Hi

Here is a problem I am trying.

Suppose $\displaystyle A\neq \varnothing$ and

$\displaystyle f:\;A\longrightarrow A$ and

$\displaystyle \forall g[(g:A\longrightarrow A)\Rightarrow (f\circ g = f)]\cdots (1)$

I have to prove that f is a constant function.

I will give the outline of what I did.

Since $\displaystyle \because A\neq \varnothing \Rightarrow \exists a \in A$

$\displaystyle \therefore a \in A$

Now consider the set

$\displaystyle g=\{(x,a)\vert \; x\in A \; \}$

$\displaystyle \because g\subseteq A\times A $

g is a relation from A to A. Then I proved that g is also a function by proving

that

$\displaystyle \forall x \in A \exists ! b \in A ((x,b) \in g)$

next , I proved that g is a constant function by proving that

$\displaystyle \exists b \in A \forall x\in A (g(x)=b)$

and finally I used the given (1) for g , which implies that

$\displaystyle f\circ g =f$

To prove that f is a constant function, I have to prove

$\displaystyle \exists b\in A \forall x\in A (f(x)=b)$

since $\displaystyle a\in A$ , I can associate some $\displaystyle c\in A$ such that

$\displaystyle f(a)=c$

so Let $\displaystyle b=c=f(a)$

so the goal now is

$\displaystyle \forall x\in A (f(x)=b)$

Let x be arbitrary , $\displaystyle \therefore x\in A$

$\displaystyle f(x)=f(g(x))=f(a)=b$

since x is arbitrary

$\displaystyle \therefore \exists b\in A \forall x\in A(f(x)=b)$

so f is a constant function..................

is the proof too detailed ? since this is from Velleman's "how to prove it" , I think,

author expects me to use all the logical machinery that I can use.

correct ?