we're doing 'formal math', but in a way my question is ontological.

>>12204938
Could you restate your second sentence.

>>12204950
Wasn't the definition of a subset a set that contains an element of the set being subsetted? And it being an element itself, as I said, the power set P(A) = {{}}

>>12204967
>P(A) is the power set of A. So what is the power set of the power set of A? The power set is also a set, so we must be able to take the power set of it.
yea we can make power set of power set, cant see where are you going with this

>>12204967
Only requirement of a set to be a subset is that it must not have any elements that aren’t in the set it is subsetting.

hot take

there is no such thing as an empty set

>>12204995
What do you mean. How does the definition of a set prevent there from being nothing side of it

>the null element

>>12204986
I'm getting at an infinite sequence of empty sets. {{{{{{{{}}}}}}}}}}
etc.

>>12204987
So we are treating the empty set as an element itself? Since the empty set doesn't contain any elements that are not in the set being subsetted then we can include the empty set in the list of sets that are a subset of a given set.

>>12205019
Read what I said again and take note that I was taking the voice of those who say an empty set is a subset of every set. I do not believe in a 'null element'.

>>12205000
my hand has a set of five fingers, with a number of subsets containing two or more but less than five of those fingers

it does not have a set of fingers that don't exist as a subset of those fingers, and if it did that set would be superfluous

>a well-defined collection of distinct objects, considered as an object in its own right

so emptiness is not distinct idk

>>12205024
Okay let’s do a proof by contradiction
Assume
A is an empty set.
B is an arbitrary set
Now assume A is not a subset of B
If A is a not subset of B then there exists an element in A that is not in B. Because A is an empty set and has no elements then this cannot be true thus A not being a subset of B is contradictory.
There you go OP simple.

>>12205032
What’s the compliment of a set that contains everything.

>>12205024
>I'm getting at an infinite sequence of empty sets. {{{{{{{{}}}}}}}}}}
Yes, this is isomorphic to the natural numbers.

>>12204678
"Ø" og "O" er forskellige.
Den anden er brugt i matematik for at symbolisere den tomme mæangde.

>>12205024
>I'm getting at an infinite sequence of empty sets. {{{{{{{{}}}}}}}}}}
>etc.
thats ok

A = {}, n = 0
P(A) = { {} }; n1 = 2^n = 2^0 = 1
P(P(A)) = { {}, {{}} }; n2 = 2^n1 = 2^1 = 2
P(P(P(A))) = { {}, {{}}, {{{}}}, {{}, {{}}} }; n3 = 2^n2 = 2^2 = 4

>>12205049
>compliment of a set that contains everything

What is the compliment of a set that contains two things, or everything except one thing? The question has no meaning.

>>12205067
What? You don’t know you can take a complement of a set? Look up complements then come back before making hot takes about things you know nothing about.

>>12205046
Fallacious argument. By suggesting that by A not being a subset of B there must contain a element in A not in B is to assume the existence of an element which is not necessary for a set.

The negation of being a subset is not "there must be a element in A that is not in B", it is "there isn't a element in A that is in B". If an element must be in the superset to be the subset the nonexistence of an element does not satisfy the condition.

>>12205108
Anon you don’t know the definition of subset. If a set is not a subset of another set then there must exist an element in that set that does exist in the other This is a fact. Cry me a river about how definitions are wrong or whatever but this is true.

>>12205072
I say it's meaningless because talking about a set's compliment is just a roundabout way of talking about the set.

In reality a set and the set that supposedly "compliments" it are just two different sets. Since the universal set contains everything, there is no compliment since you can't have a set with nothing in it as explained above.

In that case you're just talking about one set and pretending there's another set that exists when it really doesn't.

>>12205131
Likewise, if you say the compliment of the universal set is a subset of the universal set you're contradicting yourself because you're saying the universal set contains everything it doesn't contain.

>>12205108
What you are saying is that A,B are disjoint you dingus.

