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 $E\subseteq X$ such that there do not exist open sets $A$ and $B$ $\text{(}A,B\in 𝒯\text{)}$ with $$A\cap E\ne \varnothing \text{and}B\cap E\ne \varnothing \text{and}A\cup B\supseteq E\text{and}(A\cap B)\cap E=\varnothing \text{.}$$ 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 $$U\ne \varnothing \text{and}V\ne \varnothing \text{and}U\cup V=E\text{and}U\cap V=\varnothing \text{.}$$

Let $(X,{𝒯}_X)$ and $(Y,{𝒯}_Y)$ 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, $$f^{-1}\left(V\right)=\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)\in V\}\text{.}$$

Recall that, by definition, a function $f:X\to Y$ is a subset $\mathrm{\Gamma }\subseteq X\times Y$ such that if $x\in X$ then there exists a unique $y\in Y$ such that $(x,y)\in \mathrm{\Gamma }\text{.}$ Use the notation $f\left(x\right)$ so that $$\mathrm{\Gamma }=\left\{(x,f\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}\text{.}$$

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 $$A\ne \varnothing \text{and}B\ne \varnothing \text{and}A\cup B\supseteq f\left(E\right)\text{and}A\cap B=\varnothing \text{.}$$ Let $C=f^{-1}\left(A\right)$ and $D=f^{-1}\left(B\right)\text{.}$ Then $$C\cup D=f^{-1}\left(A\right)\cup f^{-1}\left(B\right)=f^{-1}(A\cup B)\supseteq f^{-1}\left(f\left(E\right)\right)\supseteq E$$ and $$C\cap D=f^{-1}\left(A\right)\cap f^{-1}\left(B\right)=f^{-1}(A\cap B)=f^{-1}\left(\varnothing \right)=\varnothing$$ and $$\begin{array}{c}C\ne \varnothing \hspace{0.17em}\text{since}\hspace{0.17em}A\ne \varnothing \hspace{0.17em}\text{and}\hspace{0.17em}A\subseteq f\left(E\right),\\ D\ne \varnothing \hspace{0.17em}\text{since}\hspace{0.17em}B\ne \varnothing \hspace{0.17em}\text{and}\hspace{0.17em}B\subseteq f\left(E\right)\text{.}\end{array}$$ So $E$ is not connected. This is a contradiction. So $f\left(E\right)$ is connected. $\square$

Let $(X,𝒯)$ 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 $$M\subseteq B\cup C,M\cap B\cap C=\varnothing ,M\cap B\ne \varnothing \text{and}M\cap C\ne \varnothing \text{.}$$ 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 $$x\in B\cap U\text{and}B\cap U\ne \varnothing \text{.}$$ Since $U\subseteq M$ then $$U\subseteq B\cap C\text{and}U\cap B\cap C=\varnothing \text{.}$$ This is a contradiction to $U$ be connected. So $M$ is connected. $\square$

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

		Proof.
	Proof by contradiction. Assume $\bar{A}$ is not connected. Let $M$ and $N$ be open subsets of $\bar{A}$ such that $$M\cup N=\bar{A},M\ne \varnothing ,N\ne \varnothing \text{and}M\cap N=\varnothing \text{.}$$ Then $$(M\cap A)\cup (N\cap A)=A\text{and}(M\cap A)\cap (N\cap A)=\varnothing \text{.}$$ There exists $x\in \bar{A}\cap M,$ $x$ 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 $y\in \bar{A}\cap N,$ $y$ 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 $\bar{A}$ is connected. $\square$

Let $(X,𝒯)$ be a topological space. Let $x\in X\text{.}$ The connected component of $x$ is $$C_x=\bigcup _{\underset{x\in A}{A\hspace{0.17em}\text{connected}}}A,\text{the union of the connected subsets of}X\text{containing}x.$$

Let $(X,𝒯)$ 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, $\overline{{𝒞}_x}$ is a connected set that contains $x\text{.}$ so $\overline{C_x}\subseteq C_x\text{.}$ So $\overline{C_x}=C_x\text{.}$ (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 $(X,𝒯)$ be a topological space. Then $X$ is connected if and only if there does not exist a continuous surjective function $f:X\to \{0,1\},$ where $\{0,1\}$ has the discrete topology.

		Proof.
	$\Rightarrow$ To show: If there exists a continuous surjective function $f:X\to \{0,1\}$ then $X$ is not connected. Assume that $f:X\to \{0,1\}$ is a continuous surjective function. Let $$A=f^{-1}\left(0\right)\text{and}B=f^{-1}\left(1\right)\text{.}$$ 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 $$A\cup B=f^{-1}\left(\{0,1\}\right)=X\text{and}A\cap B=f^{-1}\left(0\right)\cap f^{-1}\left(1\right)=\varnothing \text{.}$$ So $X$ is not connected. $\Leftarrow$ To show: If $X$ is not connected then there exists a continuous surjective function $f:X\to \{0,1\}\text{.}$ Assume $X$ is not connected. Then there exist open sets $A$ and $B$ such that $$A\cup B=X,A\ne \varnothing ,B\ne \varnothing \text{and}A\cap B=\varnothing \text{.}$$ Define $f:X\to \{0,1\}$ by $$f\left(x\right)=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}x\in A,\\ 1,& \text{if}\hspace{0.17em}x\in B\text{.}\end{array}$$ 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 $f:X\to \{0,1\}\text{.}$ $\square$

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 $\bar{Y}$ 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 $\overline{\left\{x\right\}}$. If $V$ is an irreducible subset of $X$ and $V=\overline{\left\{x\right\}}$ for some $x\in X$, then $x$ is a generic point of $V$. If $y\in \overline{\left\{x\right\}}$, $y$ is a specialization of $x$. The closed set $\overline{\left\{x\right\}}$ 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 $\bar{Y}$, 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.

page history