Question

P2.9.7 Let A be a set and f a bijection from A to itself. We say that f fixes an element r of A if f(x) = r. (a) Write a quantified statement, with variables ranging over A, that says "there is exactly one element of A that f does not fix." (b) Prove that if A has more than one element, the statement of part (a) leads to a contradiction. That is, if f does not fix 2, and there is another element in A besides 2, then there is some other element that f does not fix.

Answer 1

The required quantified statement is "For $f$ there exist a unique $y\in A$ such that $\forall x\in A$ with $x\neq y$, $f$ fixes $x$". Here y is that exactly one element which doesn't get fixed by f.

(b). Let the statement in (a) holds and A has more than one element. So, $f$ does not fix $y$, then $f(y)\neq y$. Let $f(y)=y_1$ (say). Clearly $y\neq y_1$ as $y\neq f(y)$. Now as $f$ is a bijection, $y\neq y_1$ implies $f(y_1)\neq f(y)=y_1$ . So $f$ does not fix $y_1$ also.