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Thread: Cardinality if infinite sets

  1. #1
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    Cardinality if infinite sets

    Let $\displaystyle A=\{1,2,3\}$ and $\displaystyle B =\{a,b,c\}$. Since they are finite sets, itís 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?
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    Quote Originally Posted by novice View Post
    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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    Quote Originally Posted by Plato View Post
    What does it mean Ďto have the same numberí?
    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 Ďsame numberí 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\}|
    $?
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    Quote Originally Posted by novice View Post
    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 $\displaystyle \left| {\left( {0,1} \right)\backslash A} \right| = \left| {\mathbb{R}\backslash B} \right|$
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