Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
countably infinite union of countably infinite sets is countable
proof that union of a sequence of countable sets is countable.

I'm a newbie who try to understand Set Theory. Is there anybody who can explain the solution for the following problem?

Assume that C is a countable set of countable pairwise disjoint sets, how can we prove that $\cup C$ is countable?

share|improve this question
    
@Asaf karagila Yes, I realized that later, so I deleted my comment accordingly. –  user22705 May 14 '12 at 10:50
add comment

marked as duplicate by Zhen Lin, Benjamin Lim, Martin Sleziak, t.b., Zev Chonoles May 14 '12 at 17:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

This requires a bit of the axiom of choice.

Since $C$ is countable we can write its members as $C_i$. For every $i$ fix $f_i\colon C_i\to\mathbb N\times\{i\}$ an injective function.

Now, since these are pairwise disjoint sets the union of he functions is a function, so let $f=\bigcup f_i$ be a function from $\bigcup C$ to $\mathbb N\times\mathbb N$.

Prove that this is an injective function, and use Cantor's pairing function to show that $\mathbb N\times\mathbb N$ is a countable set.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.