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Slightly harder introductory exercises about the power set.

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Since $P(\\left\\{\\right\\}) = 2^{\\left|\\left\\{\\right\\}\\right|} = 2^0 = 1$ we know that $P(\\left\\{\\right\\})$ is a one-element set which contains $\\left\\{\\right\\}$. There is only one possible answer

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$ \\left\\{\\left\\{\\right\\}\\right\\}$.

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We construct the answer gradually. Since $\\left|S\\right| = \\var{n}$

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$P(A)$ contains at least the elements $\\left\\{\\right\\}$ and $A$. The case where $\\left\\{\\right\\} = A$ is particularly interesting. What is $P(\\left\\{\\right\\})$?

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What is $\\left|\\left\\{\\right\\}\\right|$?

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Hence, what is $\\left|P(\\left\\{\\right\\})\\right|$?

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Hence $P(\\left\\{\\right\\})$ is a set that only contains one element. But it must at least contain the element $\\left\\{\\right\\}$. Ponder this for a moment and then answer the question below, and remember that the NUMBAS syntax for $\\left\\{\\right\\}$ is set().

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If $\\left|S\\right|=\\var{n}$ then what is $\\left|P(P(P(S)))\\right|$?

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Using the handy fact, $\\left|P(S)\\right| = 2^{\\var{n}}$.

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Using this handy fact again, we deduce that $\\left|P(P(S))\\right| = 2^{\\left|P(S)\\right|}$, which is

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Use the handy fact yet again to determine $\\left|P(P(P(S)))\\right|$.

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We use the notation $\\left|A\\right| = n$ to mean that $A$ contains $n$ elements. This is often called the cardinality of the set. Here is a handy fact about the number of elements in a power set.

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If $\\left|A\\right| = n$ then $\\left|P(A)\\right| =2^n$.

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Using this fact, answer the following questions regarding the power set.

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