What are the cases of not using (countable) induction?

In countably infinite union of countably infinite sets is countable the proof has been given, but when as a student I attempted the question, I tried using induction ( later to found it to be wrong way of going about that problem) but the reasoning was quite simple.

I.Union of 1,2 countable set is counatble ( obvious )

II.Suppose I is true up some n.

III.For case n+1 we get union of all the sets up to n, which was true by II , so we get back to case of union of two countable sets which is true.

I couldn't see any flows with the above, but it was wrong, I think it was due to the fact that proving something holds for all finite n, is not the same as proving it for the infinite case.

My question is, why proving something for all finite n ( after all any nu,ber that can be picked is finite), is not same as proving the infinite case ? Is there any other example induction failing for infinite clause?

Another question is : Is uncountable ( infinite of course :) union of countable ( finite or infinite) is countable? (The wrong induction method I used can't even be used for this one.)

Thank you

PS : Modified the title as it was sugeestive of induction failing, where is the case is it is being misused.

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For every $n$, a set with exactly $n$ elements is finite. However you're generalizing to infinite cases, this is going to fail. –  Chris Eagle Aug 2 '11 at 21:31
On your last question: every set is the union of its singleton subsets. So every uncountable set is an uncountable union of one-element sets. –  Chris Eagle Aug 2 '11 at 21:36

Countable induction doesn't fail; it just has an extra hypothesis that you aren't checking. Here's what the principle of induction looks like in general.

Suppose $P_{\alpha}$ is a collection of statements indexed by ordinals $\alpha$. Further suppose that the following condition holds: if $P_{\alpha}$ is true for all $\alpha < \beta$, then $P_{\beta}$ is true. Finally, suppose that $P_0$ is true. Then $P_{\alpha}$ is true for all $\alpha$.

(Proof: suppose otherwise. Then there is a least ordinal $\beta$ such that $P_{\beta}$ isn't true. But $P_{\alpha}$ is true for all $\alpha < \beta$ by minimality, and this contradicts the condition.)

If instead the collection is indexed by the set of all ordinals less than $\beta$, then we can conclude that $P_{\alpha}$ is true for all $\alpha < \beta$. But the details of the inductive step are important. For $\beta = \omega + 1$, the inductive step requires that $P_1, P_2, ...$ being true imply that $P_{\omega}$ is true, and this condition doesn't hold in general. (A basic counterexample, as Chris says, is the collection of statements $P_n : n \text{ is finite}$.)

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Don't we need a well-ordering property to guarantee that the set of propositions {$P_{\beta}$} has a least element? –  gary Aug 3 '11 at 0:42
In what sense do we not have such a property? –  Qiaochu Yuan Aug 3 '11 at 0:43
Never mind; I realized that ordinals are the order types of well-ordered sets. –  gary Aug 3 '11 at 0:53

First, it’s simply not true that any number that can be picked is finite. For example, $\omega$, the smallest infinite cardinal (and ordinal) number is clearly not finite, and I just picked it. There is no reason to think that something that is true of every non-negative integer $n$ is also true of $1/2$: after all, $1/2$ isn’t a non-negative integer. Similarly, there is no reason to think that something that is true of every non-negative integer $n$ is also true of $\omega$: $\omega$ isn’t a non-negative integer. In particular, the fact that the union of $n$ countable sets is countable provides no guarantee that the union of $\omega$ countable sets is countable. Your induction argument doesn’t provide that guarantee either, because it doesn’t contain any reasoning to bridge the gap between finite and infinite unions.

It may be easier to see this with an example in which the gap cannot be bridged. Let $P(n)$ be the statement every set of cardinality n has a finite number of subsets. Clearly $P(0)$ is true, since the empty set has only one subset. It’s also easy to see that $P(n) \Rightarrow P(n+1)$: If $S$ is a set of $n+1$ objects, pick one, $s_0$, and let $S' = S \setminus \{s_0\}$. $S'$ has only $n$ elements, so it has a finite number of subsets, say $A_1,\dots,A_m$. But then the subsets of $S$ are the $2m$ sets $$A_1,A_1\cup\{s_0\},A_2,A_2\cup\{s_0\}\dots,A_m,A_m\cup\{s_0\},$$ so $S$ also has a finite number of subsets. By induction, therefore, every finite set has a finite number of subsets. But you wouldn’t be tempted to conclude from this that a countably infinite set -- $\mathbb{N}$, for instance -- has a finite number of subsets: nothing in the argument makes the jump from finite to infinite, and besides, it’s obvious that an infinite set must have an infinite number of subsets.

To answer your last question, an uncountable union of pairwise disjoint non-empty sets is always uncountable, even if each of the sets has only one element.

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Maybe this will help. For each positive integer $n$, let $P_n$ be a "mathematically meaningful" statement that can be true or false. Then mathematical induction provides a means to prove each of these statements is true, but mathematical induction does not say anything about "statement $P_{\infty}$". In fact, even if $P_{\infty}$ seems meaningful in some sense (and often it isn't), mathematical induction doesn't see statement $P_{\infty}$. For any experts reading, I'm of course talking about ordinary school mathematical induction, not transfinite induction or still more general notions (e.g. "Elementary Induction on Abstract Structures" by Moschovakis).

Also, notice what happens with your I, II, and III when "countable" is changed to "finite". The union of two finite sets is finite, if the union of $n$ finite sets is finite then the union of $n+1$ finite sets is finite, so the union of countably many finite sets is finite ... oops!

Finally, an uncountable union of countable sets can be countable or uncountable, depending on how much overlap there is among the sets. Each uncountable set is the uncountable union of its singleton (1-element) subsets, and the union of $A_x$ over all real numbers $x$, where $A_x$ consists of the rational numbers belonging to $(-\infty,x)$, is the set of rational numbers and hence countable.

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As you said, induction only guarantees that P(n) is true _for all n in $\mathbb N$ , and $\infty$ is not in $\mathbb N$. Maybe the set of almost-disjoint sequences would be a good example of uncountable union of countables being countable. –  gary Aug 3 '11 at 1:07