Is (0,1) Uncountable If and Only If \Re is Uncountable?

[rbzima]

Show that (0,1) is uncountable if and only if \Re is uncountable.

I have a nice little proof showing (0,1) is uncountable, however I'm wondering how I can make implications that \Re and vice versa.

 

[sutupidmath]

Well, i think that you might want to see if you can construct a 1-1 function (correspondence) with the naturals( positive integers).
f:(0,1)-->Z (integers) . Well, you might also use the property that if a set A is uncountable, and further if this set A is a subset of B, then also B is uncountable.
So basically if you manage to show that (0,1) is uncountable, then automatically you have shown that R is uncountable, since even if we managed to put all other elements of R in an order and count them, we defenitely could not count the elements of R that are within the interval (0,1).

 

[morphism]

You can setup a bijection between (0,1) and R - try playing around with tan(x).

 

[rbzima]

f:(0,1)-->Z (integers). Well, you might also use the proberty that if a set A is uncountable, and further if this set A is a subset of B, then also B is uncountable.

I basically just wanted to say that I honestly think you should coin the term proberty because it makes this feel like a property of probabilistic outcomes!

 

[sutupidmath]

I basically just wanted to say that I honestly think you should coin the term proberty because it makes this feel like a property of probabilistic outcomes!

Well, if you really like, you can start using it from now on, i will not suit u for plagiarism!