0

Set up: Let $A,B$ be sets, and $f: A \longrightarrow B$, $g: B \longrightarrow A$ be injective functions.

We call $b \in B$ lonely, if $b \neq f(a)$ for any $a \in A$.

We call $b \in B$ a densecndent of $b_0$, if $b = (f \circ g)^n(b_0)$, where $n$ is some natural number.

Define $h: A \longrightarrow B$; it equals $g^{-1}(a)$ if $f(a)$ is a descendent of a lonely point, and it equals $f(a)$ otherwise.

I'm trying to understand this proof: https://artofproblemsolving.com/wiki/index.php/Schroeder-Bernstein_Theorem, but I'm confused by this part:

"We first note that for any $a \in A$, the point $h(a)$ is a descendent of a lonely point if and only if $f(a)$ is a descendent of a lonely point.".

Here's my attempt to verify this:

Let $h(a)$ be a descendent of a lonely point; I want to show that this implies that $f(a)$ is a descendent of a lonely point. Assume that $f(a)$ is not a descendent of a lonely point; then, by definition, $h(a) = f(a)$, and so we've a contradiction.

The other way:

Assume that $f(a)$ is a descendent of a lonely point, say $b_0$; then, by definition, $h(a) = g^{-1}(a)$. Since $f(a)$ is a descendent of $b_0$, we can write $f(a) = (f \circ g)(b_0)$, and since $f$ is injective, it follows that $a = g(b_0)$, but then $$h(a) = g^{-1}(a) = g^{-1}(g(b_0)) = b_0.$$

Hence $h(a)$ is lonely, but that means that $h(a)$ is not a descendent of any lonely point, since that would mean that $$\forall \alpha \in A: h(a) \neq f(\alpha) \\ \text{and} \\ h(a) = f(g(\beta)),$$ where $\beta$ is some lonely point in $B$; but this is a contradiction, since $g(\beta) \in A$.

I must've made a mistake; I'd appreciate pointing it out.

  • Why do you assume you must have made a mistake? – Eric Wofsey Dec 15 '20 at 20:29
  • Please give the definition of a lonely point, preferably in your question not in a comment. – coffeemath Dec 15 '20 at 20:30
  • Notation: "descendent of a lonely point" = *.

    Because if I didn't, we would've that "$h(a)$ is * implies $f(a)$ is ", and that "$f(a)$ is implies $h(a)$ is not ", all of which imply that "$h(a)$ is implies $h(a)$ is not *", which is a contradiction.

     –  Dec 15 '20 at 20:36
  • You neither defined "lonely point" [by itself, rather than "descendent of a lonely point"] nor put definition in your question. – coffeemath Dec 15 '20 at 20:52
  • 2
    I did define it, look at the second line. By "point" I mean "element", I think it's pretty common to call elements "points". –  Dec 15 '20 at 20:55
  • @jintan2357: See if this answer helps; my $Y_0$ is your set of lonely point of $B$. – Brian M. Scott Dec 15 '20 at 22:12
  • One error: That $f(a)$ is a descendent of a lonely point $b_0$ means that $f(a) = (f\circ g)^n(b_0)$. for some $n$. You assumed that $n = 1$. – Paul Sinclair Dec 16 '20 at 04:34

1 Answers1

0

You turn too much to proof by contradiction here. I think the more natural way to prove this is

  • $f(a)$ is not a lonely descendent $\implies h(a)$ is not a lonely descendent.
  • $f(a)$ is a lonely descendent $\implies h(a)$ ia a lonely descendent.

The first case is trivial. If $f(a)$ is the not a lonely descendent, then by definition, $h(a) = f(a)$ which is not a lonely descendent.

For the second case, we know that $f(a) = (f\circ g)^n(b_0)$ for some lonely $b_0$ and some $n$. We know that $n$ cannot be $0$, since that would make $b_0 = f(a)$, so $b_0$ would not be lonely. But then $a = g((f\circ g)^{n-1}(b_0))$, and $$h(a) = g^{-1}(a) = (f\circ g)^{n-1}(b_0)$$ which is a descendent of a lonely number (note that we must include $b_0$ as a descendent of itself for this to be always true: $n-1$ could be $0$).

Paul Sinclair
  • 43,643