# Thread: Cardinality if infinite sets

1. ## Cardinality if infinite sets

Let $A=\{1,2,3\}$ and $B =\{a,b,c\}$. 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 $|(0,1)| = |\mathbb{R}|$.
We know that $(0,1) \subset \mathbb{R}$, and I know there is a bijective function between the two sets, but how does one explain that $(0,1)$ and $\mathbb{R}$ have the same number of elements?

2. Originally Posted by novice
I have read the proof that the infinite sets $|(0,1)| = |\mathbb{R}|$.
We know that $(0,1) \subset \mathbb{R}$, and I know there is a bijective function between the two sets, but how does one explain that $(0,1)$ and $\mathbb{R}$ have the same number of elements?
What does it mean ‘to have the same number’?
Each time I pick a number from $(0,1)$ you can match it with a unique number from $\mathbb{R}$.
Likewise you pick any number from $\mathbb{R}$ then I can match it with a unique number from $(0,1)$.
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 $(0,1)$ you can match it with a unique number from $\mathbb{R}$.
Likewise you pick any number from $\mathbb{R}$ then I can match it with a unique number from $(0,1)$.
You and I have the ‘same number’ of elements.
Let $f0,1) \rightarrow \mathbb{R}" alt="f0,1) \rightarrow \mathbb{R}" />. Since there is a bijective function $f$ from $(0,1)$ to $\mathbb{R}$.

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

Have we lost the bijective function?

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

4. Originally Posted by novice
Let $f0,1) \rightarrow \mathbb{R}" alt="f0,1) \rightarrow \mathbb{R}" />. Since there is a bijective function $f$ from $(0,1)$ to $\mathbb{R}$.

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

Have we lost the bijective function?

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