Answer with Step-by-step explanation:
We are given that A and B are two countable sets
We have to show that if A and B are countable then [tex]A\cup B[/tex] is countable.
Countable means finite set or countably infinite.
Case 1: If A and B are two finite sets
Suppose A={1} and B={2}
[tex]A\cup B[/tex]={1,2}=Finite=Countable
Hence, [tex]A\cup B[/tex] is countable.
Case 2: If A finite and B is countably infinite
Suppose, A={1,2,3}
B=N={1,2,3,...}
Then, [tex]A\cup B[/tex]={1,2,3,....}=N
Hence,[tex]A\cup B[/tex] is countable.
Case 3:If A is countably infinite and B is finite set.
Suppose , A=Z={..,-2,-1,0,1,2,....}
B={-2,-3}
[tex]A\cup B[/tex]=Z=Countable
Hence, [tex]A\cup B[/tex] countable.
Case 4:If A and B are both countably infinite sets.
Suppose A=N and B=Z
Then,[tex]A\cup B[/tex]=[tex]N\cup Z[/tex]=Z
Hence,[tex]A\cup B[/tex] is countable.
Therefore, if A and B are countable sets, then [tex]A\cup B[/tex] is also countable.
Answer:
To remedy confusions like yours and to avoid the needless case analyses, I prefer to define X to be countable if there is a surjection from N to X.
This definition is equivalent to a few of the many definitions of countability, so we are not losing any generality.
It is a matter of convention whether we allow finite sets to be countable or not (though, amusingly, finite sets are the only ones whose elements you could ever finish counting).
So, if A and B be countable, let f:N→A and g:N→B be surjections. Then the two sequences (f(n):n⩾1)=(f(1),f(2),f(3),…) and (g(n):n⩾1)=(g(1),g(2),g(3),…) eventually cover all of A and B, respectively; we can interleave them to create a sequence that will surely cover A∪B:
(h(n):n⩾1):=(f(1),g(1),f(2),g(2),f(3),g(3),…).
An explicit formula for h is h(n)=f((n+1)/2) if n is odd, and h(n)=g(n/2) if n is even.
Hope it helps uh mate...✌
