UPSKILL MATH PLUS

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Learn more### Theory:

The set of all subsets of a set \(A\) is said to be power set \(A\). It is written as \(P(A)\).

Example:

Consider the set with two elements $A\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\{1,2\}$.

By the concept, 'Empty set is a subset of every set' and 'Every set is a subset of itself' the two obvious subsets of the set are the empty set and the whole set itself.

That is, $\varnothing \phantom{\rule{0.147em}{0ex}}\mathit{or}\phantom{\rule{0.147em}{0ex}}\left\{\right\}$ and $\{1,2\}$.

Now let us write the elements and its combination of subsets.

The singleton subsets are $\left\{1\right\}$ and $\left\{2\right\}$.

Therefore, the subsets of set \(A\) are $\left\{\right\},\left\{1\right\},\left\{2\right\},\phantom{\rule{0.147em}{0ex}}\{1,2\}$.

Thus, the set of power sets of \(A\), $P(A)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\left\{\right\{\},\{1\},\{2\},\{1,2\left\}\right\}$.

Important!

The number of elements in set \(A\) is always less than the number of elements in the power set of \(A\). That is $n(A)\phantom{\rule{0.147em}{0ex}}\le n[P(A)]$.

Let \(A\) be a set with \(m\) elements. That is, $n(A)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}m$.

- The number of elements in the power set of \(A\) is given by $n[P(A)]={2}^{m}$.
- The number of a proper subset of \(A\) is given by $n[P(A)]-1={2}^{m}-1$.

Example:

Consider the set with three elements $B\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\{a,b,c\}$.

Now we are going to find the number of elements in the power set of \(B\).

Set \(B\) has three elements. That is, \(m = 3\).

The formula for this is given by $n[P(B)]={2}^{m}$.

Substitute the value of \(m\).

$n[P(B)]={2}^{3}=8$

**Let us verify by the actual method:**

By the concept, 'Empty set is a subset of every set' and 'Every set is a subset of itself' the two obvious subsets of the set are the empty set and the whole set itself.

That is, $\varnothing \phantom{\rule{0.147em}{0ex}}\mathit{or}\phantom{\rule{0.147em}{0ex}}\left\{\right\}$ and $\{a,b,c\}$.

Now let us write the elements and its combination of subsets.

The singleton subsets are $\left\{a\right\}$, $\left\{b\right\}$ and $\left\{c\right\}$.

Let us write the subsets with two elements $\{a,b\},\phantom{\rule{0.147em}{0ex}}\{a,c\},\phantom{\rule{0.147em}{0ex}}\{b,c\}$.

Thus, the subsets of set \(B\) are $\left\{\right\},\left\{a\right\},\phantom{\rule{0.147em}{0ex}}\left\{b\right\},\phantom{\rule{0.147em}{0ex}}\left\{c\right\},\{a,b\},\phantom{\rule{0.147em}{0ex}}\{a,c\},\phantom{\rule{0.147em}{0ex}}\{b,c\}$.

Therefore, the power set of \(B\) becomes $n[P(B)]=\left\{\right\{\},\{a\},\phantom{\rule{0.147em}{0ex}}\{b\},\phantom{\rule{0.147em}{0ex}}\{c\},\{a,b\},\phantom{\rule{0.147em}{0ex}}\{a,c\},\{b,c\left\}\right\}$.

Note that the number of elements in the power set of \(B\) is \(8\).

Important!

The number of subsets of a set is equal to the number of elements in the respective power set.