Any way to prove Monotone Continuity?

Question

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Let $A \succeq B$ be defined as $A$ is at least as probable as $B$. $\succeq$ obeys the following axioms. For any $A, B, C$,

Non-negativity: $A \succeq \emptyset$

Additivity: If $A\cap C=B\cap C=\emptyset$, then $A\succeq B$ if and only if $A\cup C \succeq B\cup C$

Transitivity: If $A \succeq B$ and $B\succeq C$, then $A\succeq C$

Is there any way I can prove Monotone Continuity from these axioms?

Monotone Continuity: For $A_1,A_2,A_3,...$ and $B$, if

(i) $A_i \subseteq A_{i+1}$, (ii) $A=\bigcup_i A_i$, (iii) $A\cap B=\emptyset$, and (iv) $B\succeq A_i$, then $B\succeq A$.

Answer 1

The answer is no. E.g., let $X:=\{1,2,\dots\}$ with $P(A):=\sum_{x\in A}2^{-x}$ for all $A\in2^X$. For any $A$ and $B$ in $2^X$, let $A \succeq B$ be defined as ($P(A)>P(B)$ or $A=B$). Then the relation $\succeq$ satisfies all your axioms.

However, it does not have the monotone continuity property. E.g., let $B:=\{1\}$ and $A_i:=\{2,\dots,i+1\}$ for $i=1,2,\dots$. Then $A_i \subseteq A_{i+1}$ and $B\succeq A_i$ for all $i$, and $A\cap B=\emptyset$ for $A:=\bigcup_i A_i$. Yet, $B\not\succeq A$.