Question

Prove that [0,1] X [0,1] ~ [0,1]

Answer 1

what operation is \(\times\)?
what equivalence is ~?

Answer 2

Well, X is cartesian product. Im not sure what kind of equivalence ~ is. Its just equivalence? I don't think we've discussed different kinds of equivalences. I assumed the idea was to show that if you found the cartesian product given, that it would map to [0,1] and be one-to-one and onto. The idea before has seemingly been to invent a function to try and do a proof like this, but not sure how to go about it. Sorry if that isn't specific enough, im not sure what else I could add to that.

Answer 3

Writing down an explicit bijection might be a little rough, but it is not so hard to create an injection that goes from [0,1]x[0,1] to [0,1] (which implies that |[0,1]x[0,1]|<=[0,1]), and another injection that goes from [0,1] to [0,1]x[0,1] (which implies |[0,1]|<=|[0,1]x[0,1]|). Thus the cardinalities of the two sets are the same, and thats enough to show a bijection must exist.

Answer 4

I guess I'm not confident of my understanding of [0,1]x[0,1]. I can create a function, f(x), that might map [0,1] to [0,1], that makes sense to me. But if I were to use [0,1]x[0,1], does that require a function f(x,y) that maybe maps to some z and then vice versa, a f(z) that maps to (x,y)? Once I'm sure of that, creating two injections does sound much easier.

Answer 5

Yes your idea is correct. The set [0,1]x[0,1] is defined as: $$\{(x,y)\mid x,y\in [0,1]\}$$ So your function from [0,1]x[0,1] to [0,1] would need to be in two variables.
Similarly, a function from [0,1] to [0,1]x[0,1] would need to be in one variable, but spit out an ordered pair f(z)=(x,y)

Answer 6

Alright, Ill see what I can do, thanks :)