I am looking for an example which illustrates that the following doesn't have to be an equality:
$$\bigcup _{i \in I} \bigcap _{k \in K} A_{ik} \subset \bigcap _{k \in K} \bigcup _{i \in I} A_{ik} $$.
It is quite easy to prove the inclusion: $x \in \bigcup _{i \in I} \bigcap _{k \in K} A_{ik} \iff \exists i \in I : \ \forall k \in K \ : \ x \in A_{ik} \implies \forall k \in K: \ \exists i \in I : x \in A_{ik}$
We generally have that $$\exists x : \ \forall y \ : \phi(x, y) \implies \forall y: \ \exists x : \phi(x, y)$$ and here the counterexample for the reverse implication is taking $\phi(x, y)$ to be the relation $x<y, \ x, y \in \mathbb{R}$.
But taking the set $A_{ik} = \{ i \in \mathbb{R} \ : \ i<k\}$ doesn't work here:
$$\bigcap_{k \in K} A_{ik} = \bigcap_{k \in K} \{ i \in \mathbb{R} \ : \ i<k\} = \emptyset \ \ \ \text{as} \ \ \ k \to – \infty.$$
And if we take the union $$\bigcup_{i \in I} A_{ik} = \bigcup_{i \in I} \{ i \in \mathbb{R} \ : \ i<k\} $$ we don't get anything new, we just get $A_{ik}$.
Am I doing something wrong? Could you please point that mistake out for me?
Best Answer
For integers $i, k \ge 0$, let $A_{ik} = [\frac {k+1} {i+k+1}, 1]$.
Then for fixed $i$, $\bigcap_k A_{ik} = \{1\}$, so $$\bigcup_i \bigcap_k A_{ik} = \{1\}. \tag{UI}$$
However, for fixed $k$, $\bigcup_i A_{ik} = (0,1]$, so $$\bigcap_k \bigcup_i A_{ik} = (0,1]. \tag{IU}$$
Thus (UI) is strictly included in (IU).