Cardinalities of Sets: Prove |(0, 1)| = |(0, 2)| and |(0, 1)| = |(a, b)|

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How to prove the open intervals (0,1) and (0,2) have the same cardinalities? |(0, 1)| = |(0, 2)|

Let a, b be real numbers, where a<b. Prove that |(0, 1)| = |(a, b)|

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|(0,1)| = |R| = c by Theorem
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I know that we need to construct a function f: (0,1)->(0,2) and prove f is bijection so that |(0, 1)| = |(0, 2)|

same process of proving |(0, 1)| = |(a, b)|

but how to construct a function f: (0,1)->(0,2)
and how to construct a function g: (0,1)->(a,b) where a<b and a,b are real numbers?

I know how to construct a function f: (0,1)->R
by define a function f(x)=(1-2x)/[x(x-1)] where x cannot be 0 and 1 and when the middle domain(f)=1/2, f(1/2)=0

How can I expand this knowledge and to define a function that the domain(f) is within (0,1) and the range(f) falls into (0,2) or any close interval (a,b)?
 

Answers and Replies

  • #2
radou
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Try constructing a linear function from (0, 1) to (0, 2).
 
  • #3
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Yeah, my bet! a, b are real numbers

I've constructed a linear function f: (0,1)->(0,2) defined by f(x)=2x
such that f(1/2)=1, when x=1/2 (mid point of domain), y=1 (mid point of range)
This linear function is certainly bijection, therefore |(0,1)|=|(0,2)|

But how to prove |(0,1)|=|(a,b)| where a, b are real numbers and a<b???
 

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