# Thread: Cardinality if infinite sets

1. ## Cardinality if infinite sets

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?

2. Originally Posted by novice
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.

3. Originally Posted by Plato
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 $\displaystyle f0,1) \rightarrow \mathbb{R}$. 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 $\displaystyle |(0,1)-\{0.25\}|\not=|\mathbb{R}-\{5\}|$?

4. Originally Posted by novice
Let $\displaystyle f0,1) \rightarrow \mathbb{R}$. 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 $\displaystyle \left| {\left( {0,1} \right)\backslash A} \right| = \left| {\mathbb{R}\backslash B} \right|$