Showing that two intervals have the same cardinality ?

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SUMMARY

The discussion centers on demonstrating that the intervals [0,1] and [0,2] have the same cardinality by establishing a bijective function. The function f(x) = 2x is presented as a valid bijection, illustrating a systematic approach to finding such functions for similar intervals. Additionally, a non-linear function, f(x) = x(3-x), is mentioned as another option, emphasizing that while there is no universal formula for finding bijections, intuition and experience play crucial roles in this process.

PREREQUISITES
  • Understanding of cardinality in set theory
  • Knowledge of bijective functions
  • Familiarity with interval notation
  • Basic algebraic manipulation skills
NEXT STEPS
  • Research the concept of cardinality in set theory
  • Study bijective functions and their properties
  • Explore examples of bijections between different intervals
  • Learn about non-linear functions and their applications in establishing bijections
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Mathematics students, educators, and anyone interested in understanding set theory and the concept of cardinality through bijective functions.

SMA_01
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Homework Statement



I need to show that [0,1] and [0,2] have the same cardinality by giving a formula for a function that is bijective. Aren't there a number of functions that can fit this description? Can I then use any one? I'm a little confused, my teacher didn't really elaborate much upon this. Is there a specific formula to find the function, not only in this case, but generally?


Edit: I have used y=f(x)=2x, I'm guessing this process is pretty systematic when dealing with such intervals?
 
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SMA_01 said:

Homework Statement



I need to show that [0,1] and [0,2] have the same cardinality by giving a formula for a function that is bijective. Aren't there a number of functions that can fit this description? Can I then use any one? I'm a little confused, my teacher didn't really elaborate much upon this. Is there a specific formula to find the function, not only in this case, but generally?Edit: I have used y=f(x)=2x, I'm guessing this process is pretty systematic when dealing with such intervals?

You can use any bijection. f(x) = 2x is fine. So is a non-linear one like f(x) = x(3-x), which is bijective over [0,1]. It's your choice. But the former has the advantage of being immediately and intuitively obvious.
 
SMA_01 said:

Homework Statement


Is there a specific formula to find the function, not only in this case, but generally?

There's definitely no "formula" here. Intuition and experience are your best friends when seeking for Bijections.
 

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