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.
Define x and y
>>8836620
You can't pick x and y. The relation has to be true for all x and y in some group
>>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.
>>8836620
>false theorems
.999...=1
>>8838369
Shit, technically this is a false proof of a true theorem.
>>8838369
how is that a false proof?
>>8836620
Existentialism is for brainlets
Text: ignored
>>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
>>8838480
I guess you can find curves that fill the sphere
>>8838391
>>8838820
Just to put this to rest before it starts.
>>8839016
tank youu desu
everyone save this and use it
>>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