## Connected sets and connected components

Let $\left(X,𝒯\right)$ be a topological space.

A connected set is a subset $E\subseteq X$ such that there do not exist open sets $A$ and $B$ $\text{(}A,B\in 𝒯\text{)}$ with $A∩E≠∅and B∩E≠∅and A∪B⊇Eand (A∩B)∩E=∅.$ Perhaps it is better to think of $E$ with the subspace topology ${𝒯}_{E}\text{.}$ Then $E$ is connected if there do not exist $U$ and $V$ open in $E$ $\text{(}V,V\in {𝒯}_{E}\text{)}$ such that $U≠∅andV≠∅ andU∪V=E andU∩V=∅.$

Let $\left(X,{𝒯}_{X}\right)$ and $\left(Y,{𝒯}_{Y}\right)$ be topological spaces. Let $f:X\to Y$ be a function.

The function $f:X\to Y$ is continuous if $f$ satisfies: if $V\in {𝒯}_{Y}$ then ${f}^{-1}\left(V\right)\in {𝒯}_{X}\text{.}$

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

Recall that, by definition, a function $f:X\to Y$ is a subset $\mathrm{\Gamma }\subseteq X×Y$ such that if $x\in X$ then there exists a unique $y\in Y$ such that $\left(x,y\right)\in \mathrm{\Gamma }\text{.}$ Use the notation $f\left(x\right)$ so that $Γ={(x,f(x)) | x∈X}.$

Let $f:X\to Y$ be a continuous function. Let $E\subseteq X\text{.}$ If $E$ is connected
then $f\left(E\right)$ is connected.

 Proof. Assume $E$ is connected. To show: $f\left(E\right)$ is connected. Proof by contradiction. Assume $f\left(E\right)$ is not connected. Let $A$ and $B$ be open in $f\left(E\right)$ such that $A≠∅and B≠∅and A∪B⊇f(E) andA∩B=∅.$ Let $C={f}^{-1}\left(A\right)$ and $D={f}^{-1}\left(B\right)\text{.}$ Then $C∪D=f-1(A) ∪f-1(B)= f-1(A∪B)⊇ f-1(f(E)) ⊇E$ and $C∩D=f-1(A) ∩f-1(B)= f-1(A∩B)= f-1(∅)=∅$ and $C≠∅ since A≠∅ and A⊆f(E), D≠∅ since B≠∅ and B⊆f(E).$ So $E$ is not connected. This is a contradiction. So $f\left(E\right)$ is connected. $\square$

Let $\left(X,𝒯\right)$ be a topological space. Let $𝒞$ be a collection of connected subsets of $X$ such that $\bigcap _{A\in 𝒞}A\ne \varnothing \text{.}$
Prove that $\bigcup _{A\in 𝒞}A$ is connected.

 Proof. Let $M=\bigcup _{A\in 𝒞}A$ and let $x\in \bigcap _{A\in 𝒞}A\text{.}$ Proof by contradiction. Assume $M$ is not connected. Let $B$ and $C$ be open sets such that $M⊆B∪C, M∩B∩C=∅, M∩B≠∅and M∩C≠∅.$ Then $x\in B$ or $x\in C\text{.}$ Assume $x\in B\text{.}$ Let $U\in 𝒞$ be such that $C\cap U\ne \varnothing \text{.}$ Since $x\in \bigcap _{A\in 𝒞}A$ then $x\in U,$ so that $x∈B∩UandB∩U ≠∅.$ Since $U\subseteq M$ then $U⊆B∩Cand U∩B∩C=∅.$ This is a contradiction to $U$ be connected. So $M$ is connected. $\square$

Let $\left(X,𝒯\right)$ be a topological space and let $A\subseteq X$ be connected. Show that $\stackrel{‾}{A}$ is connected.

 Proof. Proof by contradiction. Assume $\stackrel{‾}{A}$ is not connected. Let $M$ and $N$ be open subsets of $\stackrel{‾}{A}$ such that $M∪N=A‾, M≠∅, N≠∅and M∩N=∅.$ Then $(M∩A)∪(N∩A) =Aand(M∩A) ∩(N∩A)=∅.$ There exists $x\in \stackrel{‾}{A}\cap M,$ $x$ is a close point of $A$ and, since $M$ is open, $M$ is a neighbourhood of $x\text{.}$ So $M\cap A\ne \varnothing \text{.}$ There exists $y\in \stackrel{‾}{A}\cap N,$ $y$ is a close point of $A$ and, since $N$ is open, $N$ is a neighbourhood of $y\text{.}$ So $N\cap A\ne \varnothing \text{.}$ This is a contradiction to $A$ is connected. So $\stackrel{‾}{A}$ is connected. $\square$

Let $\left(X,𝒯\right)$ be a topological space. Let $x\in X\text{.}$ The connected component of $x$ is $Cx= ⋃A connectedx∈A A, the union of the connected subsets of X containingx.$

