Consider the sequence of sets $S(n)=\{1,2,3,\ldots,n\}$. It's common to write:


Which I think is the same as:

$$\lim_{n\to \infty}\bigcup_{k=1}^{n}S(k)=N$$

Right? It doesn't make any difference if $k$ starts from $1$ or from any natural number $m$. What if we choose $m=n-1$?

$$\lim_{n\to \infty}\bigcup_{k=n-1}^{n}S(k)=N$$

Is it still true? Now why would we need the $S(n-1)$ when it's contained in $S(n)$. So It comes down to:

$$\lim_{n\to \infty}S(n)=N$$

Now it looks like a meaningless formula. How to make sense of this process?

The nested unions and intersections are widespreadly used. They just look unnecessary by the following process.

  • $\begingroup$ The problem is that you chose $m$ depending on $n$. That makes the difference. $\endgroup$
    – Crostul
    Oct 11, 2018 at 18:41
  • $\begingroup$ All of the statements you make are true for this specific situation. Consider a different example where instead $T(n) = \{n\}$. Here you have $\bigcup\limits_{k=1}^\infty T(n)=\Bbb N$ as well and you have $\bigcup\limits_{k=1}^n T(n) = S(n)$, but here you actually don't have any useful meaning you can give to $\lim\limits_{n\to\infty} T(n)$. You could give a useful meaning to $\lim\limits_{n\to\infty} S(n)$ however. $\endgroup$
    – JMoravitz
    Oct 11, 2018 at 18:42
  • $\begingroup$ Read about Set-theoretic limits on wikipedia. $\endgroup$
    – JMoravitz
    Oct 11, 2018 at 18:43
  • $\begingroup$ @JMoravitz, I've had a look on that. The problem is that this format is used often in Stein's real analysis without those liminf and limsup stuff. Thanks for your insight though. $\endgroup$
    – asmani
    Oct 11, 2018 at 18:48

1 Answer 1


Your last formula is absolutely meaningful! In fact for any natural number $n$ we have$$n\in S(m)\qquad,\qquad m\ge n$$so we may conclude that$$\lim_{n\to \infty}S(n)$$contains all positive integers i.e. $$\lim_{n\to \infty}S(n)=\Bbb N$$


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