Recent content by Romono

How would you disprove if z ∈ (f(X) ∩ f(Y)) then z ∈ f(X ∩ Y)? (Where f: A -> B, if X, Y ⊆ A.) I'm just not sure how to approach this.

Just to be clear in this example, {a,b} would then be the image, wouldn't it? I think I'm understanding it now...

Hi Euge, Thanks for your reply, but maybe I should rephrase my question: Could you explain what "the image of a set A' ⊆ A is the set: f(A') = {b | b = f(a) for some a ∈ A'}" actually means? Could you break it down? I don't understand what an image of a set is even after reading the definition...

Could someone please explain how the image of a set A' ⊆ A is the set: f(A') = {b | b = f(a) for some a ∈ A'}. And how can the complement of A be a subset of A? Forgive my ignorance here, I'm a beginning student of set theory. Edit: Maybe I should rephrase my question: Could you explain what...

Could someone please explain how the image of a set A' ⊆ A is the set: f(A') = {b | b = f(a) for some a ∈ A'}. And how can the complement of A be a subset of A? Forgive my ignorance here, I'm a beginning student of set theory.