>The empty set is a subset of every set
this is formally true, as evidenced by dozens of fully formal implementation capturing this basic piece

whether the "set with no elements" makes sense is another story - one that's not very mathematical but rather philosophical

The rhetoric in this thread is tiresome, when it goes on for 50 posts

>>12204650
Thats the symbol for a half diminished chord brainlet

>>12205024
>I'm getting at an infinite sequence of empty sets. {{{{{{{{}}}}}}}}}}
no you're not, youre only taking a finite number of power sets
you can get arbitrarily many brackets, but never infinitely many
the set S = {S} does not exist, it contradicts the axiom of regularity

>>12205024
>I do not believe in a 'null element'.
you should know that even referencing something nonexistent will lead to nonsense, dont entertain it
calling it _ is stupid, i could call it A and then the empty set contains A, which is clearly shit
the empty set contains nothing, giving a name to nothing is nonsensical

>>12205108
>The negation of being a subset is not "there must be a element in A that is not in B", it is "there isn't a element in A that is in B". If an element must be in the superset to be the subset the nonexistence of an element does not satisfy the condition.
A is a subset of B iff
Forall x in A, x in B
the negation is
Exists x in A, x not in B

which is read, There is an element of A that is not in B

so empty is not a subset of B if
Exists x in empty, x not in B
but x cant be in empty, there is nothing in empty, so
empty is not a subset of B is false, empty is a subset of B

im not Texing that, eat dick for being a retard

>>12204650
of course, [math]\emptyset \subseteq X[/math] for every set [math]X[/math]. is there a discussion about this going on?

>>12205409
Why can't you have a countably infinite amount of bracket pairs? Surely it's a prerequisite in defining the naturals through sets.

>>12205676
But does the empty set exist without the axiom that it exists?

>>12205681
the limit of the naturals isnt a natural
we dont need an infinite element to have infinitely many elements

I'm a brainlet but I think I can prove this:

prove: [math]\varnothing \subseteq X [/math]


[math]\varnothing = \left \{ \right \} \mathbf{} [/math]

[math]\Rightarrow \forall z\left ( z\in \varnothing \Rightarrow z\in X \right ) [/math] since the empty set has no elements this statement is always true no matter what X is.

[math]\therefore \varnothing \subseteq X, \forall X [/math] by the definition of the subset relation

[math]mathbf{QED} [/math]

I'm kinda learning by myself no bully pls

>>12204650
yes

>>12205687
it's imlied by Separation or Replacement

Damn I wish I was not retarded so I studied math to shitpost in my thesis with stuff like this...

>>12204721
> 'is a subset of the power set of every set'
Yes, this follows from >>12204650 which also is true. But did you really mean?

"Is the empty set is an element of the power set of every set?"?
The answer is yes.

I also wonder if OP really meant

"Is the empty set an element of every set?"
The answer is no.

>>12204650
you're overthinking it. the null set is a subset of every set by definition. intuition is less important than having an additive identity.

>>12205108
>can’t even do a basic negation of a definition of a subset then accuses Anon of being wrong when he him self is wrong
Sums up this thread pretty much.

>>12204650
What remains of every set when you take away every element of it?

>>12204660
Because universal quantification can be trivially satisfied.

Is the empty set a subset of the set that does not contain the empty set?

>>12206376
Finish high school

>>12204650
It's an assumption of ZFC, feel free to use different definitions and arrive at different conclusions.

>>12204650
>the zero vector is orthogonal to every vector
I'm sensing a pattern here.

What exactly do you call the compliment of the set of a set and it's compliment?

>>12206478
>the set of a set and it's compliment
That wouldn't be a set in ZFC

>>12206490
really? Cause it sounds like it's just the universal set to me.

So it's compliment would be the empty set, if there were such a thing. And there isn't.

It's already been established ITT that the empty set cannot be a subset of the universal set seeing that containing everything it doesn't contain is a contradiction of terms.

>>12205974
But does the empty set exist without the axiom that there exists a non-empty set?

>>12206552
The way you described the object you said "the set of a set and it's complement", which isn't the universe but an object containing two sets. Either way, neither this nor the universe are sets, which is what I was pointing out. For that matter, the complement of a set is never a set. But you're correct that the complement of the universe is the empty set. Likewise, the complement of the class containing a set and its complement will be the subtraction of both objects from the universe.

