11.0 Laws of algebra of sets

If $A$ and $B$ are any two sets then

$$\begin{equation} \begin{aligned} i.\;(A \cup B)' = A' \cap B' \\ ii.\;(A \cap B)' = A' \cup B' \\\end{aligned} \end{equation} $$

i. Let $x$ be an arbitrary element of $(A \cup B)'$ and $y$ be an arbitrary element of $A' \cap B'$

Then,

$$\begin{equation} \begin{aligned} Considering\;element\;x \\ \Rightarrow x \in (A \cup B)' \\ \Rightarrow x \notin (A \cup B) \\ \Rightarrow x \notin \;A\;and\;x \notin \;B \\ \Rightarrow x \in \;A'\;and\;x \in \;B' \\ \Rightarrow x \in \;A'\; \cap \;B' \\ \Rightarrow (A \cup B)' \subseteq A'\; \cap \;B' \\ Now,\;considering\;element\;y \\ \Rightarrow y \in A' \cap B' \\ \Rightarrow y \in \;A'\;and\;y \in \;B' \\ \Rightarrow y \notin \;A\;and\;y \notin \;B \\ \Rightarrow y \notin \;A\; \cup \;B \\ \Rightarrow y \in \;(A\; \cup \;B)' \\ \Rightarrow A'\; \cap \;B' \subseteq (A \cup B)' \\ Hence, \\ (A \cup B)' = A' \cap B' \\\end{aligned} \end{equation} $$

ii. Let $x$ be an arbitrary element of $(A \cap B)'$ and $y$ be an arbitrary element of $A' \cup B'$

Then,

$$\begin{equation} \begin{aligned} Considering\;element\;x \\ \Rightarrow x \in (A \cap B)' \\ \Rightarrow x \notin (A \cap B) \\ \Rightarrow x \notin \;A\;or\;x \notin \;B \\ \Rightarrow x \in \;A'\;or\;x \in \;B' \\ \Rightarrow x \in \;A'\; \cup B' \\ \Rightarrow (A \cap B)' \subseteq A'\; \cup \;B' \\ Now,\;considering\;element\;y \\ \Rightarrow y \in A' \cup B' \\ \Rightarrow y \in \;A'\;or\;y \in \;B' \\ \Rightarrow y \notin \;A\;or\;y \notin \;B \\ \Rightarrow y \notin \;A\; \cap \;B \\ \Rightarrow y \in \;(A\; \cup \;B)' \\ \Rightarrow A'\; \cup \;B' \subseteq (A \cap B)' \\ Hence, \\ (A \cap B)' = A' \cup B' \\\end{aligned} \end{equation} $$

Illustration 24. If $A$ and $B$ are two sets, prove that, if $A \cap B = A \cup B$ then, $A = B$.

Solution: Let, $$A = B$$

Then,$$\begin{equation} \begin{aligned} A \cup B = A\;\;\;\;\;\;\;\;\;\;(i) \\ A \cap B = A\;\;\;\;\;\;\;\;\;\;(ii) \\\end{aligned} \end{equation} $$

Thus from $(i)$ and $(ii)$, we get,

$$\begin{equation} \begin{aligned} A \cup B = A \cap B = A = B \\ A = B \Rightarrow A \cap B = A \cup B\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(iii) \\\end{aligned} \end{equation} $$

Conversely,

Let, $$A \cap B = A \cup B$$

and,

$$x \in A$$

$$\begin{equation} \begin{aligned} x \in A \cup B \\ x \in A \cap B \\ x \in A\;\;and\;x \in B \\ A \subset B\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(iv) \\\end{aligned} \end{equation} $$

Now, let

$$y \in B$$

$$\begin{equation} \begin{aligned} y \in A \cup B \\ y \in A \cap B \\ y \in A\;\;and{\mkern 1mu} \;y \in B \\ B \subset A\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(v) \\\end{aligned} \end{equation} $$

From $(iv)$ and $(v)$,

we get, $$A = B$$

Thus, $$A \cap B = A \cup B \Rightarrow A = B\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(vi)$$

From, $(iii)$ and $(vi)$

$$A \cap B = A \cup B \Leftrightarrow A = B$$

Illustration 25: Let $A$, $B$ and $C$ be the sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$.

