Let $A$ be an algebra over $[0,1]$, whose operations are all unary monotone (increasing or decreasing) bijections, except that $A$ also includes the infimum operation over finite or countably many arguments. Let $\Phi$ be a system of equations over $A$, where the left hand side of each equation is an isolated variable, and no variable occurs as the left hand side of two distinct equations. Must $\Phi$ have a solution?
This problem can be viewed as a fixed point problem, since we can consider $\Phi$ as defining a function $\Psi : [0,1]^V \rightarrow [0,1]^V$ ($V$ the set of variable names occurring in $\Phi$), where whenever $(p = \phi) \in \Phi$ and $I \in [0,1]^V$, $\Phi(I)(p) = I(\phi)$ (extending $I$ homomorphically to give values to formulas in the algebra). Also say that if $p \in V$ doesn't occur as the LHS of any equation, let $\Phi(I)(p) = I(p)$. A fixed point of $\Psi$ clearly gives a solution of $\Phi$.
In the case where $\Phi$ is finite or uses only finite infima, a solution exists, because in that case $\Psi$ is continuous and we can appeal to the Brouwer fixed point theorem. In the case where $\Phi$ is infinite and contains infinite infima things are harder, because in infinite dimensions the infimum function is not continuous, and so in general $\Psi$ isn't either. The question is whether the existence of solutions lifts to this infinite case. I'm also interested in just the case where $\Phi$ is countable if it makes a difference.
Context. This problem arises for me in paraconsistent set theory. I am trying to construct a model of a naive set theory in a fuzzy logic. This set theory is not covered by the usual Kripke construction, because it includes a connective $\circ\ a := a^2$, intended to be understood as a consistentizing operator. (Repeated application of this operator progressively approximates saying "$a$ has truth value 1.") The problem with this operator is that it is not monotonic on the information order, so the Kripke construction doesn't yield a fixed point. I am happy to provide further details if desired.