>>12206567
No.

>>12206587
>the complement of a set is never a set
really? Cause it sure sounds like it is, if the complement of the universal set is the empty set. Which it definitely isn't.

>The complement of a set, denoted A', is the set of all elements in the given universal set U that are not in A.

idk man I'm pretty sure that you can define the complement of a set as a set.

>>12206664
You can show the universe isn't a set using the Axiom of subsets.

[math]\textrm{If}\ B\subseteq A\ \textrm{and}\ A\ \textrm{is a set then so is}\ B[/math]

The contrapositive of the axiom is that if B isn't a set, then if B is a subclass of A then A is not a set. We know the class of classes that do not contain themselves isn't a set given Russell's paradox. But this class is a subclass of the universe so by the Axiom of subsets the universe is not a set.

You can then show using the Axiom of Union set that the complement of a set isn't a set.

[math]\textrm{If}\ A\ \textrm{is a set then so is}\ \bigcup A[/math]

From this axiom you can derive the theorem that

[math]A\cup B\ \textrm{is a set iff}\ A\ \textrm{and}\ B\ \textrm{are sets}[/math]

If A and B are both sets, we know by the Axiom of Union set that their union is a set. Similarly, if the union of A and B is a set, then A and B are sets by the Axiom of Subsets.

Having proved this theorem you can now see that the union of a set and its complement is the universe, which we established isn't a set. So by the theorem we proved, it must be that the complement of this set is not a set.

It's important you understand the goals of ZFC here. ZFC seeks to limit the size of which objects can be considered sets so as to avoid Russell's paradox.

>>12206664
>>12206834
Sorry quick adjustment to prove the theorem you need the Axiom of Pairing:

[math]\textrm{For all objects}\ a\ \textrm{and}\ b\ \textrm{the class}\ \{a,b\}\ \textrm{is a set}[/math]

You need this to prove that if A and B are sets their union is a set. Namely, if A and B are sets, the set containing A and B is a set by the Axiom of Pairing and so its union is a set by the Axiom of Union. Then we rely on the equivalence between the union of A and B and the generalized union of the set containing A and B, which is a matter of definition.

>>12204650
no

>>12204650
It's vacuously true.

It's really funny to think that everything is built upon the empty set. Upon literal nothingness.
It blows my mind that it poses people no problem that everything is built upon a literal paradox: the existence of nonexistence.

If I have set A which is a subset of B, then there is x in B such that x is in A. Assume empty set is subset of C. But for every x in C x won't be in empty set because empty set is empty. Therefore, empty set is not a subset of C.

>>12208917
I think you make it more complicated than it has to be. Set theory gives a reification of higher order properties.

What you call
>the existence of nonexistence.
translates better into the "existence" of the proposition that is true for every thing [math]x[/math], and in an equalitarian theory, you can choose [math]\neg(x=x)[/math].

What you then need is either the axiom that explicitly says that this always-false predicate can be reifeid, i.e. talked about as a "thing". That's the axiom of the empty set and I agree that it is icky.

Alternatively, however, you only need the Axiom of Separation Schema for any predicate (and [math]\neg(x=x)[/math] is one such predicate).

I imagine you can cook up a theory without emptyset by disregarding either of those, but I've never seen such a theory. I've never seen a theory that doesn't validate at least Separation for predicative predicates.

>>12208980
>If I have set A which is a subset of B, then there is x in B such that x is in A
You're just defining the word "subset" in a way that validated your conclusion.
I have nothing against this notion, but it's not the common definition.

I guess you could have Separation and throw out either Negation somehow, then you could get away with existence of emptyset being unprovable.

>>12208992
But what I wrote is just a byproduct of the definition of a subset. To apply the definition directly on empty set is void in logic because there is nothing to apply to. For all x in empty set, x is in B operates on what exactly? What is the x? If x is in B then one of elements of B is in empty set. But which is it? Or should we say it's false because there are no elements? But that's incorrect as well.

