Question

Which of the following sets Sj , (j = 1, . . . , 4) are NOT a neighbor-
hood of the given point a.
(1) S1 = (−2, 3), a = 1; (2) S2 = [−3, ∞), a = 2.
(3) S3 = {x: |x − 1| ≤ 2} , a = 3; (4) S4 = {x: |x − 3| ≥ 2} , a = 1.

Answer 1

ANSWER: $S_3,S_4$ are NOT a neighborhood of the given point a.

EXPLANATION.

Let $a\in S$ and there exist $\varepsilon >0$ such that $\left( a-\varepsilon ,a+\ \varepsilon \right) \subseteq S$ . By the definition, $S$ is a neighborhood of the point a.

Hence:

(1) $\left\{ 1 \right\} \in (-2,3)={ S }_{ 1 }$ . Let $ε=1,$ $\left( a-1,a+1 \right) =(1-1,1+1)=(0,2)\subset (-2,3)={ S }_{ 1 }\quad$ .$S_1$ is a neighborhood of the point a=1.

(2) $\left\{ 2 \right\} \in [-3,\infty)={ S }_{ 2 }$ . Let $ε=1,$ $\left( a-1,a+1 \right) =(2-1,2+1)=(1,3)\subset [-3,\infty )={ S }_{ 2 }\quad$ .$S_2$ is a neighborhood of the point a=2.

(3) ${ \left\{ 3 \right\} \in S }_{ 3 }=\left\{ x:|x-1|\le 2 \right\} =\left\{ x:-2\le x-1\ \le 2 \right\} =\left\{ x:-1\le x \ \le 3 \right\}$ .

For all $\varepsilon >0\quad \left( a-\varepsilon ,a+\varepsilon \right) \ =(3-\varepsilon ,3+\varepsilon )\supset (3\ ,3+\varepsilon )\ \ , (3\ ,3+\varepsilon )\cap \ { S }_{ 3 }=\emptyset$ . So, $S_3$ is not a neighborhood of the point a=3.

(4) ${ \left\{ 1 \right\} \in S }_{ 4 }=\left\{ x:|x-3\ |\ge 2 \right\} =\left\{ x:\ \ x-3\le -2\ \ \right\} \cup \left\{ x:2\le x-3\ \ \right\} =\left\{ x:\ \ x\quad \le 1\ \ \right\} \cup \left\{ x:x\ge 5\ \ \ \right\} .$ For all $\varepsilon >0\ \left( a-\varepsilon ,a+\varepsilon \right) \ =(1-\varepsilon ,1+\varepsilon )\supset (1\ ,1+\varepsilon )\quad$ and $(1\ ,1+\varepsilon )\cap \left( 1,5 \right)$ is not an empty set. Therefore, $(1\ ,1+\varepsilon )$ is not completely contained in $S_4$ . So, $S_4$ is not a neighborhood of the point a=1.