Why induction cannot be used for infinite sets?

Question

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

Answer 1

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.

Answer 2

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 $N$.

Answer 3

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?