Let $\left(X,𝒯\right)$ be a topological space and let $x\in X\text{.}$

 (a) ${C}_{x}$ is connected. (b) ${C}_{x}$ is closed. (c) If $y\in {C}_{x}$ then ${C}_{y}={C}_{x}\text{.}$ (d) If $x,y\in X$ then ${C}_{x}={C}_{y}$ or ${C}_{x}\cap {C}_{y}\cap \varnothing \text{.}$

 Proof. (a) This follows from Example 1. (b) By Example 2, $\stackrel{‾}{{𝒞}_{x}}$ is a connected set that contains $x\text{.}$ so $\stackrel{‾}{{C}_{x}}\subseteq {C}_{x}\text{.}$ So $\stackrel{‾}{{C}_{x}}={C}_{x}\text{.}$ (c) Assume $y\in {C}_{x}\text{.}$ Then ${C}_{x}$ is a connected set containing $y\text{.}$ So ${C}_{x}\subseteq {C}_{y}\text{.}$ Then ${C}_{y}$ is a connected set containing $x\text{.}$ So ${C}_{y}\subseteq {C}_{x}\text{.}$ so ${C}_{x}={C}_{y}\text{.}$ (d) Assume $x,y\in X$ and ${C}_{x}\cap {C}_{y}\ne \varnothing \text{.}$ Let $z\in {C}_{x}\cap {C}_{y}\text{.}$ So $z\in {C}_{x}$ and $z\in {C}_{y}\text{.}$ By (c), ${C}_{x}={C}_{z}={C}_{y}\text{.}$ So ${C}_{x}={C}_{y}\text{.}$ $\square$

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

 Proof. $⇒$ To show: If there exists a continuous surjective function $f:X\to \left\{0,1\right\}$ then $X$ is not connected. Assume that $f:X\to \left\{0,1\right\}$ 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, ${f}^{-1}\left(0\right)=A\ne \varnothing$ and ${f}^{-1}\left(1\right)=B\ne \varnothing \text{.}$ Then $A∪B=f-1 ({0,1})=X and A∩B=f-1 (0)∩f-1 (1)=∅.$ So $X$ is not connected. $⇐$ To show: If $X$ is not connected then there exists a continuous surjective function $f:X\to \left\{0,1\right\}\text{.}$ Assume $X$ is not connected. Then there exist open sets $A$ and $B$ such that $A∪B=X, A≠∅, B≠∅and A∩B=∅.$ Define $f:X\to \left\{0,1\right\}$ by $f(x)= { 0, if x∈A, 1, if x∈B.$ Since $X=A\cup B$ and $A\cap B=\varnothing ,$ $f$ is well defined. Since $A\ne \varnothing$ and $B\ne \varnothing ,$ $f$ is surjective. Since $A={f}^{-1}\left(0\right)$ and $B={f}^{-1}\left(1\right)$ are open, $f$ is continuous. So there exists a continuous surjective function $f:X\to \left\{0,1\right\}\text{.}$ $\square$

## 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 ${F}_{i}$ ($1\le i\le n$) are closed subsets which cover $X$, then $X={F}_{i}$ 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 ${\left({F}_{i}\right)}_{1\le i\le n}$ is a finite closed covering of a space $X$, and if $Y$ is an irreducible subset of $X$, then $Y\subseteq {F}_{i}$ for some $i$.
(ii)   If $X$ is irreducible, every non-empty open subset of $X$ is irreducible.
(iii)   Let ${\left({U}_{i}\right)}_{1\le i\le n}$ be a finite open covering of a space $X$, the ${U}_{i}$ being non-empty. Then $X$ is irreducible if and only if each $Ui$ is irrducible and meets each ${U}_{j}$.
(iv)   If $Y$ is a subset of $X$, then $Y$ is irreducible if and only if $\stackrel{‾}{Y}$ 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 $x\in X$, then $\left\{x\right\}$ is irreducible and therefore (by (iv) above) so is $\stackrel{‾}{\left\{x\right\}}$. If $V$ is an irreducible subset of $X$ and $V=\stackrel{‾}{\left\{x\right\}}$ for some $x\in X$, then $x$ is a generic point of $V$. If $y\in \stackrel{‾}{\left\{x\right\}}$, $y$ is a specialization of $x$. The closed set $\stackrel{‾}{\left\{x\right\}}$ 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 $\stackrel{‾}{Y}$, or equivalently again if every $y\in Y$ has an open neighborhood ${U}_{y}$ in $X$ such that $Y\cap {U}_{y}$ is closed in ${U}_{y}$.

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

• The open sets in $X$ satisfy the ascending chain condition;
• Every open subset of $X$ is quasi-compact (i.e. compact but not necessarily Hausdorff);
• Every subset of $X$ is quasi-compact.

(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 ${\left({X}_{i}\right)}_{1\le i\le n}$ be a finite covering of $X$. If the ${X}_{i}$ 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.