To be explicit, if [math]A[/math] is a set and [math]T(x)[/math] is a proposition that is always true, such as [math]x=x[/math], and if [math]F(x)[/math] is a proposition that is always false, such as [math]\neg T(x)[/math], then if you're allowed to separate sets in the sense of
[math]\{x\in A\mid F(x)\}[/math]
then you don't get around having the empty set.

I guess you could have Separation and throw out Negation, then you could get away with existence of emptyset being unprovable.

>>12206834
>there is no universal set
nah that's some bs, there is no set that contains itself
the empty set is fake news

>>12209025
If we take classical propositional logic, then the symbols are
[math]\implies, \bot[/math]
where the first is binary and the second nullary.

We can already form propositons, such as [math] (\bot\implies\bot) [/math] and we may add some more letters and rules to form propositions, as we do below. We may take some propositions as given (axioms).
In the metalogic, we use greek letters for propositional variables. Given any proposition [math]\phi[/math] and another [math]\phi\implies\psi[/math], then we also take [math]\psi[/math] as given.

If you got an equalitarian theory, then you also have
[math]=[/math]
If you got a first-order theory, then you add a quantifier and some alphabet for variables
[math]\exists, x,y,z,u,v..[/math]
and you allow for propositions to be indexed by those (we may use notation [math]\phi(x)[/math] to be explicit about that)
If you got a set theory, then you also add
[math]\in[/math]
Finally we add brackets in the standard sense that everybody can parse out.

I draw a line here.
Because this is already the full name of the game. We can make abbreviations
[math]\neg \phi[/math] for [math]\phi\implies \bot[/math]
[math]\top[/math] for [math]\neg\bot[/math]
[math]\phi\lor\psi[/math] for [math](\neg\phi)\implies\psi[/math]
[math]\phi\land\psi[/math] for [math](\neg\phi)\lor(\neg\psi)[/math]
[math](\forall x)\psi[/math] for [math]\neg((\exists x)(\neg\psi))[/math]

Similarly, there is no curly brackets in formal set theory, but we just like to write
[math]x\in\emptyset[/math] for any false proposition, such as [math]\neg(x=x)[/math].
And we write
[math] x\in \{x\in A\mid \phi(x)\} [/math] for [math] x\in A \land \phi(x) [/math]

Any set is some class any class is a property. From that perspective, set theory (a colleciton of axioms to start the game) is about turning class into something you can quantify over - do higher order logic in a first-order theory.

And the emptyset is already there before you can even define a subset.

>>12209025
If we take classical propositional logic, then the symbols are
[math]\implies, \bot[/math]
where the first is binary and the second nullary.

We can already form propositons, such as [math] (\bot\implies\bot) [/math] and we may add some more letters and rules to form propositions, as we do below. We may take some propositions as given (axioms).
In the metalogic, we use greek letters for propositional variables. Given any proposition [math]\phi[/math] and another [math]\phi\implies\psi[/math], then we also take [math]\psi[/math] as given.

If you got an equalitarian theory, then you also have
[math]=[/math]
If you got a first-order theory, then you add a quantifier and some alphabet for variables
[math]\exists, x,y,z,u,v..[/math]
and you allow for propositions to be indexed by those (we may use notation [math]\phi(x)[/math] to be explicit about that)
If you got a set theory, then you also add
[math]\in[/math]
Finally we add brackets in the standard sense that everybody can parse out.

I draw a line here.
Because this is already the full name of the game. We can make abbreviations

[math]\neg \phi[/math] for [math]\phi\implies \bot[/math]
[math]\top[/math] for [math]\neg\bot[/math]
[math]\phi\lor\psi[/math] for [math](\neg\phi)\implies\psi[/math]
[math]\phi\land\psi[/math] for [math](\neg\phi)\lor(\neg\psi)[/math]
[math](\forall x)\psi[/math] for [math] \neg ((\exists x)(\neg\psi) ) [/math]

Similarly, there is no curly brackets in formal set theory, but we like to write
[math]y\in\emptyset[/math] for any false proposition such as [math]\neg(y=y)[/math].
And we write
[math] y\in \{x\in A\mid \phi(x)\} [/math] for [math] y\in A \land \phi(y) [/math]

Any set is some class any class is a property. From that perspective, set theory (a colleciton of axioms to start the game) is about turning class into something you can quantify over - do higher order logic in a first-order theory.

