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8836620 No.8836620 [Reply] [Original]

Reflexive property of an equivalence relation is redundant.
Proof. Let ~ be an equivalence relation and pick x,y such that x~y. By symmetry, y~x. By transitivity, x~x.
Anyone got any more?
Also, spot the flaw. If you can't you should go back to 11th grade.

>> No.8836771

Define x and y

>> No.8836794

>>8836620
You can't pick x and y. The relation has to be true for all x and y in some group

>> No.8838352

>>8836771
Are you stupid?
>>8836794
Brainlet. Consider the set {1,2,3} and the congruence mod 2 relation. You clearly do have to pick x and y e.g. x= 3, y = 1 since 3 mod 2 = 1 mod 2.
Why do you not have to "pick"?

>> No.8838369

>>8836620
e z p z

There may not be any elements relating to x.

My turn:

Let L be a loop in the 2-sphere, S^2. Choose a point p on the sphere that the loop doesn't hit. S^2-{p} is homeomorphic to R^2 which is simply connected. Thus you can contract L in S^2-{p}, and so you can contract L in S^2. Therefore S^2 has trivial fundamental group.

>> No.8838371

>>8838352
>>8838352
What if you pick x and y from an empty set? You have to say what your objects belong to and assume they are. Pick,let,choose are not good enough.

>> No.8838391

>>8836620
>false theorems

.999...=1

>> No.8838429

>>8838369

Shit, technically this is a false proof of a true theorem.

>> No.8838480

>>8838369
how is that a false proof?

>> No.8838516

>>8836620
Existentialism is for brainlets
Text: ignored

>> No.8838820

>>8838391

.99999... is more the limit of a number that is growing infinitely closer to 1, or in other words 1. There aren't any integers that you can divide to get .999999.... so honestly its hardly a number but rather a theoretical geometric sum

>> No.8838853

>>8838480
I guess you can find curves that fill the sphere

>> No.8839016
File: 30 KB, 675x1127, someone made another 0.999... = 1 thread.png [View same] [iqdb] [saucenao] [google]
8839016

>>8838391
>>8838820
Just to put this to rest before it starts.

>> No.8839083

>>8839016
tank youu desu
everyone save this and use it

>> No.8839640

>>8838853
that occurred to me, but in arguments like this one, it's standard to implicitly assume that the loops are smooth, based on results like this

https://math.stackexchange.com/questions/298873/approximate-continuous-mapping-by-smooth-mappings-on-manifoldbott-tu-book

and i don't believe a smooth map from S1 to S2 can be surjective