Abstract/Discrete/Algebraic Mathematics.

[hugo28]

Show/Prove that if f: X → Y is one-to-one on X, and A subset of X, then f^-1(f(A))<= subset A.

If you wouldn’t mind, please check whether I did it correctly. Thanks in advance.

Suppose x Є A
Then, f(x) Є f(A)
By image function y =f(x)
Thus, y = f(x) Є f(A)
And by inverse image, f^-1(y) = f-1(f(x)) Є f^-1(f(A)) <= subset A

Therefore: Є f^-1(f(A)) <= subset A.

 

[Eynstone]

That's right.

 

[adriank]

Your proof doesn't make sense.

You want to prove that f^-1(f(A)) ⊆ A.
Thus, suppose x ∈ f^-1(f(A)). You want to prove x ∈ A.
Since x ∈ f^-1(f(A)), this means f(x) ∈ f(A).
Thus there is some x' ∈ A such that f(x) = f(x').
Since f is one-to-one, we have x = x', and thus x ∈ A.

This is really the only way to prove it.

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Aside: For any function f: X → Y and any subset A of X, we have A ⊆ f^-1(f(A)). This is because for any x ∈ A, f(x) ∈ f(A), so x ∈ f^-1(f(A)). Thus, if f is one-to-one, you in fact have f^-1(f(A)) = A.

Aside 2: If f: X → Y is a function such that f^-1(f(A)) ⊆ A for every subset A of X, then f is one-to-one. Proof: If x, x' ∈ X, let A = {x}. If f(x') = f(x), then f(x') ∈ f(A), so y ∈ f^-1(f(A)) ⊆ A. But then x' = x, because A only contains x. Thus:

A function f: X → Y is one-to-one if and only if for every A ⊆ X, f^-1(f(A)) ⊆ A.​