Do Intervals [0, 2) and [5, 6) U [7, 8) Have the Same Cardinality?

[KOO]

Prove that the interval A = [0 , 2) has the same cardinality as the set B = [5 , 6) U [7 , 8) by constructing a bijection between the two sets

Attempt:

x ↦ x + 5 for x ∈ [0 ; 1)
x ↦ x + 6 for x ∈ [1 ; 2)

What to do next?

 

[Fernando Revilla]

What to do next?

Prove that $f:[0,2)\to [5\color{red},\color{\black}6)\cup [7,8)$
$$f(x)=\left \{ \begin{matrix} x+5& \mbox{ if }& x\in [0,1)\\x+6 & \mbox{ if }&x\in [1\color{red},\color{\black}2)\end{matrix}\right.$$
is injective and surjective.

 

[KOO]

Prove that $f:[0,2)\to [5.6)\cup [7,8)$
$$f(x)=\left \{ \begin{matrix} x+5& \mbox{ if }& x\in [0,1)\\x+6 & \mbox{ if }&x\in [1.2)\end{matrix}\right.$$
is injective and surjective.

Did you mean [5,6) and not [5.6)?

Also, [1,2) and not [1.2)?

Thanks!

 

[Fernando Revilla]

Did you mean [5,6) and not [5.6)?
Also, [1,2) and not [1.2)?

Of course, my fingers were clumsy. :)

 

[blue_raver22]

To construct a bijection between the two sets, we need to show that every element in set A is paired with a unique element in set B and vice versa.

To do this, we can define a function f: A → B as follows:

f(x) =
x + 5 if x ∈ [0, 1)
x + 6 if x ∈ [1, 2)

This function maps every element in set A to a unique element in set B, and since it covers the entire range of both sets, it is surjective.

To show that it is also injective, we need to show that no two elements in set A map to the same element in set B. Since the intervals in set A and B do not overlap, this is true.

Therefore, we have shown that the function f is both surjective and injective, making it a bijection. This means that the two sets have the same cardinality.