Let $F$ be a family of sets. It is possible to define a topology $T$ generated by $F$ by letting it the intersection of all topologies containing $F$. If $F$ is known it is also possible to construct $T$ as follows:

(1) add $F$, $\varnothing$ and whole space to $T$

(2) add all finite intersections of sets in (1)

(3) add all unions of sets in (2)

Steps (2) and (3) can't be interchanged: adding unions first and taking intersections afterwards does not yield the topology $T$. I have been trying to prove this by providing a counter example. But I am unsuccessful so far. Here is my work:

Let the whole space $X=\mathbb R$ and assume we want $T$ to be the standard topology. I defined $F$ to be the collection of all intervals $(-\infty,a)$ and $(b,\infty)$ with $a,b \in \mathbb R$. I verified that if the steps are executed in order the result is the standard topology. Now I am stuck in the other case: After adding unions and then taking intersections. For a counter example, a set that is open but not in this collection I considered $(1,2) \cup (3,4)$. I tried to write it as (finite) intersection of unions of $(-\infty,a)$ and $(b,\infty)$ but failed. Now it seems this could be the example I am looking for but:

How can I prove that it is not possible to write $(1,2) \cup (3,4)$ as (finite) intersection of unions of $(-\infty,a)$ and $(b,\infty)$?


The unions of sets of the form $(-\infty,a)$ and $(b,+\infty)$ are sets of the forms

  1. $(-\infty, a)$, where $a \in (-\infty,+\infty]$,
  2. $(b,+\infty)$, where $b \in [-\infty,+\infty)$, and
  3. $(-\infty,c) \cup (d,+\infty)$, where $-\infty < c \leqslant d < +\infty$.

Thus $(1,2)\cup (3,4)$ is a finite intersection of such sets:

$$(1,2)\cup (3,4) = (1,4) \cap \bigl((-\infty,2)\cup(3,+\infty)\bigr).$$

You need an open set with infinitely many components to get something you can't write as a finite intersection of unions.

$$A = \bigcup_{k=1}^\infty (2k-1,2k)$$

is not an intersection of finitely many such sets, you need infinitely many. Since $A$ contains arbitrarily large real numbers, all unions of elements of $F$ containing $A$ must have a nonempty part of the form $(d_m,+\infty)$. With $d = \max \{d_m : 1 \leqslant m \leqslant M\}$, the intersection of $M$ such unions always contains a nonempty part of the form $(d,+\infty)$.

  • $\begingroup$ Thank you! Sorry: why do you restrict to only considering sets containing $A$? In the example $(1,4)$ does not contain the set $(1,2)\cup(3,4)$. $\endgroup$ – blue Feb 17 '14 at 14:37
  • $\begingroup$ If we want to write $A = U_1 \cap \dotsc \cap U_M$, then every $U_m$ must contain $A$, otherwise the intersection couldn't contain $A$. In the example, we have $\bigl((1,2)\cup (3,4)\bigr) \subset (1,4)$, so it contains $A$, as it must. $\endgroup$ – Daniel Fischer Feb 17 '14 at 14:40
  • $\begingroup$ Thank you, you are right it is contained in $(1,4)$. Now I understand your proof. $\endgroup$ – blue Feb 17 '14 at 15:04

It is possible.


But I doubt that you can write an infinite union of disjoint open intervals as a finite intersection of sets of the form $(-∞,a)\cup (b,∞)$.


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