Let \(X\) be a set and \(\ba \) be a nonempty family of topologies on \(X\). Then \(\bigcap \ba \) is a topology on \(X\).
Proof. Both \(X\) and \(\emp \) belong to \(\st \) for every \(\st \in \ba \). Thus \(X,\emp \in \bigcap \ba \).
Assume that \(\sa \sub \bigcap \ba \). Then \(\sa \sub \st \) for every \(\st \in \ba \) so \(\bigcup \sa \in \st \) for every \(\st \in \ba \). Thus \(\bigcup \sa \in \bigcap \ba \).
Assume that \(\sa \sub \bigcap \ba \) is nonempty and finite. Then \(\sa \in \st \) for every \(\st \in \ba \) so \(\bigcap \sa \in \st \) for every \(\st \in \ba \). Thus \(\bigcap \sa \in \bigcap \ba \). □
Let \(X\) be a set and \(\ss \) be any family of subsets of \(X\). Define
\[ \st \of {\ss }:=\bigcap \ba , \]
where \(\ba \) is the family of all topologies \(\st \) on \(X\) such that \(\ss \sub \st \). The set \(\ss \) is called the subbasis of the topological space \(\cur {X,\st \of {\ss }}\).
Note that any family of subsets of \(X\) is a subbasis of a unique topology on \(X\). For a given topology on \(X\) there are usually many possible subbases.
Let \(X\) be a set and \(\ss \) be a family of subsets of \(X\). The topology \(\st \of {\ss }\) consists of \(X,\emp \) and all unions of all possible intersections of nonempty finite subfamilies of \(\ss \).
Proof. It is clear that \(X,\emp \in \st \of {\ss }\) since \(X,\emp \) belong to any topology on \(X\). Moreover, any topology on \(X\) containing \(\ss \) contains all unions of all possible intersections of nonempty finite subfamilies of \(\ss \).
To complete the proof, it suffices to show that the family \(\st \) of all unions of all possible intersections of nonempty finite subfamilies of \(\ss \) together with \(X\) and \(\emp \) is a topology on \(X\). It is clear that \(X,\emp \in \st \).
Let \(\sa \sub \st \). If \(X\in \sa \), then \(\bigcup \sa =X\in \st \). Otherwise, \(\bigcup \sa =\bigcup \sa ’\), where \(\sa ’=\sa \sem \set {\emp }\). Every member of \(\sa ’\) is a union of intersection of nonempty finite subfamilies of \(\ss \), which implies that \(\bigcup \sa ’\) is such a union. Thus \(\bigcup \sa \in \st \).
Let \(\sa \sub \st \) be finite and nonempty. If \(\emp \in \sa \), then \(\bigcap \sa =\emp \in \st \). Otherwise, let \(\sa ’=\sa \sem \set X\). If \(\sa ’=\emp ,\)then \(\bigcap \sa =X\in \st \). If \(\sa ’\neq \emp \), then each member of \(\sa ’\) is a union of intersections of nonempty finite subfamilies of \(\ss \). Let \(\sa ’=\set {A_{1},A_{2},\ds ,A_{n}}\) with
\[ A_{i}=\bigcup _{j\in J_{i}}A_{i,j}, \]
where \(A_{i,j}\) is the intersection of some nonempty finite subfamily of \(\ss \) and \(J_{i}\) is some set for each \(i=1,2,\ds ,n\). Then
\( \seteqsection {1} \)\begin{align*} \bigcap \sa ’ & =\cur {\bigcup _{j\in J_{1}}A_{1,j}}\cap \cur {\bigcup _{j\in J_{2}}A_{2,j}}\cap \ds \cap \cur {\bigcup _{j\in J_{n}}A_{n,j}}\\ & =\bigcup _{f\in J}\cur {A_{1,f\of 1}\cap A_{2,f\of 2}\cap \ds \cap A_{n,f\of n}}, \end{align*} where \(J\) is the set of all functions \(f\) on \(\set {1,2,\ds ,n}\) with \(f\of i\in J_{i}\) for every \(i\). Thus \(\bigcap \sa \) is a union of intersections of nonempty finite subfamilies of \(\ss \) and so \(\bigcap \sa \in \st \). □
Let \(X\) be a set and \(\le \) be a binary relation on \(X\). We say that \(<\) is a linear order on \(X\) if
1. \(\le \) is reflexive (\(x\le x\) for each \(x\in X\)).
2. \(\le \) is transitive (\(x\le y\) and \(y\le z\) implies that \(x\le z\) for each \(x,y,z\in X\)).
3. \(\le \) is antisymmetric (\(x\le y\) and \(y\le x\) implies that \(x=y\) for every \(x,y\in X\)).
4. \(\le \) is total (\(x\le y\) or \(y\le x\) for every \(x,y\in X\)).
The standard order on \(\br \) is a linear order.
