- #1
jessicaw
- 56
- 0
Let [tex] B_n=\cup_{i=1}^n A_i [/tex].
[tex] \overline{B_n}[/tex] is the smallest closed subset containing [tex] B_n[/tex].
Note that
[tex]\cup_{i=1}^n \overline{A_i}[/tex] is a closed subset containing [tex] B_n[/tex].
Thus,
[tex] \overline{B_n}\supset \cup_{i=1}^n \overline{A_i}[/tex]Isn't the truth should be that
[tex] \overline{B_n}[/tex] is the smallest?
How come claim that
[tex]\cup_{i=1}^n \overline{A_i}[/tex] is even smaller?
[tex] \overline{B_n}[/tex] is the smallest closed subset containing [tex] B_n[/tex].
Note that
[tex]\cup_{i=1}^n \overline{A_i}[/tex] is a closed subset containing [tex] B_n[/tex].
Thus,
[tex] \overline{B_n}\supset \cup_{i=1}^n \overline{A_i}[/tex]Isn't the truth should be that
[tex] \overline{B_n}[/tex] is the smallest?
How come claim that
[tex]\cup_{i=1}^n \overline{A_i}[/tex] is even smaller?