Recent content by Scienticious

  1. S

    Show that if f: A→B is surjective and and H is a subset of B, then f([

    O I understand now I think, thank you :3 f is surjective, so for any y in H we have f(x) = y. since f^(-1)(H) is in the domain of f by definition of inverse functions, y = f(f^(-1)(y)) for any y in H and thus H is a subset of f(f^(-1)(H)). This line of reasoning checks out, right?
  2. S

    Show that if f: A→B is surjective and and H is a subset of B, then f([

    [f]^{}[/2]Homework Statement Show that if f: A→B is surjective and and H is a subset of B, then f(f^(-1)(H)) = H. Homework Equations The Attempt at a Solution Let y be an element of f(f^(-1)(H)). Since f is surjective, there exists an element x in f^(-1)(H) such that f(x) =...
  3. S

    Suppose that f is an injection. Show that f-1(f(x)) = x for all x in D

    Homework Statement Suppose that f is an injection. Show that f-1(f(x)) = x for all x in D(f). Homework Equations The Attempt at a Solution Let z be in f-1(f(x)). Then f(z) is in f(x) by definition of inverse functions. Since f is injective, z = x for some x in D(f). Thus...
  4. S

    Show that if f: A → B is injective and E is a subset of A, then f −1(f

    Simpler than I expected! Thanks , I understand now :3
  5. S

    Show that if f: A → B is injective and E is a subset of A, then f −1(f

    O of course, since f(x) is in of f(E), then x must be an element of E because injectivity implies that each value of f(x1) corresponds to a unique element of A. I think this works, but at any rate I'm still stuck on the other half of the proof.. if I try to suppose x is in E I get back to the...
  6. S

    Show that if f: A → B is injective and E is a subset of A, then f −1(f

    Intuitively I can tell that it's not obvious and that step isn't allowed, but in that case I've run into a break wall.
  7. S

    Show that if f: A → B is injective and E is a subset of A, then f −1(f

    O dear I see my mistake; I confused the definitions of injective and bijective :/ Then for the part of the proof you mentioned, x is in f-1(f(E)). Thus f(x) is in f(E). Then it seems obvious that x is in E at this point, but I'm not sure if it's trivial or not. So if this step works we...
  8. S

    Show that if f: A → B is injective and E is a subset of A, then f −1(f

    Homework Statement Show that if f: A → B is injective and E is a subset of A, then f −1(f(E) = E Homework Equations The Attempt at a Solution Let x be in E. This implies that f(x) is in f(E). Since f is injective, it has an inverse. Applying the inverse function we see that...
  9. S

    Show that if f: A → B and E, F are subsets of A, then f(E ∪ F) = f(E)

    Thank you for your reply :3 I think I get it now; here's my new proof: Let x be an element of E. Then f(x) is an element of f(E). But if x is an element of E then x is an element of E ∪ F Thus f(x) is an element of f(E ∪ F). Therefore f(E) is a subset of f(E ∪ F). Now let x be an element of...
  10. S

    Show that if f: A → B and E, F are subsets of A, then f(E ∪ F) = f(E)

    Homework Statement Show that if f: A → B and E, F are subsets of A, then f(E ∪ F) = f(E) ∪ f(F). Homework Equations The Attempt at a Solution My attempt: Suppose x is an element of E. Then f(x) is an element of f(E), which means f(x) is a subset of f(E). But x is in E...
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