Proof Function (1-1, Restriction)

• Feb 18th 2007, 08:36 PM
tbyou87
Proof Function (1-1, Restriction)
Hi,
I have to prove that a restriction of a one-to-one function is one-to-one.
Intuitively this makes sense to me, but I don't see how to approach this proof formally. Any help would be appreciated.
Thanks
• Feb 19th 2007, 07:10 AM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Hi,
I have to prove that a restriction of a one-to-one function is one-to-one.

What does it mean a restriction of a one-to-one function?
• Feb 19th 2007, 10:04 AM
tbyou87
1-1 restriction function
f: A -> B is 1-1 iff whenever f(x) = f(y), then x = y.

Let f: A -> B and let D subset of A. The restriction of f to D is the function
f|D = {(x, y): y = f(x) and x element D}.

The theorem makes sense. For example, f: R -> R where f(x) = x is 1-1. If we restirct the domain to D subset of R F: D -> R where f(x) = x is still 1-1.

I just don't know how to formally prove this.

Thanks
• Feb 19th 2007, 12:09 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
f: A -> B is 1-1 iff whenever f(x) = f(y), then x = y.

Let f: A -> B and let D subset of A. The restriction of f to D is the function
f|D = {(x, y): y = f(x) and x element D}.

The theorem makes sense. For example, f: R -> R where f(x) = x is 1-1. If we restirct the domain to D subset of R F: D -> R where f(x) = x is still 1-1.

I just don't know how to formally prove this.

Thanks

Let A and B be non-empty sets.

And f:A --> B be a one-to-one map.

Let D be a non-trivial subset of A.

Then g: D --- > B defined as g(x)=f(x).

We need to show that g(a)=g(b) implies a=b.

Let us assume that it fails, i.e. their this collapsing. Meaning we can find "a" and "b" distinct such that, f(a)=f(b). But then that means f(a)=f(b). But that cannot be because f is one-to-one. Thus, a contradiction.