Let \(\cur {X,\st }\) be a topological space and \(Y\sub X\). Let
\[ \st ’:=\set {U\cap Y:U\in \st }. \]
Then \(\st ’\) is a topology on \(Y\).
Proof. Since \(\emp \in \st \), we have
\[ \emp =\emp \cap Y\in \st ’. \]
Since \(X\in \st \) and \(Y=X\cap Y\), it follows that \(Y\in \st ’\).
Let \(\sa \sub \st ’\). Then for each \(U\in \sa \), there is \(V_{U}\in \st \) with \(U=Y\cap V_{U}\). Since
\[ \bigcup _{U\in \sa }V_{U}\in \st \]
and
\[ \bigcup \sa =\bigcup _{U\in \sa }\cur {Y\cap V_{U}}=Y\cap \cur {\bigcup _{U\in \sa }V_{U}}, \]
it follows that \(\bigcup \sa \in \st ’\).
Let \(U,V\in \st ’\). There are \(U’,V’\in \st \) with \(U=Y\cap U’\) and \(V=Y\cap V’\). Then
\[ U\cap V=Y\cap \cur {U’\cap V’} \]
and \(U’\cap V’\in \st \) so \(U\cap V\in \st ’\). □
Let \(\cur {X,\st }\) be a topological space and \(Y\sub X\). The topology
\[ \st ’:=\set {U\cap Y:U\in \st } \]
is called the relative topology (or subspace topology) on \(Y\).
Let \(\cur {Y,\st ’}\) be a subspace of a topological space \(\cur {X,\st }\). If \(Z\sub Y\) then the subspace topology on \(Z\) induced by \(\st \) is the same as the subspace topology induced by \(\st ’\).
Let \(X\) be a topological space and \(Y\sub X\). Consider \(Y\) as a topological space with the subspace topology. Then \(C\sub Y\) is closed in \(Y\) if and only if there is a closed \(C’\) in \(X\) with \(C=C’\cap Y\).
Proof. Assume that \(C\) is closed in \(Y\). Then \(Y\sem C\) is open in \(Y\) so there is open \(U\) in \(X\) with
\[ Y\sem C=U\cap Y. \]
Let \(C’:=X\sem U\). Then \(C’\) is closed in \(X\) and \(C=C’\cap Y\).
Now assume that there is a closed \(C’\) in \(X\) with \(C=C’\cap Y\). Then \(X\sem C’\) is open in \(X\) so
\[ Y\sem C=\cur {X\sem C’}\cap Y \]
is open in \(Y\) implying that \(C\) is closed in \(Y\). □
Let \(\cur {X,d}\) be a metric space and \(Y\sub X\). Then the restriction \(d’\) of \(d\) to \(Y\tm Y\) is a metric that induces the relative topology on \(Y\).
Proof. Let \(\st \) be the topology on \(X\) induced by \(d\), let \(\st ’\) be the corresponding subspace topology on \(Y\). We show that \(\st ’\) is induced by the metric \(d’\). Let \(U\in \st ’\) and \(x\in U\). There is \(V\in \st \) with \(U=V\cap Y\). Since \(x\in V\), there is \(\eps >0\) with \(B_{d}\of {x,\eps }\sub V\). Then
\[ B_{d’}\of {x,\eps }=B_{d}\of {x,\eps }\cap Y\sub U, \]
implying that \(U\) is open in the metric space \(\cur {Y,d’}\).
Now assume that \(U\) is open in the metric space \(\cur {Y,d’}\). For each \(x\in U\) there is \(\eps _{x}>0\) with
\[ B_{d’}\of {x,\eps _{x}}\sub U. \]
Since
\[ B_{d’}\of {x,\eps _{x}}=B_{d}\of {x,\eps _{x}}\cap Y, \]
it is open in the subspace topology on \(Y\). Thus
\[ U=\bigcup _{x\in U}B_{d’}\of {x,\eps _{x}}\in \st ’ \]
as required. □
Let \(X\) be a topological space and \(Y\sub X\) be a subspace of \(X\). If \(Y\) is open in \(X\), then for any \(A\sub Y\), the set \(A\) is open in \(Y\) if and only if it is open in \(X\). If \(Y\) is closed in \(X\), then any \(A\sub Y\) is closed in \(Y\) if and only if it is closed in \(X\).
Proof. Assume that \(Y\) is open in \(X\). Let \(A\sub Y\). If \(A\) is open in \(X\), then \(A=A\cap Y\) is open in \(Y\). If \(A\) is open in \(Y\), then \(A=U\cap Y\) for some \(U\sub X\) that is open in \(X\). Then \(A\) is open in \(X\).
If \(Y\) is closed in \(X\), the proof is similar. □
Let \(X\) be a topological space and \(Y\sub X\) be a subspace.
