Equivalent conditions for set-valued function

Question

Let $f:\mathcal P(X) \rightarrow\mathcal P(X)$. Suppose that

(B1) $f(\emptyset) = \emptyset$

(B2) $f(A) = f(A^\complement)$

(B3) $f(A) \subseteq B\cup f(B)$ if $A\subseteq B$

(B4) $f(f(A))\subseteq f(A)$

(B5) $f(A\cup B)\subseteq f(A)\cup f(B)$

for $A,B\in\mathcal P(X)$.

Proposition 1. Suppose that (B4) and (B5) hold. Then

(X4) $f(A\cup f(A)) \subseteq A\cup f(A)$

follows.

Proposition 2. Suppose that (B2), (B3), and (X4) hold. Then (B4) follows

Proposition 1 and Proposition 2 tell me that I can replace (B4) with (X4) and vice versa. My goal is now to prove Proposition 3 and Proposition 4:

Proposition 3. Suppose that (B1), (B2), (B3), (B4), and (B5) hold. Then

(X5) $A\cap B\cap f(A\cap B) = A\cap B\cap (f(A)\cup f(B))$

follows.

Proposition 4. Suppose that (B1), (B2), (B4) and (X5) hold. Then (B3) and (B5) follow.

By (B2) and (B5) it follows that \begin{align*} f(A \cap B) &= f(A^\complement \cup B^\complement) \\ &\subseteq f(A^\complement) \cup f(B^\complement) \\ &= f(A) \cup f(B). \end{align*} So the $\subseteq$-inclusion for Proposition 3 is easy to show. However, I failed to prove the $\supseteq$-inclusion. For Proposition 4 I was unable to prove either inclusion. I hope that someone here has an idea.

This is a follow-up question of my earlier question Axiomatizations of the boundary operator. Oliver Diaz answer shows that $f$, given axioms (B1), (B2), (B3), (B4), and (B5), yields the same topology as $f$ given the axioms (B1), (B2), (B4), and (X5). Note that $f$ here is the boundary operator $\partial$; see my other question for more details. Now I want to show that both axiomatizations not only yield the same topology but that I can replace (B3) and (B5) with (X5) (without refering to the topology in the proof). Is this possibe? Or do I have to take the "detour" via the topology in the proof? The lemma stated in Wang's master's thesis gives me the feeling that the additional condition $A\cap f(A) = \emptyset$ is necessary, which suggests using the "detour". However, these are only guesses. I hope someone here has a definite answer.

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Proof of Proposition 1. Let $A\in\mathcal P(X)$. By (B5) and (B4) it follows that \begin{align*} f(A\cup f(A)) &\subseteq f(A) \cup f(f(A)) \\ &\subseteq f(A) \cup f(A) \\ &= f(A).\end{align*} Since $f(A) \subseteq A\cup f(A)$, it follows that $f(A\cup f(A)) \subseteq A\cup f(A)$.

$\square$

Proof of Proposition 2. Trivially, it holds that $f(A)\subseteq A\cup f(A)$. By (B3) and (X4) it follows that \begin{align*} f(f(A)) &\subseteq f(A)\cup f(A)\cup f(A\cup f(A)) \\ &\subseteq f(A) \cup f(A)\cup A\cup f(A) \\ &= A\cup f(A). \end{align*} Likewise, \begin{align*} f(f(A^\complement)) &\subseteq f(A^\complement)\cup f(A^\complement)\cup f(A^\complement\cup f(A^\complement)) \\ &\subseteq f(A^\complement)\cup f(A^\complement)\cup f(A^\complement) \cup A^\complement \\ &=A^\complement\cup f(A^\complement). \end{align*} By (B2) it follows that \begin{align*} f(f(A)) &\subseteq (A\cup f(A)) \cap (A^\complement \cup f(A^\complement)) \\ &= (A\cap f(A))\cup A^\complement\cap f(A)) \\ &= (A\cap A^\complement)\cup (A\cap f(A)) \cup (A^\complement\cap f(A)) \cup (f(A)\cap f(A)) \\ &\subseteq \emptyset \cup f(A) \cup f(A) \cup f(A) \\ &= f(A). \end{align*}

$\square$

Answer 1

Summary of solution : When the inclusions you have do not go in the direction you want (e. g. when the wished inclusion seems to say that something big is included in something small), use rule $(B_2):f(A^c)=f(A)$ to "transform" big sets into small ones and vice-versa.

