Let \(X\) be a topological space and \(A\sub X\). The interior of \(A\) is denoted by \(A^{\circ }\) or by \(\inter \of A\) and is defined by
\[ A^{\circ }:=\bigcup \set {U\sub A:U\text { is open}}. \]
If \(x\in A^{\circ }\), then \(x\) is an interior point of \(A\).
Note that \(A^{\circ }\) is open and it is the largest open subset of \(A\). Moreover, \(x\) is an interior point of \(A\) if and only if \(A\) is a nbhd of \(x\). It is also clear that \(A\) is open if and only if \(A=A^{\circ }\).
In \(\br \) we have \(\bq ^{\circ }=\emp \). If \(A\) is the closed interval \(\bra {a,b}\), then \(A^{\circ }\) is the open interval \(\cur {a,b}\).
Let \(X\) be a topological space and \(A,B\sub X\).
1. \(\cur {A^{\circ }}^{\circ }=A^{\circ }\),
2. \(A\sub B\) implies that \(A^{\circ }\sub B^{\circ }\),
3. \(A^{\circ }\cap B^{\circ }=\cur {A\cap B}^{\circ }\),
4. \(A^{\circ }\cup B^{\circ }\sub \cur {A\cup B}^{\circ }.\)
If \(X:=\br \), \(A:=\bra {0,1}\) and \(B:=\bra {1,2}\), then \(1\nin A^{\circ }\cup B^{\circ }\), but \(1\in \cur {A\cup B}^{\circ }\).
Let \(X\) be a topological space and \(A\sub X\). The closure of \(A\) is denoted by \(\cl \of A\) or \(\ob A\). It is defined by
\[ \ob A:=\bigcap \set {A\sub C:C\text { is closed}}. \]
If \(x\in \ob A\), then \(x\) is an adherent point of \(A\).
Note that \(\ob A\) is the smallest closed set containing \(A\) and that \(A\) is closed if and only if \(\ob A=A\).
If \(X=\br \), then \(\ob {\bq }=\br \). If \(A\) is the open interval \(\cur {a,b}\) with \(a<b\), then \(\ob A\) is the closed interval \(\bra {a,b}\).
Let \(X\) be a topological space and \(A,B\sub X\).
1. \(\ob {\ob A}=\ob A\),
2. \(A\sub B\) implies that \(\ob A\sub \ob B\),
3. \(\ob A\cup \ob B=\ob {A\cup B}\),
4. \(\ob A\cap \ob B\sur \ob {A\cap B}.\)
If \(X:=\br \), \(A:=\cur {0,1}\) and \(B:=\cur {1,2}\), then \(1\in \ob A\cap \ob B\), but \(1\nin \ob {A\cup B}\).
Let \(X\) be a topological space and \(A\sub X\). Then \(x\in \ob A\) if and only if \(U\cap A\neq \emp \) for every open nbhd \(U\) of \(x\).
Proof. Assume that \(x\in \ob A\). Let \(U\) be an open nbhd of \(x\) and suppose, for a contradiction, that \(U\cap A=\emp \). Then \(A\sub X\sem U\) and \(X\sem U\) is closed. Thus \(x\in \ob A\sub X\sem U\), which contradicts \(x\in U\).
Now assume that \(U\cap A\neq \emp \) for every open nbhd \(U\) of \(x\). Let \(C\) be closed with \(A\sub C\). Then \(X\sem C\) is open and disjoint with \(A\) so \(x\nin X\sem C\). Thus \(x\in C\). Since \(x\) belongs to every closed set containing \(A\), it follows that \(x\in \ob A\). □
Let \(X\) be a topological space and \(A\sub X\). A point \(x\in X\) is a limit point (cluster point) of \(A\) if every open nbhd \(U\) of \(x\) contains at least one point of \(A\sem \set x\). The set \(A’\) of all limit points of \(A\) is called the derived set of \(A\).
In \(\br \) if \(A:=\cur {0,1}\cup \set 2\), then \(A’=\bra {0,1}\).
Let \(X\) be a topological space and \(A\sub X\). Then \(\ob A=A\cup A’\).
Proof. If \(x\in A\), then \(x\in \ob A\). If \(x\in A’\), then every open nbhd \(U\) of \(x\) contains at least one point of \(A\sem \set x\) so \(A\cap U\neq \emp \), which implies that \(x\in \ob A\).
Now assume that \(x\in \ob A\sem A\). We show that \(x\in A’\). Let \(U\) be an open nbhd of \(x\). Since \(x\in \ob A\), we have \(U\cap A\neq \emp \). Since \(A\sem \set x=A\), we have
\[ U\cap \cur {A\sem \set x}\neq \emp \]
as required. □
A set is closed if and only if it contains all its limit points.
Proof. \(A\) is closed if and only if \(\ob A=A\), which holds if and only if \(A=A\cup A’\), which is equivalent to \(A’\sub A\). □
Let \(X\) be a topological space and \(A\sub X\). The boundary (also called the frontier) of \(A\) is denoted by \(\pt A\) and is defined by
\[ \pt A:=\ob A\cap \ob {X\sem A}. \]
In \(\br \) if \(A=\bra {0,1}\), then \(\pt A=\set {0,1}\).
Let \(X\) be a topological space and \(A\sub X\). Then \(\ob A=A\cup \pt A\).
Proof. We have \(A\sub \ob A\) and the definition of \(\pt A\) implies that \(\pt A\sub \ob A\). Thus \(\ob A\sur A\cup \pt A\).