1. If \(\ss \) is a subbasis of the topology on \(X\) then
\[ \ss ’:=\set {S\cap Y:S\in \ss } \]
is a subbasis for the topology on \(Y\).
2. If \(\sb \) is a basis of the topology on \(X\) then
\[ \sb ’:=\set {B\cap Y:B\in \sb } \]
is a basis for the topology on \(Y\).
3. If \(\sb _{x}\) is a nbhd basis at \(x\in X\) in \(X\) and if \(x\in Y\), then
\[ \sb _{x}’:=\set {B\cap Y:B\in \sb _{x}} \]
is a nbhd basis at \(x\) in \(Y\).
Let \(X\) be a topological space and \(Y\sub X\). For \(A\sub Y\), let \(\ob A_{X}\) and \(A’_{X}\) denote the closure and the derived set of \(A\), respectively, relative to the topology on \(X\) and let \(\ob A_{Y}\) and \(A’_{Y}\) denote the closure and the derived set of \(A\), respectively, relative to the topology on \(Y\). Then
\[ \ob A_{Y}=\ob A_{X}\cap Y \]
and
\[ A’_{Y}=A’_{X}\cap Y, \]
for every \(A\sub Y\).
Proof. Let \(A\sub Y\). Assume that \(x\in \ob A_{Y}\). Then \(x\in Y\). To show that \(x\in \ob A_{X}\), let \(U\) be an open nbhd of \(x\) in \(X\). Then \(U’:=U\cap Y\) is an open nbhd of \(x\) in \(Y\) so \(U’\cap A\neq \emp \). It follows that \(U\cap A\neq \emp \) as required.
Assume that \(x\in \ob A_{X}\cap Y\). Let \(U\) be an open nbhd of \(x\) in \(Y\). Then there is open \(U’\) in \(X\) with \(U=U’\cap Y\). Since \(x\in \ob A_{X}\), we have \(U’\cap A\neq \emp \). Since \(A\sub Y\), it follows that \(U\cap A\neq \emp \). Thus \(x\in \ob A_{Y}\).
The equality for derived sets is proved in a similar way. □
Let \(X\) be a topological space and \(Y\sub X\). For \(A\sub Y\), let \(A_{X}^{\circ }\) and \(\pt A_{X}\) denote the interior and the boundary of \(A\), respectively, relative to the topology on \(X\) and let \(A_{Y}^{\circ }\) and \(\pt A_{Y}\) denote the interior and the boundary of \(A\), respectively, relative to the topology on \(Y\). Then
\[ A_{Y}^{\circ }\sur A_{X}^{\circ }\cap Y \]
and
\[ \pt A_{Y}\sub \pt A_{X}\cap Y, \]
for every \(A\sub Y\).
Proof. Let \(A\sub Y\). The inclusion \(A_{Y}^{\circ }\sur A_{X}^{\circ }\cap Y\) holds since if \(x\in A_{X}^{\circ }\cap Y\), then there is an open \(U\) in \(X\) with \(x\in U\sub A\). Since \(U\) is open in \(Y\), it follows that \(x\in A_{Y}^{\circ }\).
Let \(x\in \pt A_{Y}\). Then \(x\in \ob A_{Y}\) and \(x\in Y\). Since
\[ \ob {Y\sem A}_{Y}=\ob {Y\sem A}_{X}\cap Y\sub \ob {X\sem A}_{X}\cap Y \]
and \(x\in \ob {X\sem A}_{X}\), it follows that \(x\in \ob {Y\sem A}_{Y}\). □
Let \(X=\br \) and \(Y=\set 0\) with \(A=\set 0\). Then
\[ A_{Y}^{\circ }=\set 0\neq A_{X}^{\circ }\cap Y=\emp \]
and
\[ \pt A_{Y}=\emp \neq \pt A_{X}\cap Y=\set 0. \]
Let \(X\) be a topological space with an order topology induced by a linear order \(\le \) on \(X\) and let \(Y\sub X\). Let \(\st \) be the subspace topology on \(Y\) and \(\st ’\) be the order topology on \(Y\) induced by the restriction \(\le ’\) of \(\le \) to \(Y\tm Y\). Then \(\st ’\sub \st \). If moreover \(Y\) is an interval, then equality holds.
Proof. Let \(V\in \st ’\). If \(x\in V\) then there are \(a,b\in Y\cup \set {-\ity ,\ity }\) with
\[ x\in V_{x}=\cur {a,b}_{Y}\sub V, \]
where
\[ \cur {a,b}_{Y}:=\set {y\in Y:a<y<b}. \]
Let
\[ U_{x}:=\cur {a,b}_{X}:=\set {y\in X:a<y<b} \]
and \(U=\bigcup _{x\in V}U_{x}\). Then \(U\) is open in \(X\). Since \(V_{x}=U_{x}\cap Y\) for each \(x\in V\), we have
\[ V=\bigcup _{x\in V}V_{x}=\cur {\bigcup _{x\in V}U_{x}}\cap Y=U\cap Y, \]
so \(V\in \st \) as required.
