R2isaneighborhoodofP,sinceP∈R2,R2isopenB1=B(O,4)isnotaneighborhoodofP:∥(3,4)−(0,0)∥=5>4⇒(3,4)∈/B1B2=B(O,6)isaneighborhoodofP:∥(3,4)−(0,0)∥=5<6⇒(3,4)∈B2,B2isopenB3=R2∖B2isnotaneighborhoodofP:P∈B2⇒P∈/B3B4=B(O,1)∪B2=B2isaneighborhoodofP(seeabove)B5=B(P0,11)isnotaneighborhoodofP:∥(3,4)−(5,0)∥=22+42>11⇒(3,4)∈/B5
Comments