And the emptyset is already there before you can even define a subset.

Maybe again but more to the point. If we adopt the Axiom of the Empty set, this means to take as given the statement

[math] (\exists x)(\forall y) \neg (y\in x) [/math]
In words
>"there is some x with the property that for all y, it's not the case that y is in x"

With my above way of introducing things, this would be the statement
[math] (\exists x)(((\exists y)(y\in x))\implies \bot) [/math]
In words
>"there is some x with the property that if there were an y that is in it, we immediatenly get an absurdity"

Now take e.g. natural deduction
https://en.wikipedia.org/wiki/Natural_deduction
which formalizes more precisely what we can do with existential equanfiers during a proof (logical axioms).

Well, once a statement of existence is proven, we can instantiate elements during a proof
In particular, given e.g.
[math] (\exists x)(\forall y) \neg (y\in x) [/math]
we can instantiate [math]x[/math] as some "thing", call it [math]e[/math] (also called [math]\emptyset[/math])
and we can, during a prove, work with this [math]e[/math] assuming this letter is a thing with the property
[math] (\forall y) \neg (y\in e) [/math]

This is just to make the case that it's hard to get around this object - at least in this first-order logic conceptualization of mathematics.

There's an axiom of infinity.
Wouldn't it make more sense to have an axiom of finitude instead?
Why can't we have infinite sets right from the get go

nothing contains nothing
if follows from ZFC (axiom of foundation) that there is no empty set

>>12209243
You can leave out the axiom of infinity and be left with a nice and consistent theory that doesn't prove infinite sets exists. Indeed, there's hands on models for it, like
https://en.wikipedia.org/wiki/Hereditarily_finite_set

If you axiomize that model directly, you'll find that all sets are finite and this should be provable even in a constructive fashion.

In weak theories, you can even have it that everything is countable, see
https://en.wikipedia.org/wiki/Kripke%E2%80%93Platek_set_theory

If you drop LEM and Powerset and rewrite Regularity, iirc you can adopt as axiom that every set is subcountable.

>>12208980
>If I have set A which is a subset of B, then there is x in B such that x is in A
wrong. complete non-sequitur

>>12204650
isnt that like saying "nothing exists everywhere/anywhere"?

>>12209345
>nothing exists

nope

>>12209309
KP says nothing the size of its infinite sets. It is consistent with KP all sets are finite. Since [math]L_{\omega}[/math] is model. It is consistent with KP that there are uncountable sets, since [math]L_{\omega_2}[/math] is a model of KP. It is consistent that every set is countable since [math]L_{\omega_1}[/math] is a model of KP. If you want a re. theory in which every set is countable, you could add the axiom to KP that says for every set x there is an injective function from x into [math]\omega[/math]. The class of these models would be exactly [math]L_{\omega_1^{\text{CK}(x)}}[/math], for a real x, and [math]L_{\omega_1}[/math].

>>12209345
No, it's saying for every set X, if x is a member of the empty set then x is a member of X. This is trivially true.

>>12209379
how can x be a member of the empty set if the empty set is empty?

If a false statement implies all statements, couldn't our whole reality merely be an implication of a single false statement?

>>12209523
>a false statement implies all statements
nope

>>12204650
Think of a subset as writing a 0 or a 1 on top of every element in a main. Each subset will be uniquely defined by such a binary string. The empty set is the string of all 0s

>>12209431
who says x is element of the empty set?

>>12209551
could you be more explicit?
maybe I should have said proposition instead of statement

>>12204650
A way to test if something is a subset of a bigger set is by seeing if it's possible to append it to another set in order to get a bigger set. In this case, appending an empty set to any other set will just result in that other set, thus it is by the law we just established true that the empty set is a subset of all other sets. It's like saying that 0 is smaller than all positive numbers, because adding 0 to any positive number does not exceed said number.

