Minimal Constructible Sets

Clearly if $A\in\mathcal{U}_{n}$ then $A=A\cup A,$ hence $A\in\mathcal{U}_{n+1}.$ The second part is obvious as anything that can be constructed from  $\mathcal{H}$  in $n$ steps can also be constructed from  $\mathcal{U}$  in $n$ steps because  $\mathcal{H}$  is part of  $\mathcal{U}$, hence  $\mathcal{H}_{n}\subseteq\mathcal{U}_{n}$. Taking unions in this collection of sets over all $n$ gives  $\mathcal{H}_{\infty}\subseteq\mathcal{U}_{\infty}.$ Finally, applying the second part to the collections  $\mathcal{U}\subseteq\mathcal{U}\cup\mathcal{H},\ \mathcal{H}\subseteq\mathcal{U}\cup\mathcal{H}$      $\mathcal{H}\cap\mathcal{U}\subseteq\mathcal{H},\ \,\mathcal{H}\cap\mathcal{U}\subseteq\mathcal{U}$ and taking the corresponding operation gives the desired result. ∎

If  $\mathcal{U}$ has $N$ elements, to build $\mathcal{U}_{1}$ we need $\binom{N}{2}+N$ unions (of two elements in  $\mathcal{U}$), $\binom{N}{2}$ intersections and $N$ complements. This gives a total of at most $2N+2\binom{N}{2}=N^{2}+N$ elements in $\mathcal{U}_{1}.$ An inductive argument shows that $\mathcal{U}_{n}$ is finite. Now, consider $n_{0}$ with $|\mathcal{U}_{{n_{0}}}|=|\mathcal{U}_{{n_{0}+1}}|$. By Lemma 2, $\mathcal{U}_{{n_{0}}}\subseteq\mathcal{U}_{{n_{0}+1}}.$ Then,  $\mathcal{U}_{{n_{0}}}=\mathcal{U}_{{n_{0}+1}}$ and hence for all $n\geq n_{0},\ \mathcal{U}_{n}=\mathcal{U}_{{n_{0}}},$ hence  $\mathcal{U}_{\infty}=\mathcal{U}_{{n_{0}}}$. By Lemma 1,  $\mathcal{U}_{\infty}$ is an algebra of sets. ∎

$C_{1}\left(\mathcal{U}_{\infty}\right)=\mathcal{U}_{\infty}$ comes from Lemma 3 and Lemma 1. If  $\mathcal{U}$ is finite, then $\mathcal{U}_{\infty}$ is also finite with at most $2^{|X|}$ elements. Therefore any countable (or arbitrary) union of elements in $\mathcal{U}_{\infty}$ really becomes a finite union of elements in $\mathcal{U}_{\infty}$, therefore $\mathcal{U}_{\infty}$ is also closed under countable (or arbitrary) unions, thus $\mathcal{U}_{\infty}$ is also a sigma algebra and a topology for $X.$ ∎

We will only prove the first part by induction on $n$, the other part is analogous.  When $n=1$, it clear that

where $\hat{G}_{1}=G_{1},\ \hat{F}_{1}=\emptyset,\ \hat{G}_{2}=\emptyset$ and $\hat{F}_{2}=F_{1}$. Assume now that the results holds for $n=k$ and consider the case $n=k+1.$ Using our induction hypothesis, we conclude that there exists $\hat{G}_{1},...,\hat{G}_{m}$ open sets and $\hat{F}_{1},...,\hat{F}_{m}$ closed sets such that

which is overall an intersection of sets of the form $G\cup F$ where $G$ is an open set and $F$ is a closed set, as we wished. ∎

This is an immediate consequence of Baire’s Theorem as $A_{n}$ being nowhere dense makes $\overline{A_{n}}$ nowhere dense too and hence $\mathbb{R}$ would be a set of the first category, a contradiction according to Baire’s Theorem. ∎

Assume otherwise. Since $\mathbb{Q}\in\mathcal{U}_{\infty},$ there is $n\geq 1$ such that $\mathbb{Q}\in\mathcal{U}_{n}.$ By Lemma 8, there exists $G_{1},...,G_{N}$ open sets and $F_{1},...,F_{N}$ closed sets such that $\mathbb{Q}=\bigcup_{k=1}^{N}\left(G_{k}\cap F_{k}\right)$. Since $\mathbb{Q}$ is a dense subset of $\mathbb{R}$ we must have

