Let $X$ be a set and let $\cal A$ be a non-empty subset of $P(X)$ with the property that whenever $A\_1 \subseteq A\_2 \subseteq \cdots $ is an increasing chain of elements of $\cal A$ then $\cup\_i A\_i \in \cal A$. Let 
$f : \cal A\longrightarrow \mathbb{R}$ be increasing and bounded from above. Is it true that there must exists $M\in\cal A$ such that for any element $C$ of $\cal A$ containing $M$, $f(M)= f(C)$ ? 



A weak kind of fixed point
If not, then we can find an increasing and continuous sequence $(A\_\alpha: \alpha < \omega\_1)$ of elements of $\mathcal{A}$ such that $\alpha < \beta \implies f(A\_\alpha) < f(A\_\beta)$. This is a contradiction.




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**Edit**. Let me add more details. By contradiction, the following holds:


**$(\*)$**: For any $M \in \mathcal{A}$, there exists $M \subset C \in \mathcal{A}$ with $f(M) < f(C).$


Let $A\_0\in \mathcal{A}$ be arbitrary. Given $A\_\alpha \in \mathcal{A},$ use **$(\*)$** to find $A\_{\alpha+1} \in \mathcal{A}$ with $A\_\alpha \subset A\_{\alpha+1}$ and $f(A\_\alpha) < f(A\_{\alpha+1}).$
Now suppose $\alpha<\omega\_1$ is limit and we have defined $(A\_\beta: \beta< \alpha)$ which is increasing and continuous and for all $\beta\_1 < \beta\_2 < \alpha, f(A\_{\beta\_1}) < f(A\_{\beta\_2}).$ Let $A\_\alpha=\bigcup\_{\beta<\alpha}A\_\beta.$ Note that $A=\bigcup\_{n<\omega}A\_{\beta\_n},$ where $(\beta\_n: n<\omega)$ is increasing cofinal in $\alpha,$ so by assumption $A\_\alpha \in \mathcal{A}.$
It is also clear that for all $\beta<\alpha, f(A\_\beta) < f(A\_\alpha)$ (as $f$ is increasing by assumption).


But this is impossible, as otherwise we can find $q\_\alpha \in \mathbb{Q} \cap (f(A\_\alpha), f(A\_{\alpha+1})), \alpha < \omega\_1,$ and for $\alpha \neq \beta, q\_\alpha \neq q\_\beta,$ so $\mathbb{Q}$ is uncountable which is impossible. 


