Connected, Irreducible and Noetherian topological spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 29 July 2014

Connected sets and connected components

Let (X,𝒯) be a topological space.

A connected set is a subset EβŠ†X such that there do not exist open sets A and B (A,Bβˆˆπ’―) with A∩Eβ‰ βˆ…and B∩Eβ‰ βˆ…and AβˆͺBβŠ‡Eand (A∩B)∩E=βˆ…. Perhaps it is better to think of E with the subspace topology 𝒯E. Then E is connected if there do not exist U and V open in E (V,Vβˆˆπ’―E) such that Uβ‰ βˆ…andVβ‰ βˆ… andUβˆͺV=E andU∩V=βˆ….

Let (X,𝒯X) and (Y,𝒯Y) be topological spaces. Let f:Xβ†’Y be a function.

The function f:Xβ†’Y is continuous if f satisfies: if Vβˆˆπ’―Y then f-1(V)βˆˆπ’―X.

Recall that, by definition, f-1(V)= {x∈X | f(x)∈V}.

Recall that, by definition, a function f:Xβ†’Y is a subset Ξ“βŠ†XΓ—Y such that if x∈X then there exists a unique y∈Y such that (x,y)βˆˆΞ“. Use the notation f(x) so that Ξ“={(x,f(x)) | x∈X}.

Let f:Xβ†’Y be a continuous function. Let EβŠ†X. If E is connected
then f(E) is connected.

Proof.

Let (X,𝒯) be a topological space. Let π’ž be a collection of connected subsets of X such that β‹‚Aβˆˆπ’žAβ‰ βˆ….
Prove that ⋃Aβˆˆπ’žA is connected.

Proof.

Let (X,𝒯) be a topological space and let AβŠ†X be connected. Show that Aβ€Ύ is connected.

Proof.

Let (X,𝒯) be a topological space. Let x∈X. The connected component of x is Cx= ⋃A connectedx∈A A, the union of the connected subsets of X containingx.

Let (X,𝒯) be a topological space and let x∈X.

(a) Cx is connected.
(b) Cx is closed.
(c) If y∈Cx then Cy=Cx.
(d) If x,y∈X then Cx=Cy or Cx∩Cyβˆ©βˆ….

Proof.

Let (X,𝒯) be a topological space. Then X is connected if and only if there does not exist a continuous surjective function f:Xβ†’{0,1}, where {0,1} has the discrete topology.

Proof.

Noetherian spaces

A non-empty topological space X is irreducible if every pair of non-empty open sets in X intersect (thus X is as far as possible from being Hausdorff). Equivalent conditions:

(a)   X is not the union of two proper closed subsets.
(b)   If Fi (1≀i≀n) are closed subsets which cover X, then X=Fi for some i.
(c)   Every non-empty open set is dense in X.
(d)   Every open set in X is connected.

Examples.

(1)   Let X be an infinite set, and topologize X by taking the closed subsets to be X itself and all finite subsets of X. Then X is irreducible.
(2)   Any irreducible algebraic variety with the Zariski topology.

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 (Fi) 1≀i≀n is a finite closed covering of a space X, and if Y is an irreducible subset of X, then YβŠ†Fi for some i.
(ii)   If X is irreducible, every non-empty open subset of X is irreducible.
(iii)   Let (Ui) 1≀i≀n be a finite open covering of a space X, the Ui being non-empty. Then X is irreducible if and only if each Ui is irrducible and meets each Uj.
(iv)   If Y is a subset of X, then Y is irreducible if and only if Yβ€Ύ is irreducible.
(v)   The image of an irreducible set under a continuous map is irreducible.
(vi)   X has maximal irreducible subsets; they are all closed and they cover X. (Use Zorn's lemma for (vi).)

The maximal irreducible substes of X are called the irreducible components of X. Irreducibility is in some ways analogous to, but stronger than, connectedness.

If x∈X, then {x} is irreducible and therefore (by (iv) above) so is {x}β€Ύ. If V is an irreducible subset of X and V= {x}β€Ύ for some x∈X, then x is a generic point of V. If y∈ {x}β€Ύ , y is a specialization of x. The closed set {x}β€Ύ is the locus of x.

A subset Y of a space X is locally closed if Y is the intersection of an open set and a closed set in X, or equivalently if Y is open in its closure Yβ€Ύ, or equivalently again if every y∈Y has an open neighborhood Uy in X such that Y∩Uy is closed in Uy.

A topological space is Noetherian if the closed subsets of X satisfy the descending chain condition. Equivalent conditions:

(i)   A Noetherian space is quasi-compact.
(ii)   Every subset of a Noetherian space (with the induced topology) is Noetherian.
(iii)   Let X be a topological space and let (Xi) 1≀i≀n be a finite covering of X. If the Xi are Noetherian, then so is X.
(iv)   If X is Noetherian, the number of irreducible components of X is finite.

The proofs are straightforward.

Notes and References

These notes are taken from [Mac].

References

[Mac] I.G. Macdonald, Algebraic Geometry: Introduction to Schemes, W.A. Benjamin, New York, 1968.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

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