Question

Let\({A_i} = \{ ....., - 2, - 1,0,1,......,i\} \)Find

(a) \(\bigcup\limits_{i = 1}^n {{A_i}} \)

(b)\(\bigcap\limits_{i = 1}^n {{A_i}} \)

Short answer

(a) \(\bigcup\nolimits_{i = 1}^n {{A_i}} = {A_n} = \{ ....., - 2, - 1,0,1,......,n\} \)

(b) \(\bigcap\nolimits_{i = 1}^n {{A_i}} = {A_1} = \{ ....., - 2, - 1,0,1\} \)

Step-by-step solution

Step 1

Union\(A \cup B\): all elements that are either in\(A\)OR in\(B\)

Intersection\(A \cap B\): all elements that are both in\(A\)AND in\(B\)

\(X\)is a subset of\(Y\)if every element of\(X\)is also an element of\(Y\)

Notation\(X \subseteq Y\)

Idempotent Law

\(\begin{aligned}{l}A \cap A = A\\A \cap A = A\end{aligned}\)

Step 2

Given:

\({A_i} = \{ ....., - 2, - 1,0,1,......,i\} = \{ x \in Z\left| {x \le i} \right.\} \)

1. If\(i \le n\), then we note that\({A_i}\)is a subset of\({A_n}\):

\({A_i} \subset {A_n}\)

Let us take the union of all these sets\({A_i}\)with\(i \le n\):

\(\bigcup\limits_{i = 1}^n {{A_i}} \subseteq \bigcup\limits_{i = 1}^n {{A_n}} \)

Use the idempotent law:

\(\bigcup\limits_{i = 1}^n {{A_i}} \subseteq \bigcup\limits_{i = 1}^n {{A_n}} = {A_n}\)

By the definition of the union, we also know that\({A_n} \subseteq \bigcup\nolimits_{i = 1}^n {{A_i}} \)

Since,\(\bigcup\nolimits_{i = 1}^n {{A_i}} \subseteq {A_n}\)and\({A_n} \subseteq \bigcup\nolimits_{i = 1}^n {{A_i}} \), the two sets then have to be equal:

\(\bigcup\limits_{i = 1}^n {{A_i}} = {A_n}\)

Step 3

(b) If \(i \ge n\), then we note that \({A_1}\) is a subset of \({A_i}\):

\({A_1} \subset {A_i}\)

Let us take the intersections of all these sets\({A_i}\)with\(i \le n\):

\(\bigcap\limits_{i = 1}^n {{A_1}} \subseteq \bigcap\limits_{i = 1}^n {{A_i}} \)

Use the idempotent law:

\({A_1} = \bigcap\limits_{i = 1}^n {{A_1}} \subseteq \bigcap\limits_{i = 1}^n {{A_i}} \)

By the definition of the intersection, we also know that \(\bigcap\nolimits_{i = 1}^n {{A_i}} \subseteq {A_1}\)

Since \(\bigcap\nolimits_{i = 1}^n {{A_i}} \subseteq {A_1}\)and \({A_1} \subseteq \bigcap\nolimits_{i = 1}^n {{A_i}} \), the two sets then have to be equal:

\(\bigcap\limits_{i = 1}^n {{A_i}} = {A_1}\)

Step 4

We conclude that

(a) \(\bigcup\nolimits_{i = 1}^n {{A_i}} = {A_n} = \{ ....., - 2, - 1,0,1,......,n\} \)

(b) \(\bigcap\nolimits_{i = 1}^n {{A_i}} = {A_1} = \{ ....., - 2, - 1,0,1\} \)