# P(A)..Power set of A.

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• March 16th 2011, 07:51 PM
Oiler
P(A)..Power set of A.
Hey all,

Given $A = \{\emptyset, 1, \{2,3\}\}$
Is
$P(A) = \{\emptyset, 1, \{1,\{2,3\}\},\{2,3\}\}$ ?

And also how can I prove that for all sets $X$ and $Y$ we have $P(X) \cup P(Y) \subset P(X \cup Y)$
• March 16th 2011, 08:00 PM
Tinyboss
What is the cardinality of A (denoted |A|)? The cardinality of P(A) is 2^|A|, so what you wrote is missing a few. Also, the elements of P(A) are subsets of A, so there's no way that 1 is an element of P(anything), since it's not a set.

I think once you get the first part straightened out, the second part won't give you much trouble.