Proof of proposition 4. Let $A,B$ be arbitrary in ${\cal P}(X)$. Let $U=A\cap B\cap f(A\cap B)$ and $V=A\cap B\cap (f(A)\cup f(B))$. You already showed that $U\subseteq V$.

Conversely, taking $(A^c,A^c \cup B^c)$ in place of $(A,B)$ in $(B_3)$, we obtain $f(A) \subseteq A^c \cup B^c \cup f(A\cap B)$, or

$$f(A)\cap A \cap B \subseteq f(A\cap B). \tag{1}$$

By symmetry, we similarly have

$$f(B)\cap A \cap B \subseteq f(A\cap B). \tag{2}$$

Combining (1) with (2), we obtain $V \subseteq f(A\cap B)$ and hence $V\subseteq U$ as wished. This finishes the proof of proposition 4.

Proof of proposition 5. First, let $A\subseteq B$. Taking $(A^{c},B^{c})$ in place of $(A,B)$ in $(X_5)$, we obtain $A^c \cap B^c \cap f(A^c \cap B^c)=A^c \cap B^c \cap (f(A^c) \cup f(B^c))$. But since $A\subseteq B$, we have $B^c \subseteq A^c$ and hence $A^c \cap B^c=B^c$. We deduce that $B^c \cap f(A^c \cap B^c)= B^c \cap (f(A^c) \cup f(B^c))$, or equivalently $B^c \cap f(A \cup B)= B^c \cap (f(A) \cup f(B))$. Now $A\subseteq B$, so $B^c \cap f(B)= B^c \cap (f(A) \cup f(B))$, or (using the distributivity of $\cap$ over $\cup$), $B^c \cap f(B)= (B^c \cap f(A)) \cup (B^c \cap f(B))$. If we put $P=B^c \cap f(A)$ and $Q=B^c \cap f(B)$, we can rewrite this as $Q=P \cup Q$, or equivalently $P \subseteq Q$. Hence

$$f(A)=(B^c \cup B)\cap f(A)=P\cup(B \cap f(A)) \subseteq Q\cup(B \cap f(A)) \subseteq f(B) \cup B,$$ where the last inclusion follows from $Q\subseteq f(B)$ and $B \cap f(A) \subseteq B$. We have thus shown $(B_3)$.

Let us now proceed to the proof of $(B_5)$ ; let $A,B$ be arbitrary in ${\cal P}(X)$. We wish to show that $L\subseteq M$ where $L=f(A\cup B)$ and $M=f(A)\cup f(B)$.

Taking $(A,A\cup B)$ in place of $(A,B)$ in $(X_5)$, we obtain $A \cap f(A)=A \cap (f(A) \cup f(A\cup B))$. It follows that $L\cap A \subseteq M\cap A$ ; by symmetry, we similarly have $L\cap B \subseteq M\cap B$. So :

$$L\cap A \subseteq M, L\cap B \subseteq M. \tag{3}$$

Finally, taking $(A^c,B^c)$ in place of $(A,B)$ in $(X_5)$, we obtain $A^c\cap B^c \cap f(A\cup B)=A^c\cap B^c \cap (f(A) \cup f(B))$. It follows that

$$L\cap A^c\cap B^c \subseteq M \tag{4}$$

Combining (3) with (4), we obtain $L\subseteq M$ as wished. This finishes the proof of proposition 5.