Notation
Let
- $(X, \mathcal{T})$ be a topological space.
- $P(\mathcal{T}) = \{U \subset X : (U, \mathcal{T}|U) \text{ is locally connected}\}$, where $\mathcal{T}|U$ is the subspace topology on $U$.
- $\mathcal{T}^* = \{U \subset X : \forall C \in P(\mathcal{T}): U \cap C \in \mathcal{T}|C\}$, called here a coherence-refinement of $\mathcal{T}$.
Prove or disprove
Coherence-refinement is increasing. That is, $\mathcal{S}^* \subset \mathcal{T}^*$ for any topologies $\mathcal{S}$ and $\mathcal{T}$ on $X$ such that $\mathcal{S} \subset \mathcal{T}$.
Background
The following can be proved about coherence-refinement:
- $\mathcal{T}^*$ is a topology on $X$.
- $\mathcal{T} \subset \mathcal{T}^*$.
- $\mathcal{T}^*$ is locally connected.
- $(\mathcal{T}^*)^* = \mathcal{T}^*$.
- $\mathcal{T}^*|C = \mathcal{T}|C$ for each $C \in P(\mathcal{T})$.
- if $\mathcal{T}$ is locally connected, then $\mathcal{T}^* = \mathcal{T}$.
- $P(\mathcal{T}) \subset P(\mathcal{T}^*)$.
Consider the following intersection-refinement:
$$\mathcal{T}^+ = \bigcap \{\mathcal{T}^* \text{ is a topology in X} : \mathcal{T} \subset \mathcal{T}^* \land \mathcal{T}^* \text{ is locally connected} \}.$$
It can be shown that intersection-refinement has the same properties as coherence-refinement listed above; see here for local connectedness.
Then $\mathcal{T} \subset \mathcal{T}^+ \subset \mathcal{T}^*$. If one can show that coherence-refinement is increasing, then $\mathcal{T}^* \subset (\mathcal{T}^+)^* \subset (\mathcal{T}^*)^*$ and so $\mathcal{T}^* = \mathcal{T}^+$. On the other hand, intersection-refinement is clearly increasing, so if $\mathcal{T}^* = \mathcal{T}^+$, then $\mathcal{T}^*$ is increasing. In this sense, the problem can also be interpreted as asking whether $\mathcal{T}^* = \mathcal{T}^+$. This would then answer my previous question about a more convenient way of proving theorems about $\mathcal{T}^+$.
The union result for locally connected subsets I give here may be useful.
