Let A, B, and C be disjoint denumerable sets. Show that A∪B∪C is also denumerable.
Given: A≈N, B≈N, and C≈N. Goal:A∪B∪C≈N
Suppose A,B, and C are non-empty disjoints sets such that A≈N, B≈N, and C≈N. Then, there exists functions f:A→N, g:B→N, and h:C→N that are one-to-one and onto N.
This is where my understanding kinda falls apart...Do have to set some knda of restriction? I can't find many examples in my book of this kind of stuff. Actaully, I'm probably just not making the connection.
Suppose Φ(x)=Φ(y). We see that x and y must have same parity for or else Φ(x)≠Φ(y). So, suppose x,y∈A. Then, plugging into the definition Φ(x)=2^∝(x)=2^∝(y)=Φ(y). ln(2^∝(x))=ln(2^∝(y)) ⇒ ∝ln(2)x=∝ln(2)y ⇒ x=y. Therefore, Φ(x) is injective when x∈A.
Let y be an element of the codomain. By our definition y=2^∝(x). Solving for x we get x=ln(y)/∝ln(2) when y>0. Now, f(ln(y)/∝ln(2))=2^∝(ln(y)/∝ln(2))=y for y∈N. Thus, Φ(x) is surjective.
Then I proceed to go down the list?
Plato's method works, of course, but I had a different idea. Since A is countable, there exists a function N->A that is one-to-one and countable so each member of A can be "labeled" . Similarly, members of B and C can be "labeled" and . Now define a function from N to the union of A, B, and C by , , and . It is easy to show that function is bijective.