Assume that \(Y\) is an interval and \(V\in \st \). Let \(U\) be open in \(X\) with \(V=U\cap Y\). If \(x\in V\), then there are \(a,b\in X\cup \set {-\ity ,\ity }\) with \(x\in \cur {a,b}_{X}\sub U\). Define \(a’:=a\) if \(a\in Y\) and \(a’:=-\ity \) otherwise. Let \(b’:=b\) if \(b\in Y\) and \(b:=\ity \) otherwise. Since \(Y\) is an interval, it follows that
\[ V_{x}:=\cur {a’,b’}_{Y}=\cur {a,b}_{X}\cap Y\sub V, \]
so
\[ V=\bigcup _{x\in V}V_{x}\in \st ’ \]
as required. □
Let
\[ X:=\set 0\cup \cur {1,2}\sub \br . \]
Then \(\set 0\) is open in subspace topology on \(X\) induced from the topology on \(\br \). However, if \(X\) is equipped with the order topology induced by restricting the linear order on \(\br \) to \(X\), then \(\set 0\) is not open.
A subset \(Y\) of a topological space is called discrete if the relative topology on \(Y\) is discrete. Prove that every subset of a discrete space is discrete. Prove that the subset \(\set {1/n:n\in \bn }\) of the real line \(\br \) with the standard topology is discrete and the subset \(\set 0\cup \set {1/n:n\in \bn }\) is not discrete.
Solution. Let \(X\) be a discrete topological space and \(Y\sub X\). If \(A\sub Y\), then \(A\) is open in \(X\) and \(A=A\cap Y\) so \(A\) is open in \(Y\). Since every subset of \(Y\) is open in \(Y\), the relative topology on \(Y\) is discrete.
Let \(A:=\set {1/n:n\in \bn }\sub \br \). For each \(n\in \bn \), let \(U_{n}:=\cur {\dfrac {1}{n+1},\dfrac {1}{n-1}}\) provided \(n\ge 2\) and \(U_{1}:=\cur {\dfrac {1}{2},2}\). For each \(n\in \bn \) the set \(U_{n}\) is open in \(\br \) and \(U_{n}\cap \br =\set {\dfrac {1}{n}}\). Thus \(\set {\dfrac {1}{n}}\) is open in \(A\) for each \(n\in \bn \), which implies that the relative topology on \(A\) is discrete.
Let \(B:=\set 0\cup \set {1/n:n\in \bn }\sub \br \). To prove that the relative topology on \(B\) is not discrete, we show that the set \(\set 0\) is not open in the relative topology on \(B\). Suppose, for a contradiction that \(\set 0\) is open in \(B\). Then there is open \(U\) in \(\br \) with \(U\cap B=\set 0\). Since \(U\) is open in \(\br \), there are is an open interval \(\cur {a,b}\sub \br \) with \(x\in \cur {a,b}\sub U\). Since \(b>0\), there is \(n\in \bn \) with \(\dfrac {1}{n}<b\). Thus \(\dfrac {1}{n}\in U\cap B\), which is a contradiction. □
Let \(a,b\in \br \sem \bq \) with \(a<b\). Prove that \(\bra {a,b}\cap \bq \) is both open and closed in the relative topology on \(\bq \).
Solution. Since \(a,b\nin \bq \), we have \(\bra {a,b}\cap \bq =\cur {a,b}\cap \bq \). Since \(\cur {a,b}\) is open in \(\br \), it follows that \(\cur {a,b}\cap \bq \) is open in the relative topology on \(\bq \). Thus \(\bra {a,b}\cap \bq \) is open in the relative topology on \(\bq \).
Since \(\bra {a,b}\) is closed in \(\br \), it follows that \(\bra {a,b}\cap \bq \) is closed in the relative topology on \(\bq \). □
Let \(X\) be a topological space such that every finite subspace of \(X\) has the trivial relative topology. Prove that the topology on \(X\) is trivial.
Solution. Suppose, for a contradiction, that the topology of \(X\) is non-trivial. Let \(U\sub X\) be open in \(X\) with \(U\nin \set {\emp ,X}\). Let \(x\in U\) and \(y\in X\sem U\). Then \(\set x=U\cap \set {x,y}\) so \(\set x\) is open in the relative topology on \(\set {x,y}\), which means that the relative topology on \(\set {x,y}\) is not trivial. This is a contradiction. □
Let \(X\) be a topological space such that every finite subspace of \(X\) has the discrete topology. Does it follow that \(X\) has the discrete topology? Give a proof or a counterexample.
Solution. No. Here is a counterexample. Let \(X\) be \(\br \) with the standard topology. If \(A\sub \br \) is finite, then the relative topology on \(A\) is discrete. However the topology on \(\br \) is not discrete. □