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Explain why induction cannot be used to conclude: $$\left(\bigcup_{n=1}^\infty A_n\right)^c = \bigcap_{n=1}^\infty A_n^c$$

Thomas Andrews
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Edie
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    Hello! Could you please tell us where the problem comes from and what have you tried to solve it? – yo' Aug 31 '14 at 20:08
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    Loosely speaking: because there is no integer $n$ s.t. $n+1 = \infty$. – Winther Aug 31 '14 at 20:10
  • I just saw it in my textbook of Understanding Analysis. I think it has no problem if we find that n=1 holds, then assume n=m holds, and to prove n=m+1 holds, even for infinite n. I have no idea how to prove, but I thought that in principle this induction tool may be used to deal with infinite sets. – Edie Aug 31 '14 at 20:18
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    $n$ is not a free variable in the above expression - it is not a statement about $n$. @Edie – Thomas Andrews Aug 31 '14 at 20:22
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    Not to mention that the more general statement $\left(\bigcup_{i\in I} A_i\right)^c = \bigcap_{i\in I} A_i^c$ holds - with an arbitrary index set $I$. – Hagen von Eitzen Aug 31 '14 at 20:24
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    Actually proving the equality is a question of basic logic. No induction necessary! Just consider what it takes to be a member of the set on the left, and of the set on the right. Then work out the equivalence of the two conditions. – Harald Hanche-Olsen Aug 31 '14 at 20:35
  • That's what I am going to do in the next step. Thanks for your advice! – Edie Aug 31 '14 at 20:45
  • The lower-case n=1 is fixed. There is nothing we can do about ∞ since we never get to ∞, it that what you mean? – Edie Aug 31 '14 at 20:46

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The title is wrong. Induction can be used in the infinite set $\Bbb N$.

For all $N\in\Bbb N$ $$\left(\bigcup_{n=1}^N A_n\right)^c = \bigcap_{n=1}^N A_n^c$$ and this can be proved by induction. But you can't "jump" from any $N$ to $\infty$.

Nitpicking: there is a stonger induction: the Transfinite induction.

Martín-Blas Pérez Pinilla
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You should look at how your text defines the set $\left( \bigcup_{n=1}^\infty A_n\right)$. My guess is that it defines as the set of all elements $x$ for which there exists an $A_n\in \{A_i\}_{i=1}^\infty$ with the property that $x\in A_n$.

To expand on the previous by Pinilla we can't conclude that $\sum_{i=1}^\infty \frac{1}{i} $ is finite even though $\sum_{i=1}^N \frac{1}{i}$ is finite for every positive integer $N$.

fred
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Theorem: If $A_i$ is finite for every $i$, then:

$$\bigcup_{i=1}^\infty A_i$$

is finite.

If you could use simple induction, wouldn't this be true?

Thomas Andrews
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