It's correct. Just as a matter of style though, it could be much more terse:
Since A is non-empty, let a be in A.
Let g = {<x a> | x is in A}.
g is a function from A into A.
Suppose x is in A.
So f(x) = f(g(x)) = f(a).
So f is a constant function.
Hi
Here is a problem I am trying.
Suppose and
and
I have to prove that f is a constant function.
I will give the outline of what I did.
Since
Now consider the set
g is a relation from A to A. Then I proved that g is also a function by proving
that
next , I proved that g is a constant function by proving that
and finally I used the given (1) for g , which implies that
To prove that f is a constant function, I have to prove
since , I can associate some such that
so Let
so the goal now is
Let x be arbitrary ,
since x is arbitrary
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 ?
It's correct. Just as a matter of style though, it could be much more terse:
Since A is non-empty, let a be in A.
Let g = {<x a> | x is in A}.
g is a function from A into A.
Suppose x is in A.
So f(x) = f(g(x)) = f(a).
So f is a constant function.