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 $(X,\mathcal{T})$ be a topological space.
A connected set is a subset $E\subseteq X$ such that there do not exist open sets $A$ and
$B$ $\text{(}A,B\in \mathcal{T}\text{)}$ with
$$A\cap E\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}B\cap E\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}A\cup B\supseteq E\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}(A\cap B)\cap E=\varnothing \text{.}$$
Perhaps it is better to think of $E$ with the subspace topology ${\mathcal{T}}_{E}\text{.}$
Then $E$ is connected if there do not exist $U$ and $V$ open in $E$
$\text{(}V,V\in {\mathcal{T}}_{E}\text{)}$ such that
$$U\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}V\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}U\cup V=E\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}U\cap V=\varnothing \text{.}$$
Let $(X,{\mathcal{T}}_{X})$ and $(Y,{\mathcal{T}}_{Y})$
be topological spaces. Let $f:X\to Y$ be a function.
The function $f:X\to Y$ is continuous if $f$ satisfies:
if $V\in {\mathcal{T}}_{Y}$ then ${f}^{1}\left(V\right)\in {\mathcal{T}}_{X}\text{.}$
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
$$A\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}B\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}A\cup B\supseteq f\left(E\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}A\cap B=\varnothing \text{.}$$
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 $(X,\mathcal{T})$ be a topological space. Let $\mathcal{C}$ be a collection of connected subsets of $X$ such that
$\bigcap _{A\in \mathcal{C}}A\ne \varnothing \text{.}$
Prove that $\bigcup _{A\in \mathcal{C}}A$ is connected.


Proof. 

Let $M=\bigcup _{A\in \mathcal{C}}A$ and let
$x\in \bigcap _{A\in \mathcal{C}}A\text{.}$
Proof by contradiction.
Assume $M$ is not connected.
Let $B$ and $C$ be open sets such that
$$M\subseteq B\cup C,\phantom{\rule{2em}{0ex}}M\cap B\cap C=\varnothing ,\phantom{\rule{2em}{0ex}}M\cap B\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}M\cap C\ne \varnothing \text{.}$$
Then $x\in B$ or $x\in C\text{.}$
Assume $x\in B\text{.}$
Let $U\in \mathcal{C}$ be such that $C\cap U\ne \varnothing \text{.}$
Since $x\in \bigcap _{A\in \mathcal{C}}A$ then
$x\in U,$ so that
$$x\in B\cap U\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}B\cap U\ne \varnothing \text{.}$$
Since $U\subseteq M$ then
$$U\subseteq B\cap C\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}U\cap B\cap C=\varnothing \text{.}$$
This is a contradiction to $U$ be connected.
So $M$ is connected.
$\square $

Let $(X,\mathcal{T})$ be a topological space and let $A\subseteq X$
be connected. Show that $\stackrel{\u203e}{A}$ is connected.


Proof. 

Proof by contradiction.
Assume $\stackrel{\u203e}{A}$ is not connected.
Let $M$ and $N$ be open subsets of $\stackrel{\u203e}{A}$ such that
$$M\cup N=\stackrel{\u203e}{A},\phantom{\rule{2em}{0ex}}M\ne \varnothing ,\phantom{\rule{2em}{0ex}}N\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}M\cap N=\varnothing \text{.}$$
Then
$$(M\cap A)\cup (N\cap A)=A\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}(M\cap A)\cap (N\cap A)=\varnothing \text{.}$$
There exists $x\in \stackrel{\u203e}{A}\cap M,$
$x$ 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 $y\in \stackrel{\u203e}{A}\cap N,$
$y$ 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 $\stackrel{\u203e}{A}$ is connected.
$\square $

Let $(X,\mathcal{T})$ be a topological space. Let $x\in X\text{.}$
The connected component of $x$ is
$${C}_{x}=\bigcup _{\underset{x\in A}{A\hspace{0.17em}\text{connected}}}A,\phantom{\rule{2em}{0ex}}\text{the union of the connected subsets of}\phantom{\rule{0.5em}{0ex}}X\phantom{\rule{0.5em}{0ex}}\text{containing}\phantom{\rule{0.5em}{0ex}}x.$$
Let $(X,\mathcal{T})$ 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, $\stackrel{\u203e}{{\mathcal{C}}_{x}}$ is a connected set that contains $x\text{.}$
so $\stackrel{\u203e}{{C}_{x}}\subseteq {C}_{x}\text{.}$
So $\stackrel{\u203e}{{C}_{x}}={C}_{x}\text{.}$
(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 $(X,\mathcal{T})$ 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
$$A={f}^{1}\left(0\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}B={f}^{1}\left(1\right)\text{.}$$
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
$$A\cup B={f}^{1}\left(\{0,1\}\right)=X\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}A\cap B={f}^{1}\left(0\right)\cap {f}^{1}\left(1\right)=\varnothing \text{.}$$
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
$$A\cup B=X,\phantom{\rule{2em}{0ex}}A\ne \varnothing ,\phantom{\rule{2em}{0ex}}B\ne \varnothing \phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}A\cap B=\varnothing \text{.}$$
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 nonempty topological space $X$ is irreducible
if every pair of nonempty 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 nonempty 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 nonempty 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 nonempty. 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 $\stackrel{\u203e}{Y}$ 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
$\stackrel{\u203e}{\left\{x\right\}}$.
If $V$ is an irreducible subset of $X$ and
$V=\stackrel{\u203e}{\left\{x\right\}}$ for some $x\in X$, then $x$
is a generic point of $V$. If
$y\in \stackrel{\u203e}{\left\{x\right\}}$, $y$ is a specialization of $x$.
The closed set
$\stackrel{\u203e}{\left\{x\right\}}$
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 $\stackrel{\u203e}{Y}$, 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 quasicompact (i.e. compact but not
necessarily Hausdorff);
 Every subset of $X$ is quasicompact.
 (i)
A Noetherian space is quasicompact.
 (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.
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