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Prove that number of subsets of a set containing n distinct elements is 2n, for all n ϵ N.

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To prove; Number of subsets of a set containing n distinct elements is 2n.

For a null set there is only one element Φ and therefore only one subset.

⇒ at n = 0 number of subsets = 1 = 20

By considering a set with 1 element there will be 2 subsets with 1 and Φ

⇒ at n = 0 number of subsets = 2 = 22

By considering a set Sk with k element

Let at n = k number of subsets = 2k be true.

For a set Sk+1 containing k+1 element

The extra one element in set Sk+1 when compared with Sk will form an extra collection of subsets by combing with the existing 2k subsets and as a result an extra 2k subsets are formed.

⇒ at n = k+1 number of subsets = 2k + 2k = 2.2k = 2k+1

⇒ P(k+1) is true.

∴ By Mathematical Induction number of subsets of a set containing n distinct elements is 2n, for all n ϵ N.

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