A

Reduction to absurdity:

Suppose there is no such finite harmonious set, and:

A, b∈S, where a+b ∈ s, a-b ∈ s.

Namely: a+b= either A or B.

A-b = a or b

Therefore, if an element of s is 0 and b=0, then the set S' = {a, 0} must not be a harmonic set.

But:

a+0 = a∈S’；

a-0=a∈S '

That is, s' is a harmonious set.

conflict

So, suppose there is an error and there is a finite set S.

B

Prove:

k 1，k2∈Z，k 1≠k2

Then: k 1a+k2a=(k 1+k2)a, where (k1+k2) ∈ z.

∴(k 1+k2)a∈S

Similarly: (k 1-k2)a∈S

That is, s is a harmonious set.

C

We can know that 0∈S from A's reduction to absurdity.

∴S 1∩S2 ≠ empty set is established.

D

By reducing to absurdity, suppose s1∪ S2 = R.

According to a, we can know that S 1 = {a, 0} and S2 = {b, 0} are harmonic sets, where a ≠ b.

Then there must be: s1∪ S2 = R.

This is obviously wrong and contradictory.

Therefore: S 1∪S2=R is not valid.

Option d