Abstract Algebra: extra help with sets, maps and equivalence: 8-29-17

[BillShillito]

Here's my proof for §0.3 #11. I'd love some feedback if you're so inclined! :)

Let α, β∈M. Suppose Φ(α)=Φ(β); this means that (α(1), α(2)) = (β(1), β(2)). These ordered pairs are equal iff their first and second coordinates are equal, so α(1)=β(1) and α(2)=β(2). Now we can conclude that α=β, since α and β have the same actions on all elements of the domain {1, 2}. Therefore Φ is one-to-one.

Now, let x, y∈B. Since M is the set of all maps from {1, 2} to B, there is a map α∈M such that α(1)=x and α(2)=y. Then Φ(α)=(x, y), and Φ is onto.

Thus we have shown that Φ is a bijection and therefore must be invertible. The inverse Φ⁻¹ takes in an ordered pair (x, y) and finds the map α∈M such that α(1)=x and α(2)=y.

[alanlang]

There is indeed a "pointless" argument for section 0.2 #8.

A = A n (A U X)
= A n (B U X)
= (A n B) U (A n X)
= (A n B) U (B n X)
= (B n A) U (B n X)
= B n (A U X)
= B n (B U X)
= B

[ritesharora]

What is recomended textbook fot this course?

[chrisschmidt]

Love the prayer before the lectures! It's a wonderful opportunity for reflection of purpose.