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Our set is defined as $$A_n :=\left]0,\tfrac{1}{n}\right[$$Where n∈$N$. What is? $$\bigcap_{n+1}^\infty A_n$$

Solution : Let’s examine $A_1,\,A_2,\,A_3$:

$$A_1:=\left]0,1\right[$$ $$A_2:=\left]0,\tfrac{1}{2}\right[$$$$A_3:=\left]0,\tfrac{1}{3}\right[$$ It's clear that $A_{n+1}\subset A_n$. Let's define this multiple intersection through an index set $I$:$$\bigcap_{i\in I}^\infty A_i:=\{x∣\forall i\in I:x\in A_{i+1}\to x\in A_n\}, I\subseteq \Bbb N,$$

where $\Bbb N$ is the set of natural numbers.

J.G.
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In the blind
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1 Answers1

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Just to expand on @KaviRamaMurthy's comment: for any $x>0$, there exists $n\in\Bbb N$ with $\frac{1}{n}<x$, so $x\notin A_n$. Hence $\bigcap_nA_n=\emptyset$.

J.G.
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  • Is this the same as saying "The only common element across all these sets is the empty set"? – In the blind Oct 11 '19 at 07:49
  • @juhani no, the empty set is not an element of any of the intervals. What it says is that there is no common element across all the sets. – Vsotvep Oct 11 '19 at 07:51
  • @juhani No, it's saying their common elements are the elements of the empty set, i.e. none at all. – J.G. Oct 11 '19 at 07:51
  • I'm still confused. Looking at these sets, we could say we are constantly taking a smaller and smaller interval on the reals. Doesn't that mean there is always at least some element which is common across all these sets, i.e " the smallest possible interval is the common element" – In the blind Oct 11 '19 at 07:58
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    @juhani $\bigcup_{n=1}^kA_n$ is nonempty for any finite $k$, but $\bigcup_{n=1}^\infty A_n$ is empty. – J.G. Oct 11 '19 at 07:59