Let \(X\) be a set with a linear order \(\le \). For \(x\in X\), let
\[ \cur {-\ity ,x}=\set {y\in X:y\le x\text { and }y\neq x} \]
and
\[ \cur {x,\ity }=\set {y\in X:x\le y\text { and }y\neq x}. \]
The order topology on \(X\) induced by \(\le \) is the topology \(\st \of {\ss }\), where
\[ \ss =\set {\cur {-\ity ,x}:x\in X}\cup \set {\cur {x,\ity }:x\in X}. \]
The standard topology on \(\br \) is the order topology on \(\br \) induced by the standard order.
Consider \(\bn \) with the standard order. Prove that the resulting order topology on \(\bn \) is discrete.
Solution. Let \(n\in \bn \). If \(n=1\), then \(\set n=\cur {-\ity ,2}\) is open in the order topology. If \(n\ge 2\), then
\[ \set n=\cur {-\ity ,n+1}\cap \cur {n-1,\ity } \]
so \(\set n\) is open as well. Since every singleton \(\set n\) is open, the topology is discrete. □
Consider the set \(X:=\set {1,2}\tm \bn \) with the dictionary order, that is such that \(\ang {a,b}\le \ang {c,d}\) when \(a<c\) or (\(a=c\) and \(b\le d\)). Prove that the resulting order topology on \(X\) is not discrete.
Solution. We show that \(A:=\set {\ang {2,1}}\) is not open. Suppose, for a contradiction, that \(A\) is open. Then \(A=\bigcup \sk \) for some family \(\sk \) consisting of intersections of finite nonempty subfamilies of the subbasis
\[ \ss =\set {\cur {-\ity ,x}:x\in X}\cup \set {\cur {x,\ity }:x\in X}. \]
Let \(K\in \sk \) be such that \(A\sub K\). Then \(K\) must contain \(\ang {2,1}\) so
\( \seteqsection {1} \)\begin{align*} K & =\cur {-\ity ,\ang {2,k_{1}}}\cap \ds \cap \cur {-\ity ,\ang {2,k_{s}}}\\ & \qquad \cap \cur {\ang {1,k_{s+1}},\ity }\cap \ds \cap \cur {\ang {1,k_{t}},\ity } \end{align*} for some \(s,t\in \bn \) with \(0\le s\le t.\) Let \(\ell :=\max \set {k_{s+1},\ds ,k_{t}}+1\). Then \(\ang {1,\ell }\in K\) so \(\ang {1,\ell }\in A\), which is a contradiction. □
Describe the topology on the plane for which the family of all straight lines is a subbasis.
Solution. The intersection of two lines that are not parallel is a singleton. Any singleton on the plane can be represented as the intersection of two lines. Thus each singleton is open and so the topology is discrete. □
For each \(q\in \bq \), let \(A_{q}:=\set {x\in \br :x>q}\) and \(B_{q}:=\set {x\in \br :x<q}\). Prove that the set
\[ \ss :=\set {A_{q}:q\in \bq }\cup \set {B_{q}:q\in \bq } \]
is a subbasis for the standard topology on \(\br \).
Solution. Let \(\st :=\st \of {\ss }\) and \(\st ’\) be the standard topology on \(\br \). Since \(\ss \sub \st ’\), it follows that \(\st \sub \st ’\). It remains to show that \(\st ’\sub \st \).
Let \(U\in \st ’\). For each \(x\in U\) there are rational \(p,q\) with \(p<x<q\) and \(J_{x}:=\cur {p,q}\sub U\). Then
\[ U=\bigcup _{x\in U}J_{x} \]
and \(J_{x}=A_{p}\cap B_{q}\in \st \) for every \(x\in U\). Thus \(U\in \st \). □
A set \(X\) is well-ordered by \(\le \) if \(\le \) is a linear order on \(X\) and for every nonempty \(A\sub X\) there is \(a\in A\) such that \(a\le b\) for every \(b\in A\).
The axioms of set theory imply that every set can be well-ordered.
Let \(X\) be any uncountable set and \(\le \) be a well-ordering of \(X\). If the set
\[ A=\set {x\in X:\set {y\in X:y\le x}\text { is uncountable}} \]
is nonempty, let \(\ome \) be the smallest element of \(A\) and
\[ \bra {0,\ome }:=\set {x\in X:x\le \ome }. \]
Otherwise, let
\[ \bra {0,\ome }:=X\cup \set {\ome } \]
with \(\le \) extended to \(\bra {0,\ome }\) by declaring that \(x\le \ome \) for every \(x\in \bra {0,\ome }\).
We will consider \(\bra {0,\ome }\) as a topological space with the order topology.
