Question

how null set is a subset of every fainite and infainite set prove that?​

Answer 1

hii♥

here is your proof♥

Take as the definition of "finite set" that SS is finite iff SS is in bijection with a set of the form

{k∈N∣k<n}

{k∈N∣k<n}

where n∈Nn∈N. This is the definition used, e.g., on the nLab; the standard definition that SS is finite iff SS is in bijection with a set of the form

{1,…,n}

{1,…,n}

where n∈Nn∈N (used, e.g., on Wikipedia) only necessitates defining the empty set to be finite, which is probably not satisfactory to you.

Now note that ∅∅ is in bijection with the subset ∅={k∈N∣k<1}∅={k∈N∣k<1} (note that I am taking NN to not include 00). Hence, ∅∅ is finite. (If you prefer 0∈N0∈N, which there are many good reasons for, as Asaf points out below, and which many people do use, then of course we have ∅={k∈N∣k<0}∅={k∈N∣k<0} instead.)

Alternatively, you could define an "infinite set" to be a set SS for which there exists a proper subset T⊊S T⊊S which is in bijection with SS; then define a finite set to be one which is not infinite. (As Asaf warns below, this is only an equivalent definition if one accepts the axiom of choice.)

Under this definition, because ∅∅ has no proper subsets, it cannot be infinite; hence it is finite.