Proving something that is equinumerous

[ver_mathstats]

Homework Statement

Prove that (0,1) ≈ R.

Relevant Equations

arctan: R→(-π/2,π/2)
arctan(x)=y ⇔ tan(y)=x for x∈R and y∈(-π/2,π/2)

I am slightly confused as to how to prove this. I know that two sets are equinumerous if there is a bijection between them. So we are trying to find f: (0,1)→R? It told us that we may assume the inverse tangent function so that would mean
arctan: R→(-π/2,π/2). This satisfies arctan(x)=y ⇔ tan(y)=x for x∈R and y∈(-π/2,π/2). We were also given that we could assume arctan is a bijective without proof. I am having trouble moving forward from this information and how to construct a proof out of all of these.

Thank you.

 

[LCKurtz]

I am slightly confused as to how to prove this. I know that two sets are equinumerous if there is a bijection between them. So we are trying to find f: (0,1)→R? It told us that we may assume the inverse tangent function so that would mean
arctan: R→(-π/2,π/2). This satisfies arctan(x)=y ⇔ tan(y)=x for x∈R and y∈(-π/2,π/2). We were also given that we could assume arctan is a bijective without proof. I am having trouble moving forward from this information and how to construct a proof out of all of these.

Thank you.

Think about how to get a bijection between $(0,1)$ and $(-\frac \pi 2,\frac \pi 2)$ so you can combine it with the given arctan bijection to establish your 1-1 correspondence.

 

[ver_mathstats]

Think about how to get a bijection between $(0,1)$ and $(-\frac \pi 2,\frac \pi 2)$ so you can combine it with the given arctan bijection to establish your 1-1 correspondence.

I found f(x)=(1/2)+(x/π). But if we combine with arctan would it then be f(x)=(1/2)+(1/π)arctan(x)?

 

[LCKurtz]

I found f(x)=(1/2)+(x/π). But if we combine with arctan would it then be f(x)=(1/2)+(1/π)arctan(x)?

Can't you tell me instead of asking? Does that work?

 

[ver_mathstats]

Can't you tell me instead of asking? Does that work?

I am slightly confused if it works. When I input my values I get numbers (0,1) or 0<x<1, so yes unless I am not understanding but this is how I interpreted it.

 

[LCKurtz]

You have arctan which is a 1-1 function from R onto $(-\frac \pi 2,\frac \pi 2)$ and a function f which is (you need to prove) one to one from $(-\frac \pi 2,\frac \pi 2)$ onto $(0,1)$. Then you need to show or have a theorem about the compositions of bijections being a bijection.