Let S be any finite set and suppose [itex]x \notin S[/itex]. Let [itex]K = S \cup \left\{ x \right\}[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

1. Prove that P(K) is the disjoint union of P(S) and [itex]X = \left \{T \subseteq K : x \in T \right\}[/itex]. That is, show that [itex]P(K) = P(S) \cup X[/itex] and [itex]P(S) \cap X = \emptyset [/itex]

2. Prove that every element of X is the union of a subset of S with {x}, and that if you take different subsets of S you get different elements of X. Argue that, therefore, X has the same number of elements as P(S).

3. Argue that the previous two parts allow you to conclude that if S is a finite set, then P(K) has twice as many elements as P(S).

I can do 1. Can anyone help with 2 and 3?

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