R 2 i s a n e i g h b o r h o o d o f P , sin c e P ∈ R 2 , R 2 i s o p e n B 1 = B ( O , 4 ) i s n o t a n e i g h b o r h o o d o f P : ∥ ( 3 , 4 ) − ( 0 , 0 ) ∥ = 5 > 4 ⇒ ( 3 , 4 ) ∉ B 1 B 2 = B ( O , 6 ) i s a n e i g h b o r h o o d o f P : ∥ ( 3 , 4 ) − ( 0 , 0 ) ∥ = 5 < 6 ⇒ ( 3 , 4 ) ∈ B 2 , B 2 i s o p e n B 3 = R 2 ∖ B 2 i s n o t a n e i g h b o r h o o d o f P : P ∈ B 2 ⇒ P ∉ B 3 B 4 = B ( O , 1 ) ∪ B 2 = B 2 i s a n e i g h b o r h o o d o f P ( s e e a b o v e ) B 5 = B ( P 0 , 11 ) i s n o t a n e i g h b o r h o o d o f P : ∥ ( 3 , 4 ) − ( 5 , 0 ) ∥ = 2 2 + 4 2 > 11 ⇒ ( 3 , 4 ) ∉ B 5 \mathbb{R} ^2\,\,is\,\,a\,\,neighborhood\,\,of\,\,P, \sin ce\,\,P\in \mathbb{R} ^2,\mathbb{R} ^2\,\,is\,\,open\\B_1=B\left( O,4 \right) \,\,is\,\,not\,\,a\,\,neighborhood\,\,of\,\,P: \left\| \left( 3,4 \right) -\left( 0,0 \right) \right\| =5>4\Rightarrow \left( 3,4 \right) \notin B_1\\B_2=B\left( O,6 \right) \,\,is\,\,a\,\,neighborhood\,\,of\,\,P: \left\| \left( 3,4 \right) -\left( 0,0 \right) \right\| =5<6\Rightarrow \left( 3,4 \right) \in B_2,B_2\,\,is\,\,open\\B_3=\mathbb{R} ^2\setminus B_2\,\,is\,\,not\,\,a\,\,neighborhood\,\,of\,\,P: P\in B_2\Rightarrow P\notin B_3\\B_4=B\left( O,1 \right) \cup B_2=B_2\,\,is\,\,a\,\,neighborhood\,\,of\,\,P\,\,\left( see\,\,above \right) \\B_5=B\left( P_0,\sqrt{11} \right) \,\,is\,\,not\,\,a\,\,neighborhood\,\,of\,\,P: \left\| \left( 3,4 \right) -\left( 5,0 \right) \right\| =\sqrt{2^2+4^2}>\sqrt{11}\Rightarrow \left( 3,4 \right) \notin B_5 R 2 i s a n e i g hb or h oo d o f P , sin ce P ∈ R 2 , R 2 i s o p e n B 1 = B ( O , 4 ) i s n o t a n e i g hb or h oo d o f P : ∥ ( 3 , 4 ) − ( 0 , 0 ) ∥ = 5 > 4 ⇒ ( 3 , 4 ) ∈ / B 1 B 2 = B ( O , 6 ) i s a n e i g hb or h oo d o f P : ∥ ( 3 , 4 ) − ( 0 , 0 ) ∥ = 5 < 6 ⇒ ( 3 , 4 ) ∈ B 2 , B 2 i s o p e n B 3 = R 2 ∖ B 2 i s n o t a n e i g hb or h oo d o f P : P ∈ B 2 ⇒ P ∈ / B 3 B 4 = B ( O , 1 ) ∪ B 2 = B 2 i s a n e i g hb or h oo d o f P ( see ab o v e ) B 5 = B ( P 0 , 11 ) i s n o t a n e i g hb or h oo d o f P : ∥ ( 3 , 4 ) − ( 5 , 0 ) ∥ = 2 2 + 4 2 > 11 ⇒ ( 3 , 4 ) ∈ / B 5
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