From Lemma 9, it follows that one of the members of this union must not be nowhere dense. Therefore, there exists $k\geq 1$ such that $\overline{G_{k}\cap F_{k}}$ has nonempty interior; hence there is a non-empty open interval $(a,b)$ such that $\left(a,b\right)\subseteq\overline{G_{k}\cap F_{k}}$. This implies that $(a,b)\subseteq\overline{F_{k}}=F_{k}$ and also $\left(a,b\right)\subseteq\overline{G_{k}}$. Since $G$ is open in the usual topology, there are $(a_{n},b_{n})$ such that $G_{k}=\bigcup\limits_{n=1}^{\infty}(a_{n},b_{n})$. Since $(a,b)\subseteq\overline{\bigcup\limits_{n=1}^{\infty}(a_{n},b_{n})},$ there is $n$ with $(a,b)\cap(a_{n},b_{n})\neq\emptyset.$ Therefore  $G_{k}\cap(a,b)$ is non-empty. Hence $G_{k}\cap(a,b)\subseteq G_{k}\cap F_{k}\subseteq\bigcup\limits_{k=1}^{n}\left(G_{k}\cap F_{k}\right)=\mathbb{Q}$. But $\mathbb{Q}$ does not have non-empty open subsets, a contradiction. We therefore conclude that $\mathbb{Q}\notin\mathcal{U}_{\infty}$. ∎

Clearly for any real number $x$ we have that $\{x\}=(-\infty,x)^{c}\cap(x,\infty)^{c}$, hence $\{x\}$ is constructible (in fact it is 2-constructible). Now if $A$ has the form 4.1, it can be constructed in at most $n\vee(|P|+2)$ steps. We will prove the converse by induction, more precisely, we will prove that if $A\in\mathcal{U}_{{m}}$ then $A$ has the special form 4.1.

1. 1.

   When $m=1$ we have two cases. If $A=(a,b),\,(a,b)\cup(c,d)\,$ or $\,(a,b)\cap(c,d),\ \,A$ can be expressed as the union of disjoint open intervals (including $(a,\infty)$ or $(-\infty,b)$), in which case $A$ has the form 4.1 with $P=\emptyset$. If $A={(a,b)}^{\mathsf{c}}$ with $a,b\in\mathbb{R}$, then $A=\{a,b\}\cup(-\infty,a)\cup(b,\infty)$; if $a=-\infty$, then $A={(a,b)}^{\mathsf{c}}=\{b\}\cup(b,\infty)$, and if $b=+\infty$ then $A={(a,b)}^{\mathsf{c}}=\{a\}\cup(-\infty,a)$. The case $a=-\infty$ and $b=\infty$ reduces to $A=\mathbb{R}$, hence $A^{c}=\emptyset$, which certainly has the form 4.1 with $P=\emptyset$ and $n=0$. In all the cases, $A$ has the form 4.1.

2. 2.

   Suppose that statement holds for $m=k.$

3. 3.

   Consider $A\in\mathcal{U}_{{m+1}}.$ By induction hypotheses we can assume that $A$ is the union of two elements of the form 4.1 (case 1), the intersection of two elements of such form (case 2) or the complement of an element of such form (case 3). In case 1, it is clear the union of two elements of the form 4.1 has the same form, we just need to rewrite the union of the two disjoint unions as a huge disjoint union and adjust the singletons. In the second case, assume that

   $\displaystyle A=\left(P\cup\bigcup_{i=1}^{n}(a_{i},b_{i})\right)\cap\left(P^{\prime}\cup\bigcup_{i=1}^{r}(a^{\prime}_{i},b^{\prime}_{i})\right),$

