If something is true for every $n\in\Bbb{N}$, does it imply that thing is true for $\infty$? I have the following expression

$m(\bigcup\limits_{i=1}^\infty E_i)\geq \sum\limits_{i=1}^n m(E_i)$.
Since the left side of the inequality is independent of $n$, we have $m(\bigcup\limits_{i=1}^\infty E_i)\geq \sum\limits_{i=1}^\infty m(E_i)$

I don't understand this argument. Is $\infty$ a natural number? If something is true for every natural number, how does it prove that it is true for $\infty$? Is there some kind of limiting process happening here?
 A: Take the limit of both sides as $n\to\infty$. On the left-hand side, nothing happens, as $n$ does not appear. On the right-hand side, we get exactly $\sum\limits_{i=1}^{\infty}m(E_i)$.
A: You're right, it's not that simple. But it's pretty close: by definition, the infinite sum $S_\infty=\sum_{i=1}^\infty x_i$ is the limit of $S_n=\sum_{i=1}^nx_i$. If each $S_n$ is less than some fixed value $a$, then certainly the limit of the $S_n$ cannot be bigger than $a$.
A: For each $n\in\{0,1,2,3,\ldots\}$, every set of size $n$ is finite.  Therefore every set of size $\infty$ is finite.
Most certainly it is not true that if something is true for every $n\in\mathbb N$, it is true of $\infty$.  The intersection of any finite number of open sets is open; the intersection of infinitely many open sets is usually not open.
Recall that $\displaystyle \sum_{k=1}^\infty m(E_k)$ is defined as $\displaystyle\lim_{n\to\infty} \sum_{k=1}^n m(E_k)$.  So the question comes to this:
$$
\text{If }A \ge B_n\text{ for all $n$, does it follow that }A\ge\lim_{n\to\infty} B_n\text{?}
$$
Assuming the latter limit exists, the only alternative is
$$
A < \lim_{n\to\infty} B_n = L.
$$
Assuming that alternative, let $\varepsilon = \lim_{n\to\infty} B_n - A > 0$.  By the $\text{“}\varepsilon$-$N\text{''}$ definition of the limit of a sequence, there exists $N$ so large that for all $n\ge N$, $B_n>L-\varepsilon$. That implies $B_n>A$, contradicting the hypothesis.
Postscript 23 hours later:
Right: If $A\ge B_n$ for all $n$, then $A\ge \lim\limits_{n\to\infty} B_n$.
Wrong: If $A > B_n$ for all $n$, then $A > \lim\limits_{n\to\infty} B_n$.
The fact that the second statement is wrong is another instance of the fact that you cannot say that if something is true of every $n\in\mathbb N$, then it is true of $\infty$.
