For any set $A$, let $T(A)$ be the set consisting of all sets $S \subseteq \mathcal{P}(A)$ that satisfy the following conditions:

  • (i) $\emptyset \in S$
  • (ii) $A \in S$
  • (iii) $\forall X, Y \in S, (X \cup Y) \in S$
  • (iv) $\forall X, Y \in S, (X \cap Y) \in S$

Prove that:

$\forall S_1, S_2 \subseteq \mathcal{P}(\mathbb{R}), [(S_1 \in T(\mathbb{R})) \land (S_2 \in T(\mathbb{R}))] \implies [(S_1 \cup S_2) \in T(\mathbb{R})]$.


First, I noticed that $T$ is sort of like a power set of power sets except with these extra conditions.

I tried starting with property (iii) and using it on both conditions on the left hand side of the implication but reached a dead end?

  • 2
    $\begingroup$ Are you sure you mean unions? Because the statement is false. Take $S_1 = \{\varnothing, \{0\}, \mathbb{R}\}$ and $S_2 = \{\varnothing,\{1\},\mathbb{R}\}$. Then $S_1$ and $S_2$ are both in $T(\mathbb{R})$, but their union is not, because the union contains $\{0\}$ and $\{1\}$ but not $\{0\}\cup\{1\}$, so it fails (iii). Now, on the other hand, if you were trying to prove that $S_1\cap S_2\in T(\mathbb{R})$.... $\endgroup$ Nov 11 '19 at 2:39
  • $\begingroup$ Oh, thank you very much! $\endgroup$
    – Zexion12
    Nov 11 '19 at 2:43
  • 1
    $\begingroup$ What went through my mind is that something was amiss; because if you take one set from $S_1$ and one set from $S_2$, there is absolutely no reason to expect their union to be an element $S_1\cup S_2$. At that point, you want to see if you can construct a counterexample. $\endgroup$ Nov 11 '19 at 3:14
  • 2
    $\begingroup$ Here’s what my advisor, George Bergman, likes to say: try really hard to prove it; if you run into a dead end, try to see what is going wrong, and try to use it to construct a counterexample. Try really hard to build the counterexample. If you can’t, then try to figure out what is going wrong with your construction, to see if that gives you a clue to complete the proof. And so on. Try working on the problem from both ends, but not just pushing, but if you can’t push it through, take a step back and see if you can figure out why you can’t push it through. $\endgroup$ Nov 11 '19 at 3:16
  • 1
    $\begingroup$ To add on what @Arturo wrote, here is my take on how to solve your problems. $\endgroup$
    – Asaf Karagila
    Nov 11 '19 at 4:10

A true proposition is
$\forall S_1, S_2 \subseteq \mathcal{P}(\mathbb{R}), [(S_1 \in T(\mathbb{R})) \land (S_2 \in T(\mathbb{R}))] \implies [(S_1 \cap S_2) \in T(\mathbb{R})]$.


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