UPSKILL MATH PLUS

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Let \(A\) and \(B\) are two sets. If every element of \(A\) is also an element of \(B\), then \(A\) is a subset of \(B\). It is denoted as $A\subseteq B$.

We can read $A\subseteq B$ as '\(A\) is a subset of \(B\)'.

Suppose $A\subseteq B$ and $\left\{1\right\}\in A$ then $\left\{1\right\}\in B$.

If \(A\) is not a subset of \(B\), then we can write $A\not\subset B$.

Important!

- If \(A\) is a subset of \(B\), the number of elements in the set \(A\) must be less than or equal to the number of elements in the set \(B\). That is, $n(A)\le n(B)$. Since every element of \(A\) is also an element of \(B\), the set \(B\) must have at least as many elements as \(A\), thus \(n(A) ≤ n(B)\). It can be concluded that if \(A\) is a subset of \(B\), then the cardinal number of \(A\) must be less than the cardinal number of \(B\).
- If $A\subseteq B$ and $B\subseteq A$, then \(A = B\).
- An empty set is a subset of every set.
- Every set is a subset of itself.

Example:

**1.**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 the set \(A\) are $\left\{\right\},\left\{1\right\},\left\{2\right\}\phantom{\rule{0.147em}{0ex}}\mathit{and}\phantom{\rule{0.147em}{0ex}}\{1,2\}$.

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

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.

$\{a,b\},\phantom{\rule{0.147em}{0ex}}\{a,c\},\phantom{\rule{0.147em}{0ex}}\{b,c\}$.

**Therefore, the subsets of the set**\(B\)

**are**$\left\{\right\},\left\{a\right\},\phantom{\rule{0.147em}{0ex}}\left\{b\right\},\phantom{\rule{0.147em}{0ex}}\left\{c\right\},\{a,b\},\phantom{\rule{0.147em}{0ex}}\{a,c\},\phantom{\rule{0.147em}{0ex}}\{b,c\}$.