# Clues for this Set Proof?

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:

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

Attempt:

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?

• 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 $S_1\cap S_2\in T(\mathbb{R})$.... Nov 11 '19 at 2:39
• Oh, thank you very much! Nov 11 '19 at 2:43
• 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. Nov 11 '19 at 3:14
• 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. Nov 11 '19 at 3:16
• To add on what @Arturo wrote, here is my take on how to solve your problems. Nov 11 '19 at 4:10

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