DavidWhitbeck
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[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 (\cup_{n=1}^{N}A_n)^{c} = \cap_{n=1}^{N}A_n^c using induction, but I don't see why induction can't be used to say that (\cup_{n=1}^{\infty}A_n)^{c} = \cap_{n=1}^{\infty}A_n^c?
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 B_1 \supset B_2 \supset \cdots and each B_n is countably infinite, their intersection \cap_{n=1}^{\infty}B_n 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 B_m = \cup_{n=m}^{\infty}A_n or perhaps B_m = \cap_{n=1}^{m}A_n^c so that B_1 \supset B_2 \supset \cdots 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 (\cup_{n=1}^{N}A_n)^{c} = \cap_{n=1}^{N}A_n^c using induction, but I don't see why induction can't be used to say that (\cup_{n=1}^{\infty}A_n)^{c} = \cap_{n=1}^{\infty}A_n^c?
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 B_1 \supset B_2 \supset \cdots and each B_n is countably infinite, their intersection \cap_{n=1}^{\infty}B_n 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 B_m = \cup_{n=m}^{\infty}A_n or perhaps B_m = \cap_{n=1}^{m}A_n^c so that B_1 \supset B_2 \supset \cdots 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?