Undergrad Is this generalization equivalent to usual Aposyndetic

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The discussion explores the equivalence between two definitions of aposyndetic in connected metric spaces: the standard definition and the newly proposed "zero set aposyndetic." It is established that "zero set aposyndetic" implies the standard definition, but the reverse implication is questioned, with a suggestion that dense, connected subsets could serve as a counterexample. The example proposed involves using the algebraic numbers in a two-dimensional space, specifically considering subsets of X that are dense and connected. A clarification is made that the definition of a continuum should also include compactness, and it is noted that "zero set aposyndetic" applies only to non-intersecting sets. The conversation emphasizes the need for careful definitions and examples in topology.
kmitza
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I want to check if this "new" notion is equivalent to a well known one and I feel like I am missing an obvious counterexample
For some basic definitions we call connected, metric space a continuum and we say that continuum is aposyndetic if for every pair of points p,q exists a subcontinuum W such that $p \in int(W) \subset W \subset X \setminus \{q\}$ similarly I introduce a notion of "zero set aposyndetic" as:
X is aposyndetic if for every two empty interior connected subsets U,V exist W such that $U \subset int(W) \subset W \subset X\setminusV$
I want to check if the two are equivalent as it is obvious that "zero set aposyndetic" implies aposyndetic but I feel intuitively that the other direction might not be true, however I can't see the counterexample nor the proof.
 
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If we could find two dense, connected subsets, that should serve as a counterexample, since the only open set containing either of them will be the whole space X, so it must overlap the other subset.

Taking ##\mathbb A## to represent the algebraic numbers, I think if we take ##X=\mathbb R^2, U = X - \mathbb A^2, V = \mathbb A^2 - \mathbb Q^2## it might work.

Certainly U and V are dense in X. I think they may also both be connected (not path-connected, but that doesn't matter). This doesn't work with ##X=\mathbb R## but I think it works in two dimensions.
 
Hi, first of all thank you for your answer :)
Secondly I now notice I made a wrong definition of continuum... It should also be compact.
Anyways your approach works if I work a dense set and any other set to go along with it.
Also just for clarification zero set aposyndetic can only work with non intersecting sets I didn't state that in the definition as I made it up so I was being careless.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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