How can the inequality hold for an injective function?

[waht]

This is something I understood before, but for some reason I forgot it. How do you prove this inequality holds, if f is injective?

$$A_0 \subset f^{-1}(f(A_0))$$

 

[Nolen Ryba]

If x is not in A, then f(x) is not in f(A) (injectivity) so x is not in f^-1(f(A))

 

[Hurkyl]

This is something I understood before, but for some reason I forgot it. How do you prove this inequality holds, if f is injective?

$$A_0 \subset f^{-1}(f(A_0))$$

Proofs of set algebra identities tend to be rather formulaic. If you look at the definition of "subset", then two proofs should immediately suggest themselves:

Let x be an element of A_0 ... Therefore x is in f^{-1}(f(A_0))​

and

Suppose x is not an element of f^{-1}(f(A_0)) ... Therefore x is not in A_0​

And from there, you simply have to fill in the missing steps. And again, the missing steps are usually obvious from unwinding the definitions.

 

[HallsofIvy]

This is something I understood before, but for some reason I forgot it. How do you prove this inequality holds, if f is injective?

$$A_0 \subset f^{-1}(f(A_0))$$

If x is not in A, then f(x) is not in f(A) (injectivity) so x is not in f^-1(f(A))

Yes, but that doesn't say anything about what happens if x IS A, which is the whole point.

what, the standard way of proving "$A\subset B$ is to start "If x is in A" and then conclude "then x is in B".
If x is in A_0, you know that f(x) is in f(A_0). Now, what does the fact that f is injective say about x and f^-1(f(A_0)).

 

[Nolen Ryba]

HallsofIvy,

I'm not sure what you mean. I showed $$f^{-1}(f(A_0)) \subset A_0$$ which is the other half of the equality what was asking for.

 

[HallsofIvy]

HallsofIvy,

I'm not sure what you mean. I showed $$f^{-1}(f(A_0)) \subset A_0$$ which is the other half of the equality what was asking for.

Yes, I understood that. That was why I said, "If x is in A_0"- because that's the direction you want to prove.

 

[Nolen Ryba]

Yes, I understood that. That was why I said, "If x is in A_0"- because that's the direction you want to prove.

Am I reading this incorrectly?

How do you prove this inequality holds, if f is injective?

 

[fopc]

This is something I understood before, but for some reason I forgot it. How do you prove this inequality holds, if f is injective?

$$A_0 \subset f^{-1}(f(A_0))$$

I'll assume you're starting from: f:A -> B and A_0 is a subset of A.

The inclusion relation you've written holds regardless of whether f is injective or not.
However, if f is injective, then the relation can be written as an equality.
Proof is nothing more than working the definitions, as has already been suggested.

 

[HallsofIvy]

Finally, it dawns on me. I was reading the whole thing backwards. I thought the question was to prove that if f is injective, then... Sorry, everyone.

 

[waht]

Thanks I get it now,

I should have been more clearer.