- #1
DavidWhitbeck
- 351
- 1
[SOLVED] Basic sets question
It's been years since I've taken analysis, and so I thought I would have a refresher by studying Abbott's Understanding Analysis.
Anyway to the point-- there is a simple exercise in the beginning that stumps me (don't laugh I'm a physicist).
First of all I am fine with, and have proven that [tex](\cup_{n=1}^{N}A_n)^{c} = \cap_{n=1}^{N}A_n^c[/tex] using induction, but I don't see why induction can't be used to say that [tex](\cup_{n=1}^{\infty}A_n)^{c} = \cap_{n=1}^{\infty}A_n^c[/tex]?
Abbott then wants me, the reader, to prove that set equality if it's valid using another method. There was a hint given, and that was to use the fact that if [tex]B_1 \supset B_2 \supset \cdots[/tex] and each [tex]B_n[/tex] is countably infinite, their intersection [tex]\cap_{n=1}^{\infty}B_n[/tex] does not have to be.
The only natural construction of sets that I could think of that would fit with the hint would be something like [tex]B_m = \cup_{n=m}^{\infty}A_n[/tex] or perhaps [tex]B_m = \cap_{n=1}^{m}A_n^c[/tex] so that [tex]B_1 \supset B_2 \supset \cdots[/tex] is satisfied and I have an expression either way that appears in the conjecture.
It's probably a standard result, but I can't figure it out, can someone help me with this?
It's been years since I've taken analysis, and so I thought I would have a refresher by studying Abbott's Understanding Analysis.
Anyway to the point-- there is a simple exercise in the beginning that stumps me (don't laugh I'm a physicist).
First of all I am fine with, and have proven that [tex](\cup_{n=1}^{N}A_n)^{c} = \cap_{n=1}^{N}A_n^c[/tex] using induction, but I don't see why induction can't be used to say that [tex](\cup_{n=1}^{\infty}A_n)^{c} = \cap_{n=1}^{\infty}A_n^c[/tex]?
Abbott then wants me, the reader, to prove that set equality if it's valid using another method. There was a hint given, and that was to use the fact that if [tex]B_1 \supset B_2 \supset \cdots[/tex] and each [tex]B_n[/tex] is countably infinite, their intersection [tex]\cap_{n=1}^{\infty}B_n[/tex] does not have to be.
The only natural construction of sets that I could think of that would fit with the hint would be something like [tex]B_m = \cup_{n=m}^{\infty}A_n[/tex] or perhaps [tex]B_m = \cap_{n=1}^{m}A_n^c[/tex] so that [tex]B_1 \supset B_2 \supset \cdots[/tex] is satisfied and I have an expression either way that appears in the conjecture.
It's probably a standard result, but I can't figure it out, can someone help me with this?