   (4.2)

   which is the following union of singletons

   $$\left[P\cap\left(P^{\prime}\cup\bigcup_{i=1}^{r}(a^{\prime}_{i},b^{\prime}_{i})\right)\right]\cup\left[\left(\bigcup_{i=1}^{n}(a_{i},b_{i})\right)\cap P^{\prime}\right]$$

   and

   $$\left(\bigcup_{i=1}^{n}(a_{i},b_{i})\right)\cap\left(\bigcup_{i=1}^{r}(a^{\prime}_{i},b^{\prime}_{i})\right).$$

   Here we use the fact that the intersection of two finite unions of open intervals is again a disjoint union of open intervals. Making the correct singletons-adjustment we conclude that the set in Equation 4.2 has the form 4.1. Finally, in case 3

   $\displaystyle A={\left(P\cup\bigcup_{i=1}^{n}(a_{i},b_{i})\right)}^{\mathsf{c}}={P}^{\mathsf{c}}\cap\bigcap_{i=1}^{n}{(a_{i},b_{i})}^{\mathsf{c}}.$

   (4.3)

Here ${P}^{\mathsf{c}}$ is clearly a union of open intervals (hence it is of the form 4.1). Let’s assume without lost of generality that $-\infty<a_{1}<b_{1}<a_{2}<b_{2}<\cdots<a_{n}<b_{n}<\infty$. Hence

which is of the form 4.1. Therefore $A$ is the intersection of two sets of the form 4.1 and by case 2, $A$ has the form 4.1. ∎

Clearly,  $\mathcal{U}_{\infty}$  is finite. Let $a\in X$ and consider $f(a)=\min\{|A|\,:\,A\in\mathcal{U}_{\infty},\,a\in A\}$. We claim that $f(a)=1.$ Suppose that $f(a)>1$ and take $A$ a set where this minimum is achieved, that is, $A\in\mathcal{U}_{\infty}$ with $a\in A$ and $f(a)=|A|.$ Since $|A|>1$, there exist $b\in A$ such that $b\neq a.$ Since $X$ is separable in  $\mathcal{U}$  there exists $B\in\mathcal{U}$ such that $a\in B,\,b\in{B}^{\mathsf{c}}$ or $a\in{B}^{\mathsf{c}},\,b\in B.$ Define

Then clearly $C\in\mathcal{U}_{\infty},\,a\in C$ and $|C|<|A|$ as $b\in A-C,$ contradicting the selection of $A.$ Since $f(a)=1$ and $a$ was arbitrary, we conclude that  $\mathcal{U}_{\infty}$  contains all singletons, hence it contains all possible unions of such and therefore  $\mathcal{U}_{\infty}$ consists of all possible subsets of $X.$ ∎

Clearly $a\sim a$ as there is not $A\in\mathcal{U}$ such that $a\in A$ and $a\notin A.$ From the definition we can see that if $a\sim b$ then $b\sim a$. Now suppose that $a\sim b,\,b\sim c$ and suppose that $a\not\sim c$ hence $a,c$ are separable. Then there is $A\in\mathcal{U}$ such that either $a\in A,\,c\notin A$ or $a\notin A,\,c\in A$. If $a\in A,\,c\notin A$, since $a$ and $b$ are unseparable, $a,b\in A$; since $b$ and $c$ are also unseparable, $b,c\in A$; Therefore $a,b,c\in A,$ a contradiction. The other case $a\notin A,\,c\in A$ is treated in the same manner. ∎

Clearly, if $A$ has the form as in Equation 5.1, then it can be constructed by unions and is thus an element of  $\mathcal{U}_{\infty}$. By Lemma 6, $\mathcal{U}_{\infty}$ is an algebra and a sigma algebra. It is well known ([2], P.2.21) that if $A\in\mathcal{U}_{\infty}$ then $A$ is a union of some elements in the partition, hence $A$ satisfies Equation 5.1 for some $\Lambda$ subset of $\{1,2,...,n\}.$ To find $|\mathcal{U}_{\infty}|$, note that every set $A\in\mathcal{U}_{\infty}$ can be written as unions of $k$ sets in $\mathcal{U}$ and $k\in\{0,1,...,n\}$. The number of ways to select such possible sets is given by $\binom{n}{0}+\binom{n}{1}+\mathellipsis\binom{n}{n}=2^{n}.$

