How do I continue the sequence for the Schröder-Bernstein Theorem?
$begingroup$
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 $x_4$?
real-analysis elementary-set-theory
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
asked Dec 19 '18 at 1:49
K.MK.M
710413
$endgroup$
add a comment |
$begingroup$
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 $x_4$?
real-analysis elementary-set-theory
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
asked Dec 19 '18 at 1:49
K.MK.M
710413
$endgroup$
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
add a comment |
$begingroup$
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 $x_4$?
real-analysis elementary-set-theory
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
asked Dec 19 '18 at 1:49
K.MK.M
710413
$endgroup$
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
real-analysis elementary-set-theory
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
asked Dec 19 '18 at 1:49
K.MK.M
710413
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
asked Dec 19 '18 at 1:49
K.MK.M
710413
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
edited Dec 19 '18 at 2:12
Chris Custer
14.2k3827
14.2k3827
asked Dec 19 '18 at 1:49
K.MK.M
710413
asked Dec 19 '18 at 1:49
K.MK.M
710413
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
add a comment |
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
add a comment |
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$begingroup$
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