Cardinality if infinite sets

Sep 2009
502
39
Let \(\displaystyle A=\{1,2,3\}\) and \(\displaystyle B =\{a,b,c\}\). Since they are finite sets, it’s quite obvious that they have the same number of elements.

I have read the proof that the infinite sets \(\displaystyle |(0,1)| = |\mathbb{R}|\).
We know that \(\displaystyle (0,1) \subset \mathbb{R}\), and I know there is a bijective function between the two sets, but how does one explain that \(\displaystyle (0,1) \) and \(\displaystyle \mathbb{R}\) have the same number of elements?
 

Plato

MHF Helper
Aug 2006
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8,633
I have read the proof that the infinite sets \(\displaystyle |(0,1)| = |\mathbb{R}|\).
We know that \(\displaystyle (0,1) \subset \mathbb{R}\), and I know there is a bijective function between the two sets, but how does one explain that \(\displaystyle (0,1) \) and \(\displaystyle \mathbb{R}\) have the same number of elements?
What does it mean ‘to have the same number’?
Each time I pick a number from \(\displaystyle (0,1)\) you can match it with a unique number from \(\displaystyle \mathbb{R}\).
Likewise you pick any number from \(\displaystyle \mathbb{R}\) then I can match it with a unique number from \(\displaystyle (0,1)\).
And the matches are all different because of the bijection.
You and I have the ‘same number’ of elements.
 
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Sep 2009
502
39
What does it mean ‘to have the same number’?
Each time I pick a number from \(\displaystyle (0,1)\) you can match it with a unique number from \(\displaystyle \mathbb{R}\).
Likewise you pick any number from \(\displaystyle \mathbb{R}\) then I can match it with a unique number from \(\displaystyle (0,1)\).
You and I have the ‘same number’ of elements.
Let \(\displaystyle f:(0,1) \rightarrow \mathbb{R}\). Since there is a bijective function \(\displaystyle f\) from \(\displaystyle (0,1)\) to \(\displaystyle \mathbb{R}\).

Let us say \(\displaystyle f(0.25)=3\) and \(\displaystyle f(0.3)=5\). If I remove equal number of elements from both sets, say \(\displaystyle (0,1)-\{0.25\}\) and \(\displaystyle R-\{5\}\). Now \(\displaystyle 0.25 \notin (0,1)\) and \(\displaystyle 3 \in \mathbb{R}\) , and \(\displaystyle 0.3 \in (0,1)\) and \(\displaystyle 5 \notin \mathbb{R}\).

Have we lost the bijective function?

Can we say \(\displaystyle
|(0,1)-\{0.25\}|\not=|\mathbb{R}-\{5\}|
\)?
 

Plato

MHF Helper
Aug 2006
22,461
8,633
Let \(\displaystyle f:(0,1) \rightarrow \mathbb{R}\). Since there is a bijective function \(\displaystyle f\) from \(\displaystyle (0,1)\) to \(\displaystyle \mathbb{R}\).

Let us say \(\displaystyle f(0.25)=3\) and \(\displaystyle f(0.3)=5\). If I remove equal number of elements from both sets, say \(\displaystyle (0,1)-\{0.25\}\) and \(\displaystyle R-\{5\}\). Now \(\displaystyle 0.25 \notin (0,1)\) and \(\displaystyle 3 \in \mathbb{R}\) , and \(\displaystyle 0.3 \in (0,1)\) and \(\displaystyle 5 \notin \mathbb{R}\).

Have we lost the bijective function?

Can we say \(\displaystyle
|(0,1)-\{0.25\}|\not=|\mathbb{R}-\{5\}|
\)?
I have no idea what any of that says. Much less what your point is.
If \(\displaystyle A\) is any finite subset of \(\displaystyle (0,1)\) and \(\displaystyle B\) is any finite subset of \(\displaystyle \mathbb{R}\) then \(\displaystyle \left| {\left( {0,1} \right)\backslash A} \right| = \left| {\mathbb{R}\backslash B} \right|\)
 
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