# intersection of intervals from 1 to infinity

This is a continuation of my previous quesion concerned with finding $\cup^\infty_{k=1} S_k$ for $S_k = (1 − 1/k, 2 + 1/k], k \in N$.

And the answer intuitively makes sense, but what about for a relative complement with the real numbers?

$\cap_{k=1}^\infty R$\ $S_k$

The largest interval of $S_k$ is (0, 3]. So the relative complement is all elements of the real numbers not in $S_k$ so it would be $(-\infty, 0] \cup (3, \infty)$, and union is $\cup_{k=1}^\infty R$\ $S_k = (-\infty, 1]\cup[2, \infty)$ since the smallest interval is (1, 2)?

Is that correct or have I made an error? And what's the correct way to formally state this for intervals?

1) The first question is correct: $\cap_{k=1}^\infty R\backslash S_k$ in in fact $(-\infty, 0] \cup (3, \infty)$ and to formally state this there are many ways.

Here's one: Firstly observe that $S_k \supseteq S_{k+1} \ \forall \ k \in \mathbb{N}$ so $S_k \cup S_{k+1} = S_k\$ , even more; $\cup^\infty_{k=1} S_k = S_1 = (0,3]$

So $\cap_{k=1}^\infty R\backslash S_k = (\cup^\infty_{k=1} S_k)^c =(S_1)^c = (-\infty, 0] \cup (3, \infty)$

2) The second answer is wrong. The answer is $\cup_{k=1}^\infty R\backslash S_k = (-\infty, 1)\cup (2, \infty)$

Observe that $(S_k)^c=R\backslash S_k = (-\infty,1-1/k] \cup (2+1/k,+\infty)$ and $x\in \cup A_k \Leftrightarrow \exists \ k\ \ \text{so that} \ x\in A_k$

So $1\in \cup_{k=1}^\infty (S_k)^c \Leftrightarrow \exists \ k \ \ \text{so that} \ 1\in (S_k)^c=(-\infty,1-1/k] \cup(2+1/k,+\infty)$ But this is not the case for any k.

the same goes for 2.