Suchergebnisse
Suchergebnisse:
A noetherian topological space has a nite \irreducible component decomposition". Since irreducible spaces are obviously connected (as all non-empty open subsets are dense, so no separation is possible), we see that noetherian spaces have nitely many connected com-ponents, obtained from chains of irreducible components where each touches another. In
5. A topological space X X is called irreducible if given A1,A2 A 1, A 2 open sets ≠ ∅ ≠ ∅ then A1 ∩A2 ≠ ∅ A 1 ∩ A 2 ≠ ∅. The maximal irreducible topological subspaces of X X are called irreducible components. Let A A be a commutative ring with unit, X = Spec(A) X = S p e c ( A) with the Zariski topology.
1. Irreducible Scheme A topological space X is irreducible if X is nonempty and can not be a union of two proper closed subsets. A nonempty closed subset Zof Xis irreducible if Zequipped with the induced topology from Xis an irreducible topological space. A point ˘of an irreducible closed subset Zof Xis called a generic point if Z= f˘g ...
An irreducible component of a topological space is a maximal irreducible subset. My question: What do we mean with maximal here? I thought if a subset is irreducible it is automatically maximal because we can not split it up into two smaller closed sets. Can someone explain what is wrong with my thought? It would be great if someone could give ...
18. Let X be a topological space. Let Σ be the set of irreducible components of X. Let X = ∪i ∈ IXi = ∪j ∈ JYj, Xi, Yj ∈ Σ for some index set I, J. Xi 's are distinct from each other. Yj 's are distinct from each other. I want an example such that X has two distinct expression, i.e. there exist {Xi | i ∈ I} ≠ {Yj | j ∈ J ...
$(i)$ That every set with the indiscrete topology is irreducible, and every infinite set with the finite complement topology is irreducible. $(ii)$ That an irreducible Hausdorff space contains at most one point. Can you give me any suggestions on how to try this?
29. Juli 2014 · A subset Y of a space X is irreducible if Y is irreducible in the induced topology. The following facts are not hard to prove: (i) If (F i) 1 ≤ i ≤ n is a finite closed covering of a space X, and if Y is an irreducible subset of X, then Y ⊆ F i for some i. (ii) If X is irreducible, every non-empty open subset of X is irreducible.
