Countable union of disjoint sets

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!

• 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. Oct 27 '18 at 15:39
• 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 $n$. Oct 27 '18 at 15:39
• Assuming $A_n$ is non-empty, and countable means countably infinite, this is true, since one can pick one element from each $A_n$. Oct 27 '18 at 15:47
• Disjointed is not necessary. $|\bigcup_{k=1}^n A_k|\ \le \sum_{k=1}^n |A_k|$. Oct 27 '18 at 16:36

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.
• 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 Oct 28 '18 at 7:15