# 1-1 functions and Schroder-Bernstein Theorem

Using Schroder-Bernsten Theorem. Assume there exists a 1-1 function $f:X \rightarrow Y$ and another 1-1 function $g:Y \rightarrow X$. If we define $f^{(-1)}(y)=x$, then $f^{(-1)}$ is a 1-1 function from $f(X)$ onto $X$. Follow the steps to show that there exists a 1-1, onto function $h:X \rightarrow Y$.

Let $x$ be in $X$ be arbitrary. Let the chain $Cx$ be the set consisting of all elements of the form

$$....,\, f^{(-1)}(g^{(-1)}(x)),\, g^{(-1)}(x),\,x,\,f(x), g(f(x)),\,f(g(f(x))),\,....$$

Explain why the (distinct) number of elements to the left of $x$ in the above chain may be zero, finite, or infinite.

Since the functions are 1-1, say $x$ map to a distinct $y$ in $Y$. So evaluating the chain (the $x$ in bold separates the left and right sides of the chain), you get: $$....x,\, y,\, \textbf{x},\, y,\, x,\, y,\, ...$$ The left side of the chain is obviously composed of the $x$ and $y$ predefined above. I don't really understand what the question means. My interpretation is that the left side of the chain is finite because the functions, being 1-1, map a distinct $x$ to a distinct $y$, so those are finitely many elements (2 in this case). But I'm not sure if I'm right. Any input? Thanks.

Edit: Sorry I didn't mention that g^(-1) is also a 1-1 function, defined g^(-1): g(X)->Y

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The chain as currently set up doesn't make sense: there's no reason $g^{-1}$ should be defined on $x$, because $x$ needn't be in the image of $g$. – Kevin Carlson Oct 3 '12 at 5:14
Kevin- I edited my question, it is defined on x. Sorry about that. – Alti Oct 3 '12 at 5:19
OK, but you specifically said $x$ is an arbitrary element of $X$, while $g^{-1}$ is only defined on $g(Y)$, and it was not assumed that $g$ is surjective. How am I to interpret the chain for $x \notin g(Y)$? – Kevin Carlson Oct 3 '12 at 5:32
I see what you mean, I neglected the arbitrary part and fixed an x and assumed it mapped to Y. So I evaluated my chain, incorrectly. But that leaves me more confused, how I am supposed to determine the elements of the chain? OR at least know if they finite, infinite, or zero? – Alti Oct 3 '12 at 5:42
The chain doesn't even exist as you've set it up-there is no such object as $g^{-1}(x)$ for arbitrary $x$. – Kevin Carlson Oct 3 '12 at 5:45

You've already shown a case when the number of interest is finite, in particular 2: when $f(x)=g^{-1}(x)$, so that $f^{-1}g^{-1}(x)=x$ and the chain oscillates between $x$ and $f^{-1}(x)$. In the comments we've seen a case of $0$ elements to the left as well, when $x$ isn't in $g(Y)$. For the infinite case, consider something like $X=Y=\mathbb{Z}, f=g, f(n)=n+1$.