Finally, suppose that $a,b$ are unseparable in  $\mathcal{U}$. If $a,b$ were separable in  $\mathcal{U}_{\infty}$, without loss of generality, there would exists a set $A$ in  $\mathcal{U}_{\infty}$ such that $a\in A$ and $b\notin A.$ Since $A$ satisfies Equation 5.1 for some $\Lambda$ subset of $\{1,2,...,n\},$ there would exist only one $k\in\Lambda$ such that $a\in A_{k},b\notin A_{k},$ contradicting the fact that $a,b$ are unseparable in  $\mathcal{U}.$ Conversely, if $a,b$ are separable in  $\mathcal{U}$, since  $\mathcal{U}\subseteq\mathcal{U}_{\infty}$, by definition of separability, $a,b$ are necessarily separable in  $\mathcal{U}_{\infty},$ the converse of the last statement finishes the proof of this lemma. ∎

Take $b\in[a]$ and $A\in\mathcal{A}$ such that $a\in A.$ If $b\notin A$, then $a$ and $b$ are separable, contradicting the fact that $a\sim b,$ therefore $b\in A$ and hence necessarily $b$ belongs to the right hand side of Equation 5.2. Conversely, take an element $b$ in the right hand side of Equation 5.2. If $b$ and $a$ were separable then there would exist a set $H\in\mathcal{A}$ such that either $a\in H,\,b\notin H$ or $a\notin H,\,b\in H.$ In either case, by considering the possibility ${H}^{\mathsf{c}}\in\mathcal{A}$, we conclude that there exists a set $A\in\mathcal{A}$ such that $a\in A,\,b\notin A,$ contradicting the fact that $b\in\bigcap_{A\in\mathcal{A},a\in A}A.$ Therefore necessarily $b$ and $a$ are unseparable, hence $b\in[a].$   For an example where Equation 5.2 does not necessary hold, consider $X=\{1,2,3,4\},\,\mathcal{U}=\{\{1\},\{1,2\},\{1,2,3,4\}\}$. The equivalence classes are $[1]=\{1\},\,[2]=\{2\}$ and $[3]=[4]=\{3,4\}.$ In this example $\bigcap_{A\in\mathcal{U}\,:\,3\in A}A=\{1,2,3,4\}\neq[3].$ ∎

Consider $\mathcal{P}=\{A_{1},...,A_{n}\}$ be the partition induced by the relationship in Lemma 10 and take $E\in\mathcal{A}.$ Clearly

Take now $k\leq n$ such that $E\cap A_{k}\neq\emptyset.$ Consider any $a\in A_{k}.$ Since $E\cap A_{k}\neq\emptyset$, there exists $b\in E\cap A_{k}.$ Since $b\in A_{k},\,a\in A_{k}$ and $A_{k}$ is an equivalence class, we conclude that $a\sim b.$ By Lemma 12 we have

Since $a\in[a],$ Equation 5.3 implies that $a\in\bigcap_{B\in\mathcal{A},b\in B}B.$ Therefore $a\in E$ as $E\in\mathcal{A}$ and $b\in E.$   Since $a$ was arbitrary in $A_{k}$ we have that $A_{k}\subseteq E\cap A_{k}.$ Since the other containment is always true, we conclude that $A_{k}=E\cap A_{k}$, therefore

Given that each $A_{k}\in\mathcal{P}$, we conclude that $E\in\mathcal{P}_{\infty},$ hence  $\mathcal{A}\subseteq\mathcal{P}_{\infty}.$ By Lemma 11,  $\mathcal{P}_{\infty}$ is finite, therefore  $\mathcal{A}$  is finite. Conversely, take $A_{k}\in\mathcal{P}.$ Since  $\mathcal{A}$  is finite and $A_{k}$ is an equivalence class, Lemma 12 implies that $A_{k}$ is a finite union of elements of  $\mathcal{A}$, thus $A_{k}\in\mathcal{A}$, therefore  $\mathcal{P}\subseteq\mathcal{A}.$ By Lemma 4, $\mathcal{P}_{\infty}\subseteq\mathcal{A}.$ Hence $\mathcal{P}_{\infty}=\mathcal{A}.$ ∎