# Thread: Proof Function (1-1, Restriction)

1. ## 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

2. 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?

3. ## 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

4. 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.

,

### how to prove a function is one to one

Click on a term to search for related topics.