I've proved the following theorem using model-theoretic techniques, namely ultraproducts: A continuum is locally connected if every semi-monotone mapping onto it (from another continuum) is monotone. Monotone means the usual thing; semi-monotone means that every subcontinuum K of the range space is the image of a subcontinuum in the domain space, which contains the pre-image of the interior of K. The part that uses ultraproducts is where we want to prove that non-locally connected implies being the image under a semi-monotone map that isn't monotone. Basically, I'm wondering if someone has any insights into obtaining a new proof more palatable to a continuum theorist. (E.g.: start with a non-locally connected metric continuum Y and directly construct a metric continuum X and a semi-monotone f:X->Y which is not monotone.)
There are many incarnations of what is known as a continuum, e.g.,
1. continuum in set theory: the set R.
2. coninuum in topology: Hausdorff space. Distinct points belong to disjoint neighbourhoods.
I can suggest an approach in terms of a Hausdorff space. Let x and y be distinct points in a metric Hausdorff space X endowed with an Efremovic proximity relation. Let Nx, Ny be disjoint neighbourhoods of x, y, respective, where, for example,
\[
N_x = \left\{y\in X: d(x,y) < \varepsilon\right\},
\]
with $\varepsilon > 0$ and $d$ is the standard distance between x and y. Then Nx and Ny are connected, provided
\[
\mbox{cl}(A) \cap \mbox{cl}(B) \neq \emptyset.
\]
In other words, cl(Nx) and cl(Ny) have at least one point in common and cl(Nx) and cl(Ny) are then adjacent to each other. In other words, Nx and Ny are disjoint but cl(Nx) and cl(Ny) are not disjoint. If every pair of closures of disjoint neighbourhoods is connected in a subspace of X, then the subspace is locally connected.
One question to consider is this: what if Nx = {x} and Ny = {y}. Does the above line of reasoning work?
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