Question

Prove that \({\bf{P}}\left( {\bf{A}} \right) \subseteq {\bf{P}}\left( {\bf{B}} \right)\) if and only if \({\bf{A}} \subseteq {\bf{B}}\).

Short answer

\(P\left( A \right) \subseteq P\left( B \right)\) if and only if \(A \subseteq B\).

Step-by-step solution

Power of a set

Given a set S, the power set of S is the set of all subsets of the set S.

Therefore, the power set of S is denoted by P(S).

Subset:

X is a subset of Y if every element of X is also an element of Y.

Therefore, \(X \subset Y\) means that X is a subset of Y, but X is not the same set of Y.

To prove \({\bf{P(A)}} \subseteq {\bf{P(B)}}\) if and only if \({\bf{A}} \subseteq {\bf{B}}\)

Proof:

\(\left( i \right) \Leftrightarrow \left( {ii} \right)\)

Assume that \(P\left( A \right) \subseteq P\left( B \right)\).

Let \(x\) belongs to A or \(x\) is an element of A \(x \in A\).

If \(x\) is an element in a set S, then \(x \in P\left( A \right)\).

Since, A is a subset of B, i.e., \(P\left( A \right) \subseteq P\left( B \right)\).

Therefore, x belongs to B \(x \in P\left( B \right)\) Or, \(x \in B\).

Thus, every element \(x\) in A has also to be in B.

Therefore, A is also a subset of B i.e.,\(A \subseteq B\).

\(\left( {ii} \right) \Leftrightarrow \left( i \right)\)

Assume that \(A \subseteq B\).

There exist a set of the form \(\left\{ x \right\}\) in \(P\left( A \right)\) if \(P\left( A \right)\) does not contain only the empty set \(\phi \).

Let \(x\)is an element of \(P\left( A \right)\) which is written as \(x \in P\left( A \right)\) Or, \(x \in A\).

Since, A is a subset of B i.e., \(A \subseteq B\).

Therefore, x belongs to B i.e., \(x \in B\).

Now, by definition of a subset, \(P\left( A \right) \subseteq P\left( B \right)\).

If \(x\)is an element in set S, then, x belongs to B or, \(x \in P\left( B \right)\).

Therefore, it is proved that every element \(x\) in \(P\left( A \right)\)also has to be in \(P\left( B \right)\).