# Disprove counterexample for $\limsup A_n = \emptyset$

Let $(A_n)_{n=1}^{\infty}$ be a pairwise disjoint collection.

$\lim A_n = \emptyset$? (see here and there)

What about a set of extended real numbers $A_n=(n,n+1]$?

It seems that

$\limsup A_n = \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} A_n$

$= \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} (n, n+1]$

$= \bigcap_{k=1}^{\infty} ((k, k+1] \cup (k+1, k+2] \cup ...)$

$= \bigcap_{k=1}^{\infty} ((k, \infty])$

... = {$\infty$} ?

• Note that $\infty\not\in (n,n+1]$ for every $n\in\mathbb{N}$, so $\infty\not\in \bigcup_{n=k}^\infty (n,n+1]$. Jul 11, 2014 at 5:11
• Note that $\infty$ is not an element of $\cup_{n=k}^\infty(n,n+1]$ for any $k$. Jul 11, 2014 at 5:11
• @Brandon Are you saying that the penultimate step is incorrect?
– BCLC
Jul 11, 2014 at 5:15
• @Chellapillai Are you saying that the penultimate step is incorrect?
– BCLC
Jul 11, 2014 at 5:17
• @BCLC: Yes.${}$ Jul 11, 2014 at 5:20

$$\infty$$ is not in those unions (i.e., it is not in $$\bigcup\limits_{n=k}^\infty(n,n+1]$$ for any $$k$$), because it is not in any of the individual sets being unioned (i.e., it is not in $$(n,n+1]$$ for any $$n$$).
• @BCLC: Yes. In that case, every positive number is eventually in the sets, but $0$ is not in any of them, so it is not in their union. Jul 11, 2014 at 5:50