How do I continue the sequence for the Schröder-Bernstein Theorem?












1












$begingroup$



If $ X, Y$ are sets such that there are one-to-one maps $f: X rightarrow Y$ and $g : Yrightarrow X$, then there is a one-to-one, onto map $h : X rightarrow Y$.



proof. We divide $X$ into three disjoint subsets $X_X$, $X_Y$ , $X_infty$ with different mapping properties as follows.



Consider a point $x_1 in X$. If $x_1$ is not in the range of $g$, then we say $x_1 in X_X$. Otherwise there exists $y_1 in Y$ such that $g(y_1) = x_1$, and $y_1$ is unique since $g$ is one-to-one. If $y_1$ is not in the range of $f$, then we say $x_1 in X_Y$ . Otherwise there exists a unique $x_2 in X$ such that $f(x_2) = y_1$. Continuing in this way, we generate
a sequence of points



$x_1, y_1, x_2, y_2, cdots, x_n, y_n, x_{n+1}, cdots$



with $x_n in X$, $y_n in Y$ and $g(y_n) = x_n, f(x_{n+1}) = y_n$.
We assign the starting point $x_1$ to a subset in the following way: (a) $x_1 in X$ if the sequence terminates at some $x_n in X$ that isn’t in the range of $g$; (b) $x_1 in X_Y$ if the sequence terminates at some $y_n in Y$ that isn’t in the range of $f$; (c) $x_1 in X_{infty}$ if the sequence never terminates.




When we continue, does that mean that we would then say that "if $x_2$ is not in the range of $g$, then we say $x_2 in X_X$. Otherwise there exists $y_2in Y$ ..."?



Also, would we assign $x_2$ and $x_3$ to $X_X$, for example, if the sequence ends at $x_4$?










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$endgroup$








  • 1




    $begingroup$
    If a sequence starting at $x_1$ terminates at some $x_n$ then $x_1,...,x_n in X_X$... If it terminates at some $y_n$ then $x_1,..,x_nin X_Y$.... The idea is that $xin X_X$ if the sequence starting at $x$ has an odd number of terms, but if it has an even number of terms then $xin X_Y.$
    $endgroup$
    – DanielWainfleet
    Dec 19 '18 at 2:57
















1












$begingroup$



If $ X, Y$ are sets such that there are one-to-one maps $f: X rightarrow Y$ and $g : Yrightarrow X$, then there is a one-to-one, onto map $h : X rightarrow Y$.



proof. We divide $X$ into three disjoint subsets $X_X$, $X_Y$ , $X_infty$ with different mapping properties as follows.



Consider a point $x_1 in X$. If $x_1$ is not in the range of $g$, then we say $x_1 in X_X$. Otherwise there exists $y_1 in Y$ such that $g(y_1) = x_1$, and $y_1$ is unique since $g$ is one-to-one. If $y_1$ is not in the range of $f$, then we say $x_1 in X_Y$ . Otherwise there exists a unique $x_2 in X$ such that $f(x_2) = y_1$. Continuing in this way, we generate
a sequence of points



$x_1, y_1, x_2, y_2, cdots, x_n, y_n, x_{n+1}, cdots$



with $x_n in X$, $y_n in Y$ and $g(y_n) = x_n, f(x_{n+1}) = y_n$.
We assign the starting point $x_1$ to a subset in the following way: (a) $x_1 in X$ if the sequence terminates at some $x_n in X$ that isn’t in the range of $g$; (b) $x_1 in X_Y$ if the sequence terminates at some $y_n in Y$ that isn’t in the range of $f$; (c) $x_1 in X_{infty}$ if the sequence never terminates.




When we continue, does that mean that we would then say that "if $x_2$ is not in the range of $g$, then we say $x_2 in X_X$. Otherwise there exists $y_2in Y$ ..."?



Also, would we assign $x_2$ and $x_3$ to $X_X$, for example, if the sequence ends at $x_4$?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If a sequence starting at $x_1$ terminates at some $x_n$ then $x_1,...,x_n in X_X$... If it terminates at some $y_n$ then $x_1,..,x_nin X_Y$.... The idea is that $xin X_X$ if the sequence starting at $x$ has an odd number of terms, but if it has an even number of terms then $xin X_Y.$
    $endgroup$
    – DanielWainfleet
    Dec 19 '18 at 2:57














1












1








1





$begingroup$



If $ X, Y$ are sets such that there are one-to-one maps $f: X rightarrow Y$ and $g : Yrightarrow X$, then there is a one-to-one, onto map $h : X rightarrow Y$.



proof. We divide $X$ into three disjoint subsets $X_X$, $X_Y$ , $X_infty$ with different mapping properties as follows.



