Question

Let A and B be nonempty sets. Prove that if f is an injection, then f(A − B) = f(A) − f(B)

Answer #1

Let A and B be nonempty sets. Prove that if f is an injection,
then f(A − B) = f(A) − f(B)

Let A be open and nonempty and f : A → R. Prove that f is
continuous at a if and only if f is both upper and lower
semicontinuous at a.

Let A be a nonempty set. Prove that the set S(A) = {f : A → A |
f is one-to-one and onto } is a group under the operation of
function composition.

Let A, B, C be sets and let f : A → B and g : f (A) → C be
one-to-one functions. Prove that their composition g ◦ f , defined
by g ◦ f (x) = g(f (x)), is also one-to-one.

Let A⊆R be a nonempty set, which is bounded above. Let
B={a-5:a∈ A}. Prove that sup(B)=sup(A)-5

Suppose A and B are
nonempty sets of real numbers, and that for every x
∈ A, and every y ∈ B, we have x < y. Prove that A ≤
inf(B).

Let A and B be sets. Prove that A ⊆ B if and only if A − B =
∅.

Let A and B be sets. Prove that (A∪B)\(A∩B) = (A\B)∪(B\A)

Suppose S and T are nonempty sets of real numbers such that for
each x ∈ s and y ∈ T we have x<y.
a) Prove that sup S and int T exist
b) Let M = sup S and N= inf T. Prove that M<=N

let A be a nonempty subset of R that is bounded below. Prove
that inf A = -sup{-a: a in A}

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