Solution: Given,

$$A \cup B = A \cup C$$

On intersection with set $C$,

$$(A \cup B) \cap C = (A \cup C) \cap C$$

We know that,

$$(A \cup C) \cap C = C$$

Thus,

$$(A \cap C) \cup (B \cap C) = C$$

Also given that,

$$A \cap B = A \cap C$$

Hence, we can write it as,

$$(A \cap B) \cup (B \cap C) = C\;\;\;\;\;\;\;\;\;(i)$$

Now, on intersection with set $B$,

$$(A \cup B) \cap B = (A \cup C) \cap B$$

We know that,

$$(A \cup B) \cap B = B$$

Thus,

$$(A \cap B) \cup (B \cap C) = B$$

Also given that,

$$A \cap B = A \cap C$$

Hence, we can write it as,

$$(A \cap B) \cup (B \cap C) = B\;\;\;\;\;\;\;\;\;(ii)$$

From $(i)$ and $(ii)$ , we get,

$$(A \cap B) \cup (B \cap C) = C = B$$

i.e.,

$$B = C$$

Hence proved.

Illustration 26. For any two sets $A$ and $B$ prove that, $P(A \cap B) = P(A) \cap P(B)$.

Solution: To prove,

$$P(A \cap B) = P(A) \cap P(B)$$

We use the property that if two sets are subsets of each other, then they are equal sets.

i.e.

$$P(A \cap B) \subset P(A) \cap P(B)\,\;and\;\;P(A) \cap P(B) \subset P(A \cap B)$$ would imply, $$P(A \cap B) = P(A) \cap P(B)$$

Let us consider a set $X$,

$$X \in P(A \cap B)$$

This means, $X$ is a subset of the intersection of $A$ and $B$,

$$X \subset A \cap B$$

Also, if a set is a subset of intersection of two sets, then it is a subset of each of the two sets.

i.e. $$X \subset A\;\;and\;\;X \subset B$$

And, if a set is a subset of another set, then it belongs to the power set of the super-set, i.e., $$X \in P(A)\;\;and\;\;X \in P(B)$$

This implies, $$X \in P(A) \cap P(B)$$

Thus,

$$P(A \cap B) \subset P(A) \cap P(B)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,....(i)$$

Now let us consider a set $Y$,

$$Y \in P(A) \cap P(B)$$

Then, $$ \Rightarrow Y \in P(A)\;\;and\;\;Y \in P(B)$$

Thus, $Y$ is a subset of each of the two sets, $$ \Rightarrow Y \subset A\,\,and\,\,Y \subset B$$

Hence, $Y$ is a subset of their intersections and thus belongs to the power set of their intersections.

$$Y \subset A \cap B \Rightarrow Y \in P(A \cap B)$$

Thus, $$P(A) \cap P(B) \subset P(A \cap B)\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...(ii)$$

From $(i)$ and $(ii)$, it is clear that, $$P(A \cap B) \subset P(A) \cap P(B)\,\;and\;\;P(A) \cap P(B) \subset P(A \cap B)$$

i.e., $$P(A \cap B) = P(A) \cap P(B)$$

Hence proved.

Illustration 27. Suppose ${A_1}$, ${A_2}$,......, ${A_{30}}$ are thirty sets each with five elements and ${B_1}$, ${B_2}$,......, ${B_n}$ are $n$ sets each with three elements, and $\bigcup\limits_{i = 1}^{30} {{A_i}} = \bigcup\limits_{j = 1}^n {{B_j}} = S$.Assume that each element of $S$ belongs to exactly ten of the ${{A_i}}$'s and nine of the ${{B_j}}$ 's. Find $n$.

Solution: Since given that ${{A_i}}$ has exactly $5$ elements each and each element of $S$ belongs to exactly ten of the ${{A_i}}$'s,

$$\forall \;\;1 \le i \le 30,\;n({A_i}) = 5$$,

$$\begin{equation} \begin{aligned} S = \bigcup\limits_{i = 1}^{30} {{A_i}} \\ n(S) = {1 \over {10}}\sum\limits_{i = 1}^{30} {n({A_i})} \\ n(S) = {1 \over {10}}(5 \times 30) = 15\;\;\;\;\;\;\;\;...(i) \\\end{aligned} \end{equation} $$

Also given that ${{B_j}}$ has exactly $3$ elements each and each element of $S$ belongs to exactly nine of the ${{B_j}}$'s,

$$\forall \;\;1 \le j \le n,\;n({B_j}) = 3$$

$$\begin{equation} \begin{aligned} S = \bigcup\limits_{j = 1}^n {{B_j}} \\ n(S) = {1 \over 9}\sum\limits_{j = 1}^n {n({B_j})} \\ n(S) = {1 \over 9}(3n) = {n \over 3}\;\;\;\;\;\;\;\;\;\;\;\;\;...(ii) \\\end{aligned} \end{equation} $$

From $(i)$ and $(ii)$, we get,

$$15 = {n \over 3} \Rightarrow n = 45$$

Thus the number of sets in $B$ is $45$