Consider a point $x_1 in X$. If $x_1$ is not in the range of $g$, then we say $x_1 in X_X$. Otherwise there exists $y_1 in Y$ such that $g(y_1) = x_1$, and $y_1$ is unique since $g$ is one-to-one. If $y_1$ is not in the range of $f$, then we say $x_1 in X_Y$ . Otherwise there exists a unique $x_2 in X$ such that $f(x_2) = y_1$. Continuing in this way, we generate
a sequence of points



$x_1, y_1, x_2, y_2, cdots, x_n, y_n, x_{n+1}, cdots$



with $x_n in X$, $y_n in Y$ and $g(y_n) = x_n, f(x_{n+1}) = y_n$.
We assign the starting point $x_1$ to a subset in the following way: (a) $x_1 in X$ if the sequence terminates at some $x_n in X$ that isn’t in the range of $g$; (b) $x_1 in X_Y$ if the sequence terminates at some $y_n in Y$ that isn’t in the range of $f$; (c) $x_1 in X_{infty}$ if the sequence never terminates.




When we continue, does that mean that we would then say that "if $x_2$ is not in the range of $g$, then we say $x_2 in X_X$. Otherwise there exists $y_2in Y$ ..."?



Also, would we assign $x_2$ and $x_3$ to $X_X$, for example, if the sequence ends at $x_4$?










share|cite|improve this question











$endgroup$





If $ X, Y$ are sets such that there are one-to-one maps $f: X rightarrow Y$ and $g : Yrightarrow X$, then there is a one-to-one, onto map $h : X rightarrow Y$.



proof. We divide $X$ into three disjoint subsets $X_X$, $X_Y$ , $X_infty$ with different mapping properties as follows.



Consider a point $x_1 in X$. If $x_1$ is not in the range of $g$, then we say $x_1 in X_X$. Otherwise there exists $y_1 in Y$ such that $g(y_1) = x_1$, and $y_1$ is unique since $g$ is one-to-one. If $y_1$ is not in the range of $f$, then we say $x_1 in X_Y$ . Otherwise there exists a unique $x_2 in X$ such that $f(x_2) = y_1$. Continuing in this way, we generate
a sequence of points



$x_1, y_1, x_2, y_2, cdots, x_n, y_n, x_{n+1}, cdots$



with $x_n in X$, $y_n in Y$ and $g(y_n) = x_n, f(x_{n+1}) = y_n$.
We assign the starting point $x_1$ to a subset in the following way: (a) $x_1 in X$ if the sequence terminates at some $x_n in X$ that isn’t in the range of $g$; (b) $x_1 in X_Y$ if the sequence terminates at some $y_n in Y$ that isn’t in the range of $f$; (c) $x_1 in X_{infty}$ if the sequence never terminates.




When we continue, does that mean that we would then say that "if $x_2$ is not in the range of $g$, then we say $x_2 in X_X$. Otherwise there exists $y_2in Y$ ..."?



Also, would we assign $x_2$ and $x_3$ to $X_X$, for example, if the sequence ends at $x_4$?







real-analysis elementary-set-theory






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edited Dec 19 '18 at 2:12









Chris Custer

14.2k3827




14.2k3827










asked Dec 19 '18 at 1:49









K.MK.M

710413




710413








  • 1




    $begingroup$
    If a sequence starting at $x_1$ terminates at some $x_n$ then $x_1,...,x_n in X_X$... If it terminates at some $y_n$ then $x_1,..,x_nin X_Y$.... The idea is that $xin X_X$ if the sequence starting at $x$ has an odd number of terms, but if it has an even number of terms then $xin X_Y.$
    $endgroup$
    – DanielWainfleet
    Dec 19 '18 at 2:57














  • 1




    $begingroup$
    If a sequence starting at $x_1$ terminates at some $x_n$ then $x_1,...,x_n in X_X$... If it terminates at some $y_n$ then $x_1,..,x_nin X_Y$.... The idea is that $xin X_X$ if the sequence starting at $x$ has an odd number of terms, but if it has an even number of terms then $xin X_Y.$
    $endgroup$
    – DanielWainfleet
    Dec 19 '18 at 2:57








1




1




$begingroup$
If a sequence starting at $x_1$ terminates at some $x_n$ then $x_1,...,x_n in X_X$... If it terminates at some $y_n$ then $x_1,..,x_nin X_Y$.... The idea is that $xin X_X$ if the sequence starting at $x$ has an odd number of terms, but if it has an even number of terms then $xin X_Y.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 2:57




$begingroup$
If a sequence starting at $x_1$ terminates at some $x_n$ then $x_1,...,x_n in X_X$... If it terminates at some $y_n$ then $x_1,..,x_nin X_Y$.... The idea is that $xin X_X$ if the sequence starting at $x$ has an odd number of terms, but if it has an even number of terms then $xin X_Y.$
$endgroup$
– DanielWainfleet
Dec 19 '18 at 2:57










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