[imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A\cup B)[/imath]

[Ozma]

I denote with [imath]\mathcal{P}(T)[/imath] the set of subsets of a set [imath]T[/imath]. My textbook proves that for any given sets [imath]A,B[/imath] it is [imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A\cup B)[/imath] using the following argument: let [imath]X \in \mathcal{P}(A) \cup \mathcal{P}(B)[/imath], then [imath]X \in \mathcal{P}(B) \lor X \in \mathcal{P}(B)[/imath]. Hence, [imath]X \subseteq A \lor X \subseteq B[/imath] and so [imath]X \subseteq A\cup B[/imath]. That is, [imath]X \in \mathcal{P}(A \cup B)[/imath].

I came up with this other argument: since it is [imath]A \subseteq A \cup B[/imath] and [imath]B \subseteq A \cup B[/imath], it is [imath]\mathcal{P}(A) \subseteq \mathcal{P}(A \cup B)[/imath] and [imath]\mathcal{P}(B) \subseteq \mathcal{P}(A \cup B)[/imath]. Since union preserves inclusions and inclusion is idempotent, it is [imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B) \cup \mathcal{P}(A \cup B) = \mathcal{P}(A \cup B)[/imath]. Is my proof valid too?

 

[pka]

I denote with [imath]\mathcal{P}(T)[/imath] the set of subsets of a set [imath]T[/imath]. My textbook proves that for any given sets [imath]A,B[/imath] it is [imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A\cup B)[/imath] using the following argument: let [imath]X \in \mathcal{P}(A) \cup \mathcal{P}(B)[/imath], then [imath]X \in \mathcal{P}(B) \lor X \in \mathcal{P}(B)[/imath]. Hence, [imath]X \subseteq A \lor X \subseteq B[/imath] and so [imath]X \subseteq A\cup B[/imath]. That is, [imath]X \in \mathcal{P}(A \cup B)[/imath].

I came up with this other argument: since it is [imath]A \subseteq A \cup B[/imath] and [imath]B \subseteq A \cup B[/imath], it is [imath]\mathcal{P}(A) \subseteq \mathcal{P}(A \cup B)[/imath] and [imath]\mathcal{P}(B) \subseteq \mathcal{P}(A \cup B)[/imath]. Since union preserves inclusions and inclusion is idempotent, it is [imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B) \cup \mathcal{P}(A \cup B) = \mathcal{P}(A \cup B)[/imath]. Is my proof valid too?

You want to show that any element of [imath]\mathcal{P}(A)\cup\mathcal{P}(B)[/imath] is an element of [imath]\mathcal{P}(A\cup B)[/imath]
i.e. Show that any subset of [imath]A[/imath] is a subset of [imath]A\cup B[/imath] and any subset of [imath]B[/imath] is a subset of [imath]A\cup B[/imath]
NOTE that is the second line above you have [imath]\bf B[/imath] twice. One of those should be an [imath]\bf A[/imath].


[imath][/imath][imath][/imath][imath][/imath]

 

[MaxMath]

I denote with [imath]\mathcal{P}(T)[/imath] the set of subsets of a set [imath]T[/imath]. My textbook proves that for any given sets [imath]A,B[/imath] it is [imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A\cup B)[/imath] using the following argument: let [imath]X \in \mathcal{P}(A) \cup \mathcal{P}(B)[/imath], then [imath]X \in \mathcal{P}(B) \lor X \in \mathcal{P}(B)[/imath]. Hence, [imath]X \subseteq A \lor X \subseteq B[/imath] and so [imath]X \subseteq A\cup B[/imath]. That is, [imath]X \in \mathcal{P}(A \cup B)[/imath].

I came up with this other argument: since it is [imath]A \subseteq A \cup B[/imath] and [imath]B \subseteq A \cup B[/imath], it is [imath]\mathcal{P}(A) \subseteq \mathcal{P}(A \cup B)[/imath] and [imath]\mathcal{P}(B) \subseteq \mathcal{P}(A \cup B)[/imath]. Since union preserves inclusions and inclusion is idempotent, it is [imath]\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B) \cup \mathcal{P}(A \cup B) = \mathcal{P}(A \cup B)[/imath]. Is my proof valid too?

I believe your proof is correct. You may want to improve it by explicitly proving one theorem on which your proof is implicitly based—
[math]A\subseteq B\implies\mathcal{P}(A)\subseteq\mathcal{P}(B)[/math]which is straightforward. So no one can say any step is omitted.

(I agree with @pka that in your post "[imath]X \in \mathcal{P}(B) \lor X \in \mathcal{P}(B)[/imath]" should read "[imath]X \in \mathcal{P}(A) \lor X \in \mathcal{P}(B)[/imath]".)