>>12209577
>could you be more explicit?
nope

>>12209562
>the empty set is

nope

Nope party? Nope party

>>12209591
nope

>>12206552
>the universal set
>>12206664
>in the given universal set U
these are different things
if you are working with a, b, and c, your given universal set is {a, b, c}
there is no universal set for all sets

>>12209574
>who says x is element of the empty set?
>>12209379
>if x is a member of the empty set

>>12209609
>there is no universal set for all sets
There is
I will call it Ü
Its an universal set such that it contains all possible sets except itself

>>12204650
Try describing a set that doesn't have the empty set as a subset. Let me know if you come up with anything good.

>>12209616
I came up with something good

>>12209630
good for you faggot

>>12209614
>using a set in its own definition

>>12209637
i did not
the Chad set will be the set that contains all possible and impossible sets
the Chad set will be a transcendent set

>>12209616
>Try describing a set that doesn't have the empty set as a subset
Any set?
Every mathematical object only comes to exist if someone think about it or represents it in reality
If you never think or represent explicitly a set with an empty set, then you got what you needed

>>12209660
Sure, any set. Just describe one.

>>12204650
take a set. take away nothing from it and put it in a set. you have the same set plus a null set.
so, following these simple steps, yes.

Set A is a subset of B iff A intersect B is A. Prove this and then A=emptyset.

>>12209609
The universal set is the context. In the context of analysis is [math] \mathbb{R} [/math] for example. In that of number theory, the universe is either [math] \mathbb{Z} [/math] or [math] \mathbb{N} [/math].

>>12209824
yes, thank you for agreeing with me
but number theory beyond first semester usually involves much more, so the universe is typically the category of fields which itself is not a set

>>12209775
S = {1}
done

>>12210052
{} is a subset of {1}
try again?

>>12210074
>{} is a subset of {1}
prove it

>>12210149
it’s what the word means. you don’t prove that blue is blue, blue is simply a word to describe something that’s blue.

>>12210163
i meant prove that {} is the same as the empty set

>>12210180
it’s literally a symbol that means the empty set. you don’t prove that 1 is the same as one is the same as uno is the same as ichi is the same as eins. these are all just symbols describing the same thing.

>>12210180
Are you retarded?

>>12210198
>>12210192
bunch of fucking retarded niggers, you didnt prove the empty set exists

>>12210212
thats because its a fucking axiom retard

>>12210212
no one asked me to, schizo. they asked me to prove that 1=one

>>12209847
classic number theory

>>12204650
daily reminder that both infinity and absolute nothingness are a mystery - they work on paper but nowhere else

>>12209374
I just said "you can have", not that the theory proves it.
In ZF, you cannot have it.

>>12204650
>Is this true?
Well if you can make the empty set just by removing elements from another set it must be a subset, right?

>>12209431
>how can x be a member of the empty set if the empty set is empty
That's why the conditional is vacuously true.

>>12204650
Let S be the set that does not include the empty set. QED.

>>12204650
>>12204650
no. the empty set is not a member of the set of sets that cannot exist.

for instance. the set of all sets that do not contain themselves is a set that cannot exist. therefore it is a member of the set of sets that can not exist. this set dose not contain the empty set.

>sets

>>12211620
>Let n be the natural number that is not divisible by 1.
retard

>>12211624
wrong. the empty set is still a subset of the set of sets that cannot exist. if you remove all members of the set of sets that cannot exist, you get the empty set.

>>12211624
My God finish high school or hit the books. Not everything can be a set, and the class of classes that cannot exist is not a set.

>>12211681
wrong.

if the set can only contain sets that cannot exist and the empty set is a set that can exist then by definition the empty set cannot be a member of the set of sets that cannot exist.

>>12211804
it's not a member of the set, it's a member of the set's subsets

>>12211804
the set of sets (A) that cannot exist is a paradox
if a set (B) cannot exist, then it is in A, and therefore B exists. A contradiction
if a set B exists, then it is not B. A contradiction

>>12211620
Prove that such a set exists

>>12204650
>>12204660
If you can't understand this, study something else

>>12204678
back off its ours now

>>12204774
P(A) = {B | B subset of A} (power set includes every subset) and since the emptyset is a subset of every set by definition it has to be a member of the power set