(Set Theory) reall dumb question)

[cunning yew]

I'm going through analysis I by Terry Tao and I'm supposed to prove this

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[cunning yew]

But along the way I reached this

Isn't this a contradiction? having both y∈A and y∉A as property

(im using y instead of x bc im already using capital X)

[edgy cedar]

So first of all, X = {a, b, c} is never mentioned anywhere

it's a bit strange to assume that X is in said form

Do you have something along the lines of $X \setminus A = X \cap \overline{A}$ ?

[neat condor]

Rion

[cunning yew]

Its mentioned at the start of the exercise sorry

[edgy cedar]

It still doesn't seem all too needed to me to prove the property

Also the last line is wrong

y can be in the union of A, B and C without being in A at all

[cunning yew]

Ohh thats it thanks

Sorry am stupid

[stiff pond]

@cunning yew has given 1 rep to @edgy cedar

[cunning yew]

@edgy cedar Is there an "order of operations" in group theory though?

[edgy cedar]

Hello again

[cunning yew]

1 min lemme writ eit it out

[edgy cedar]

Group theory seems a bit disconnected from set theory, though

[cunning yew]

set theory* not group theory

sorry

[edgy cedar]

Ah ok

Well some operations commute/are associative

some don't

[cunning yew]

Like is $x \notin A: x \in B$ the same as $x \in B: x \notin A$

[neat condor]

TaschenRechner

[edgy cedar]

The notation is quite confusing here, but the latter would indicate $B \setminus A$, whereas the former would indicate $\overline{A} \cap B$

[neat condor]

Rion

[edgy cedar]

Which, in all technicality, should be the same thing

[cunning yew]

wha does the overline mean?

[edgy cedar]

The complement?

[cunning yew]

ye

[edgy cedar]

So in this specific case yes the sets should be the same

[cunning yew]

what if there is an element thats inboth though?

[edgy cedar]

In both what?

A and B?

[cunning yew]

ye

[edgy cedar]

well it is in neither set

you can check the specifications manually

[cunning yew]

wdym

[edgy cedar]

Sorry

It is in neither ${ x \notin A, x \in B}$ and ${ x \in B, x \notin A}$

[neat condor]

Rion

[cunning yew]

so $\notin$ takes "precedence" over $\in$?

[neat condor]

TaschenRechner

[edgy cedar]

It doesn't

I just argued those two are the same set

$B \setminus A = B \cap \overline{A} = \overline{A} \cap B$

[neat condor]

Rion

[cunning yew]

wait shit

the problem is like, in monkey brain, if there is an element thats both in A and B, it says "the elements in this set must include all that are in b but also none of them should be in a".
so what happens if there is an element both in a and b

[compact vapor]

doesn’t this only work if A is a subset of X

[cunning yew]

It is

[compact vapor]

which is generally what the word “partition” means

[cunning yew]

I just forgot t omention it

But I have another question though

Which im asking rn

[compact vapor]

oh

sorry i didn’t see that one

[compact vapor]

this?

[cunning yew]

kind of

[cunning yew]

.

[edgy cedar]

Then it is just not in B \ A

It does not meet the requirements to be in B \ A

[compact vapor]

^

[edgy cedar]

I am not sure what your doubt is

[cunning yew]

sorry i suck at set theory

[edgy cedar]

I mean, no worries, you are a beginner, I am here to answer your questions

[cunning yew]

the problem is if i write it I get ${x \notin A, x \in B$
"x must not be in a but must also be in b"

[neat condor]

TaschenRechner
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[edgy cedar]

I just need to understand where your problem lies

[cunning yew]

oops

the problem is if i write it I get ${x \notin A, x \in B}$
"x must not be in a but must also be in b"

[neat condor]

TaschenRechner

[edgy cedar]

Yes, so any x that is in both A and B is not in your aforementioned set

Because it fails to fulfill the first requirement

[cunning yew]

so, they are not perscriptive but rather descriptive?

[edgy cedar]

The set of elements that are in both A and B is commonly denoted $A \cap B$

[neat condor]

Rion

[cunning yew]

ik

[compact vapor]

sorry quick question, what do those mean in terms of sets

[edgy cedar]

You can see a set as a collection of elements that have a certain property, no matter how convoluted it is

So if an element does not have said property, it is not in the corresponding set.

The property in question is to be in B and not in A

[cunning yew]

this is the problem
but honestly its more helpful to think of it as ${x \in B:x \in \overline A}$

[neat condor]

TaschenRechner

[cunning yew]

(is that how $\notin$ is defined? sorry)

[neat condor]

TaschenRechner

[edgy cedar]

Whatever floats your boat man

[edgy cedar]

That is exactly $B \cap \overline{A}$

[neat condor]

Rion

[cunning yew]

wait sorry

is it okay if i think of $A \backslash B$ as $x \in A : x \in \overline B$ instead of ${x \in A : x \notin B}$?

[neat condor]

TaschenRechner

[edgy cedar]

Yes, because they are the same set

As said, whatever floats your boat

[cunning yew]

ok thanks a lot

@edgy cedar Oke I think i was just thinking about properties these the wrong way. Instead of "collection of elements that fulfill all these" I was thinking about "all elements that are in B that also fulfill the following:"

[edgy cedar]

This in itself is just a conjunction of two properties