Question

Union of two coutable set is countable prove it

Answer 1

Answer:

Proof 1

Let ⟨Sn⟩n∈N⟨Sn⟩n∈⁡N be a sequence of countable sets.

Define:

S=⋃n∈NSnS=⋃n∈⁡NSn

For all n∈Nn∈N, let FnFn denote the set of all injections from SnSn to NN.

Since SnSn is countable, FnFn is non-empty.

Using the axiom of countable choice, there exists a sequence ⟨fn⟩n∈N⟨fn⟩n∈⁡N such that fn∈Fnfn∈Fn for all n∈Nn∈N.

Let ϕ:S→N×Nϕ:S→N×N, where ×× denotes the cartesian product, be the mapping defined by:

ϕ(x)=(n,fn(x))ϕ(x)=(n,fn(x))

where nn is the (unique) smallest natural number such that x∈Snx∈Sn.

From the Well-Ordering Principle, such an nn exists; hence, the mapping ϕϕ exists.

Since each