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

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SUMMARY

The discussion centers on proving that the interval A = [0, 2) has the same cardinality as the set B = [5, 6) U [7, 8) by constructing a bijection. The proposed function is defined as f(x) = x + 5 for x ∈ [0, 1) and f(x) = x + 6 for x ∈ [1, 2). Participants emphasize the need to demonstrate that this function is both injective and surjective. Corrections were made regarding notation, specifically ensuring the intervals are accurately represented as [5, 6) and [1, 2).

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KOO
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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?
 
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KOO said:
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.
 
Last edited:
Fernando Revilla said:
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!
 
KOO said:
Did you mean [5,6) and not [5.6)?
Also, [1,2) and not [1.2)?

Of course, my fingers were clumsy. :)
 

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