Compare cardinalities: | (0,1) | = | (0,2] ∪ (4,7) |

[mar_f]

Hello, I was given this exercise at school:
[MATH] |(0,1)| = |(0,2] \cup (4,7)| [/MATH]Where should I start? Can I somehow use here the fact, that if two disjoint sets A,B have the same cardinality as natural numbers, then the cardinality of the union of A,B is the same as the cardinality of N?

Thanks.

 

[pka]

Hello, I was given this exercise at school:
[MATH] |(0,1)| = |(0,2> \cup (4,7)| [/MATH]Where should I start? Can I somehow use here the fact, that if two disjoint sets A,B have the same cardinality as natural numbers, then the cardinality of the union of A,B is the same as the cardinality of N?

Your notation is a bit non-standard. It may be simply a typo. Do you mean \(\displaystyle |(0.1)|=|(0,2)\cup (4,7)|~? \)
Also the set \(\displaystyle \mathbb{N} \) is the set of natural numbers and is countable. Whereas \(\displaystyle (0,1)=\{x\in\mathbb{R}: 0<x<1\} \) in uncountable and equal in cardinally to \(\displaystyle \mathbb{R} \) itself.
To answer this find to bijections \(\displaystyle (0,0.5] \Leftrightarrow (0,2)\& (0.5,1) \Leftrightarrow (4,7)\)

 

[mar_f]

Hello,
I am sorry for the notation, I always forget about that. (We use [MATH] > [/MATH] instead of [MATH] ] [/MATH] here in Europe.)
So the bijections should be:
[MATH] (0,\frac{1}{2}] \leftrightarrow (0,2] [/MATH] and
[MATH] (\frac{1}{2},1) \leftrightarrow (4,7) [/MATH] ?
Can you please explain me how did you find this solution?

Many thanks.

 

[pka]

Hello,
I am sorry for the notation, I always forget about that. (We use [MATH] > [/MATH] instead of [MATH] ] [/MATH] here in Europe.)
So the bijections should be:
[MATH] (0,\frac{1}{2}] \leftrightarrow (0,2] [/MATH] and
[MATH] (\frac{1}{2},1) \leftrightarrow (4,7) [/MATH] ?
Can you please explain me how did you find this solution?

Many thanks.

Suppose that each of \(\displaystyle \Theta~\&~\Phi\) is a bijection \(\displaystyle \Theta : (0,0.5] \leftrightarrow (0,2)\quad \Phi : (0.5,1) \leftrightarrow (4,7)\)
\(\displaystyle F=\begin{cases}\Theta(x) &: x\in(0,0.5] \\ \Phi(x) &: x\in (0.5,1)\end{cases}\)
Now show that
\(\displaystyle F(x) : (0,1)\leftrightarrow (0,2]\cup (4,7)\) is a bijection.

 

[mar_f]

Hello, is this solution correct please?
[MATH] \theta(x) = 4x[/MATH] and [MATH] \phi(x) = 6x+1[/MATH] ?
Is this the final result or do I have to find a single bijection for them?
Thank you.

 

[pka]

Hello, is this solution correct please?
[MATH] \theta(x) = 4x[/MATH] and [MATH] \phi(x) = 6x+1[/MATH] ?
Is this the final result or do I have to find a single bijection for them?

Given that you use \(\displaystyle >\) where we use \(\displaystyle ]\), it does indeed work.

 

[mar_f]

Thank you. Can I have one more question please?
Can I solve this one [MATH] |[2,3]| = |[0,1] \cup (5,8) \cup [10,11]| [/MATH] similarly like the previous one?
[MATH] \phi_{1}(x): [2,\frac{7}{3}] \leftrightarrow [0,1][/MATH][MATH] \phi_{2}(x): (\frac{7}{3},\frac{8}{3}) \leftrightarrow (5,8)[/MATH][MATH] \phi_{3}(x): [\frac{8}{3},3] \leftrightarrow [10,11][/MATH]
[MATH]\phi(x) = \begin{cases} 3x-6, x \in [2,\frac{7}{3}] \\ 9x-16 , x \in (\frac{7}{3},\frac{8}{3}) \\ 3x+2 , x \in [\frac{8}{3},3] \end{cases} [/MATH]
Thanks.

 

[pka]

Thank you. Can I have one more question please?
Can I solve this one [MATH] |[2,3]| = |[0,1] \cup (5,8) \cup [10,11]| [/MATH] similarly like the previous one?
[MATH] \phi_{1}(x): [2,\frac{7}{3}] \leftrightarrow [0,1][/MATH][MATH] \phi_{2}(x): (\frac{7}{3},\frac{8}{3}) \leftrightarrow (5,8)[/MATH][MATH] \phi_{3}(x): [\frac{8}{3},3] \leftrightarrow [10,11][/MATH]
[MATH]\phi(x) = \begin{cases} 3x-6, x \in [2,\frac{7}{3}] \\ 9x-16 , x \in (\frac{7}{3},\frac{8}{3}) \\ 3x+2 , x \in [\frac{8}{3},3] \end{cases} [/MATH]
Thanks.

There are as many ways to do these as people doing them.
Have a look HERE.