Connected, Irreducible and Noetherian topological spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updates: 29 July 2014

Connected sets and connected components

Let (X,𝒯) be a topological space.

A connected set is a subset EX such that there do not exist open sets A and B (A,B𝒯) with AEand BEand ABEand (AB)E=. 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,V𝒯E) such that UandV andUV=E andUV=.

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

The function f:XY is continuous if f satisfies: if V𝒯Y then f-1(V)𝒯X.

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

Recall that, by definition, a function f:XY is a subset ΓX×Y such that if xX then there exists a unique yY such that (x,y)Γ. Use the notation f(x) so that Γ={(x,f(x))|xX}.

Let f:XY be a continuous function. Let EX. If E is connected
then f(E) is connected.


Assume E is connected.
To show: f(E) is connected.
Proof by contradiction.
Assume f(E) is not connected.
Let A and B be open in f(E) such that Aand Band ABf(E) andAB=. Let C=f-1(A) and D=f-1(B).
Then CD=f-1(A) f-1(B)= f-1(AB) f-1(f(E)) E and CD=f-1(A) f-1(B)= f-1(AB)= f-1()= and CsinceA andAf(E), DsinceB andBf(E). So E is not connected. This is a contradiction.
So f(E) is connected.

Let (X,𝒯) be a topological space. Let 𝒞 be a collection of connected subsets of X such that A𝒞A.
Prove that A𝒞A is connected.


Let M=A𝒞A and let xA𝒞A.
Proof by contradiction.
Assume M is not connected.
Let B and C be open sets such that MBC, MBC=, MBand MC. Then xB or xC.
Assume xB.
Let U𝒞 be such that CU.
Since xA𝒞A then xU, so that xBUandBU . Since UM then UBCand UBC=. This is a contradiction to U be connected.
So M is connected.

Let (X,𝒯) be a topological space and let AX be connected. Show that A is connected.


Proof by contradiction.
Assume A is not connected.
Let M and N be open subsets of A such that MN=A, M, Nand MN=. Then (MA)(NA) =Aand(MA) (NA)=. There exists xAM, x is a close point of A and, since M is open, M is a neighbourhood of x.
So MA.
There exists yAN, y is a close point of A and, since N is open, N is a neighbourhood of y.
So NA.
This is a contradiction to A is connected.
So A is connected.

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

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

(a) Cx is connected.
(b) Cx is closed.
(c) If yCx then Cy=Cx.
(d) If x,yX then Cx=Cy or CxCy.


(a) This follows from Example 1.

(b) By Example 2, 𝒞x is a connected set that contains x. so CxCx. So Cx=Cx.

(c) Assume yCx. Then Cx is a connected set containing y. So CxCy. Then Cy is a connected set containing x. So CyCx. so Cx=Cy.

(d) Assume x,yX and CxCy. Let zCxCy. So zCx and zCy. By (c), Cx=Cz=Cy. So Cx=Cy.

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


To show: If there exists a continuous surjective function f:X{0,1} then X is not connected.
Assume that f:X{0,1} is a continuous surjective function.
Let A=f-1(0) andB=f-1(1). Since f is continuous, A and B are open.
Since f is surjective, f-1(0)=A and f-1(1)=B.
Then AB=f-1 ({0,1})=X and AB=f-1 (0)f-1 (1)=. So X is not connected.

To show: If X is not connected then there exists a continuous surjective function f:X{0,1}.
Assume X is not connected.
Then there exist open sets A and B such that AB=X, A, Band AB=. Define f:X{0,1} by f(x)= { 0, ifxA, 1, ifxB. Since X=AB and AB=, f is well defined.
Since A and B, f is surjective.
Since A=f-1(0) and B=f-1(1) are open, f is continuous.
So there exists a continuous surjective function f:X{0,1}.

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 (1in) 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.


(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) 1in is a finite closed covering of a space X, and if Y is an irreducible subset of X, then YFi for some i.
(ii)   If X is irreducible, every non-empty open subset of X is irreducible.
(iii)   Let (Ui) 1in 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 Y 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 xX, then {x} is irreducible and therefore (by (iv) above) so is {x}. If V is an irreducible subset of X and V= {x} for some xX, then x is a generic point of V. If y {x} , y is a specialization of x. The closed set {x} 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 Y, or equivalently again if every yY has an open neighborhood Uy in X such that YUy 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) 1in 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].


[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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