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First question:

Is it true or false that the countable union of disjointed finite sets is always infinite?

In symbols: let $\{A_n\}_{n\in\mathbb{N}}$ a sequence of sets such that $A_i\cap A_j=\emptyset$ for $i\ne j$, and $|A_n|<+\infty$ for all $n\in\mathbb{N}$. Then $$ \bigg|\bigcup_n A_n\bigg|=+\infty. $$

For me it is true. My problem is if this thing always happens.

Second question

Is it true or false that the countable union of a finite number of finite sets, where the remaining ones are empty, has finite cardinality?

In symbols: let $\{A_n\}_{n\in\mathbb{N}}$ a sequence of sets such that $A_i\cap A_j=\emptyset$ for $i\ne j$, and exists $\overline{n}\in\mathbb{N}$ such that $|A_n|<+\infty$ for $n=1,2,\dots,\overline{n}$ and $A_{\overline{n}+1}=\cdot\cdot\cdot=A_m=\emptyset=\cdot\cdot\cdot$, then $$ \bigg|\bigcup_n A_n\bigg|<+\infty. $$

This obviously seems true to me, but in mathematics the obvious should also be shown, because I do not understand why it is important that the sets should be disjointed.

In my opinion, it proceeds in this way: $$ \bigcup_n A_n =\bigcup_{n=1}^\overline{n} A_n, $$ then $$ \bigg |\bigcup_n A_n\bigg|=\bigg |\bigcup_{n=1}^{\overline{n}} A_n\bigg|=\sum_{n=1}^{\overline{n}}|A_n|<+\infty\quad(\text{here we use the hypotheses that are disjointed)} $$

Thanks!

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    $\begingroup$ Please edit the question to tell us what you think the answers are, and why. Then perhaps we can help. Note: (1) is obviously false if you allow the $A_i$ to be empty. $\endgroup$ Oct 27 '18 at 15:39
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    $\begingroup$ Usually also finite sets are looked at as countable sets, so you better use the expression "countably infinite". Further also the empty set is a finite set, so we could just take $A_n=\varnothing$ for each $n$. $\endgroup$
    – drhab
    Oct 27 '18 at 15:39
  • $\begingroup$ Assuming $A_n$ is non-empty, and countable means countably infinite, this is true, since one can pick one element from each $A_n$. $\endgroup$ Oct 27 '18 at 15:47
  • $\begingroup$ Disjointed is not necessary. $|\bigcup_{k=1}^n A_k|\ \le \sum_{k=1}^n |A_k|$. $\endgroup$ Oct 27 '18 at 16:36
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Yes for the first question.
At most, only one of the A's are disjoint.
So assume WLOG, they are all not empty.
For each i in N, pick some a$_i$ in A$_i$.
Show the map i -> a$_i$ is an injection from N into the union of the A's.
Thus the union is infinite.

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  • $\begingroup$ Thanks for your answer. Therefore, we suppose that infinitely many of the $A_n$ are nonempty, we consider the map $f\colon\mathbb{N}\to\cup_n A_n$ such that $i\mapsto a_i$, where $a_i\in A_i$. This map is cleary an injection, then $|\cup_n A_n|\ge |\mathbb{N}|$. It's correct? Can you also consider the same map with $i\mapsto A_i$? This is also an injection $\endgroup$
    – Jack J.
    Oct 28 '18 at 7:15
  • $\begingroup$ You gof the idea. The other injection just says the collection of A's is infinite which was already infered in the question. $\endgroup$ Oct 28 '18 at 11:13

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