Now assume that \(x\in \ob A\sem A\). Then \(x\in X\sem A\) so \(x\in \ob {X\sem A}\). Thus \(x\in \pt A\). □
A set is closed if and only if it contains it’s boundary.
Proof. \(A\) is closed if and only if \(\ob A=A\), which is equivalent to \(A=A\cup \pt A\) and to \(\pt A\sub A\). □
Let \(X\) be a topological space and \(A\sub X\). If \(x\in A\sem A’\), then \(x\) is an isolated point of \(A\).
Let \(X\) be a topological space and \(A\sub X\). We say that \(A\) is perfect if \(A\) is closed and has no isolated points.
Let
\[ J_{1}=\cur {\frac {1}{3},\frac {2}{3}} \]
\[ J_{2}=\cur {\frac {1}{9},\frac {2}{9}}\cup \cur {\frac {7}{9},\frac {8}{9}} \]
and, in general, for each \(n\in \bn \), let
\[ J_{n}=\bigcup _{k=0}^{3^{n-1}-1}\cur {\frac {1+3k}{3^{n}},\frac {2+3k}{3^{n}}}\sem \bigcup _{j=1}^{n-1}J_{j}. \]
The set
\[ C:=\bra {0,1}\sem \bigcup _{n=1}^{\ity }J_{n} \]
is called the Cantor set. It consists of those \(x\in \bra {0,1}\) that have a triadic expansion using only the digits \(0\) and \(2\).
The Cantor set is perfect since it is closed and has no isolated points. If \(x\in C\) has a triadic expansion \(0.a_{1}a_{2}\ds \), and \(U\) is any open interval containing \(x\), then we can choose \(n\in \bn \) large enough, so that changing the digit \(a_{n}\) to \(2-a_{n}\) produces a different point in \(U\cap C\).
Let \(X\) be a topological space and \(A\sub X\). We say that \(A\) is dense if \(\ob A=X\).
\(\bq \) is dense in \(\br \). If \(X\) is an infinite set with cofinite topology, then any infinite subset of \(X\) is dense.
Let \(X:=\set {a,b,c}\) with the topology \(\set {\emp ,X,\set a,\set {a,b}}\). Find the derived sets of \(\set a\), \(\set b\), \(\set c\) and \(\set {a,c}\).
Solution.
\( \seteqsection {1} \)\begin{align*} \set a’ & =\set {b,c}\\ \set b’ & =\set c\\ \set c’ & =\emp \\ \set {a,c}’ & =\set {b,c} \end{align*} □
Let \(U\) be open in a topological space \(X\). Prove that
\[ \ob U=\ob {\inter \cur {\ob U}}. \]
Solution. \(U\) is open and \(U\sub \ob U\) so \(U\sub \inter \of {\ob U}\), which implies that \(\ob U\sub \ob {\inter \of {\ob U}}\). Since \(\inter \of {\ob U}\sub \ob U\) and since \(\ob U\) is closed, it follows that \(\ob {\inter \of {\ob U}}\sub \ob U\). Thus \(\ob U=\ob {\inter \cur {\ob U}}\). □
Let \(X\) be a topological space and \(G\sub X\). Prove that \(G\) is open if and only if
\[ \ob {G\cap \ob A}=\ob {G\cap A} \]
for every \(A\sub X\).
Solution. Assume that \(G\) is open. Since \(A\sub \ob A\), it follows that \(G\cap A\sub G\cap \ob A\), which implies that \(\ob {G\cap A}\sub \ob {G\cap \ob A}\).
Let \(x\in \ob {G\cap \ob A}\). If \(U\) is an open nbhd of \(x\), then \(U\cap \cur {G\cap \ob A}\neq \emp \). Let \(y\in U\cap \cur {G\cap \ob A}\). Then \(U\cap G\) is an open nbhd of \(y\) and \(y\in \ob A\) so \(U\cap G\cap A\neq \emp \). Since any open nbhd of \(x\) has a nonempty intersection with \(G\cap A\), it follows that \(x\in \ob {G\cap A}\). Thus \(\ob {G\cap \ob A}\sub \ob {G\cap A}\). Therefore \(\ob {G\cap \ob A}=\ob {G\cap A}\).
Now assume that \(\ob {G\cap \ob A}=\ob {G\cap A}\) for every \(A\sub X\). We aim to show that \(G\) is open. Suppose, for a contradiction, that \(G\) is not open. Then \(A:=X\sem G\) is not closed, so \(\ob A\cap G\neq \emp \). Then \(\ob {G\cap A}=\emp \), but \(\ob {G\cap \ob A}\neq \emp \), which is a contradiction. □
Let \(X\) be an infinite set with the cofinite topology. Prove that if \(A\sub X\) is infinite, then every point in \(A\) is a limit point of \(A\) and that if \(A\) is finite, then it has no limit points.
Solution. Assume that \(A\sub X\) is infinite. If \(x\in A\) and \(U\) is an open nbhd of \(x\), then \(X\sem U\) is finite so \(U\cap A\) is infinite and hence contains an element of \(A\) distinct from \(x\). Thus \(x\) is a limit point of \(A\).
Assume that \(A\) is finite. If \(x\in X\) then \(U:=\cur {X\sem A}\cup \set x\) is an open nbhd of \(x\) such that \(U\cap A\sub \set x\). Thus \(x\) is not a limit point of \(A\). □