Question

X and Y be any sets. a) Prove or disprove that ℘(X)∪℘(Y)⊆℘(X∪Y), where ℘(X) is the power set of X b)Prove or disprove that ℘(X∪Y)⊆℘(X)∪℘(Y), where ℘(X) is the power set of X Based on parts a and b, can we conclude that ℘(X∪Y)=℘(X)∪℘(Y). Justify your answer.

Answer 1

Explanation:

a) To prove that ℘(X)∪℘(Y)⊆℘(X∪Y), we need to show that every element in the union of the power sets of X and Y is also an element of the power set of their union.

Let's consider an arbitrary element A in ℘(X)∪℘(Y). This means that A is either an element of ℘(X) or ℘(Y).

Case 1: A∈℘(X)

If A∈℘(X), then A is a subset of X. Since A is a subset of X, it is also a subset of X∪Y. Therefore, A∈℘(X∪Y).

Case 2: A∈℘(Y)

If A∈℘(Y), then A is a subset of Y. Similarly to Case 1, if A is a subset of Y, it is also a subset of X∪Y. Therefore, A∈℘(X∪Y).

Since A∈℘(X)∪℘(Y) implies A∈℘(X∪Y) for any arbitrary element A, we can conclude that ℘(X)∪℘(Y)⊆℘(X∪Y).

b) To disprove that ℘(X∪Y)⊆℘(X)∪℘(Y), we need to find a counterexample, i.e., an example where there exists an element in the power set of the union of X and Y that is not in the union of their power sets.

Let's consider the following example:

X = {1}

Y = {2}

℘(X) = {{}, {1}}

℘(Y) = {{}, {2}}

X∪Y = {1, 2}

℘(X∪Y) = {{}, {1}, {2}, {1, 2}}

℘(X)∪℘(Y) = {{}, {1}} ∪ {{}, {2}} = {{}, {1}, {2}}

In this example, we can see that {1, 2} is an element of ℘(X∪Y) but not an element of ℘(X)∪℘(Y). Therefore, we have found a counterexample, and we can disprove that ℘(X∪Y)⊆℘(X)∪℘(Y).

Based on parts a and b, we cannot conclude that ℘(X∪Y) = ℘(X)∪℘(Y) because we have shown that ℘(X)∪℘(Y) is a subset of ℘(X∪Y) (part a), but the reverse inclusion is not always true (part b).