The set \(\bra {0,\ome }\) is an uncountable well-ordered set such that for every \(x\in \bra {0,\ome }\) that is strictly smaller than \(\ome \), the set
\[ \set {y\in \bra {0,\ome }:y\le x} \]
is countable. Moreover, if \(X\) is any well-ordered set with the largest element \(\ome ’\) such that for every \(x\in X\) that is strictly smaller than \(\ome ’\), the set
\[ \set {y\in X:y\le x} \]
is countable, then there is a bijection \(\ph :X\to \bra {0,\ome }\) such that for every \(x,y\in X\) with \(x\le y\), we have \(\ph \of x\le \ph \of y\).
Proof. The set \(\bra {0,\ome }\) has the required properties directly from the definition. The proof of the existence of the bijection \(\ph \) is omitted. □
Let \(X\) be a topological space. A basis for the topology on \(X\) is a family \(\sb \) of open subsets of \(X\) such that for every open set \(U\) and \(x\in U\) there is \(B\in \sb \) with \(x\in B\sub U\).
\(\sb \) is a basis for the topology \(\st \) on \(X\) if and only if \(\sb \sub \st \) and every open set is a union of members of \(\sb \). Any basis for the topology on \(X\) is also a subbasis.
In a discrete space \(X\) the family of all singletons \(\set x\) is a basis. In a metric space \(X\) the collection of all open balls \(B\of {x,r}\) for \(x\in X\) and \(r>0\) is a basis.
Let \(\ss \) be a subbasis of the topology on \(X\). Then the family \(\sb \) consisting of \(X\) and all intersections of finite nonempty subfamilies of \(\ss \) is a basis for the topology on \(X\).
Let \(X\) be the topological space having the order topology induced by a linear order \(\le \). If for \(a,b\in X\) we define
\[ \cur {a,b}:=\set {x\in X\sem \set {a,b}:a\le x\le b}, \]
then
\[ \sb :=\set {\cur {a,b}:a,b\in X}\cup \set {\cur {-\ity ,a}:a\in X}\cup \set {,\cur {a,\ity }:a\in X} \]
is a basis for the topology on \(X\).
Consider the topological space \(\bra {0,\ome }\). Let \(S\) be the set of all successor elements in \(\bra {0,\ome }\), that is let \(x\in S\) if the set
\[ \set {y\in \bra {0,\ome }:y<x} \]
has the largest element. Let \(S’:=S\cup \set 0\), where \(0\) is the smallest element of \(\bra {0,\ome }\) and \(L:=\bra {0,\ome }\sem S’\). Define
\[ \sb :=\set {\set a:a\in S’}\cup \set {\ocl {a,b}:a<b,\,b\in L}. \]
Then \(\sb \) is a basis for the topology on \(\bra {0,\ome }\).
Let \(X\) be a set and \(\sb \) be a collection of subsets of \(X\). Then \(\sb \) is a basis for some topology on \(X\) if and only if the following conditions hold:
1. \(\bigcup \sb =X\) and
2. for every \(B_{1},B_{2}\in \sb \) and every \(x\in B_{1}\cap B_{2}\) there is \(B\in \sb \) with \(x\in B\sub B_{1}\cap B_{2}\).
Proof. Assume that \(\sb \) is a basis for a topology \(\st \) on \(X\). Since \(X\in \st \), it follows that 1. holds. To prove 2., assume that \(B_{1},B_{2}\in \sb \) and \(x\in B_{1}\cap B_{2}\). Since \(B_{1},B_{2}\in \st \), it follows that \(B_{1}\cap B_{2}\in \st \) so there is \(B\in \sb \) with \(x\in B\sub B_{1}\cap B_{2}\). Thus 2. holds.
Now assume that 1. and 2. hold. Let \(\st \) be the family of all unions of subfamilies of \(\sb \). Then 1. implies that \(X\in \st \) and \(\emp \) is the union of the empty family so \(\emp \in \st \). The family \(\st \) is closed under taking unions since the union of unions of subfamilies of \(\sb \) is also a union of subfamilies of \(\sb \).
Let \(U,V\in \st \). We show that \(U\cap V\in \st \). Assume
\[ U=\bigcup _{\ga \in A}B_{\ga } \]
and
\[ V=\bigcup _{\ga \in C}D_{\ga }, \]
where \(A,C\) are some sets and \(B_{\ga },D_{\gb }\in \sb \) for each \(\ga \in A\) and \(\gb \in C\). We have
\[ U\cap V=\bigcup _{\cur {\ga ,\gb }\in A\tm C}B_{\ga }\cap D_{\gb }. \]
For each \(\cur {\ga ,\gb }\in A\tm C\) and each \(x\in B_{\ga }\cap D_{\gb }\) let \(G_{\ga ,\gb ,x}\in \sb \) be such that
\[ x\in G_{\ga ,\gb ,x}\sub B_{\ga }\cap D_{\gb }. \]
Then
\[ U\cap V=\bigcup _{\cur {\ga ,\gb }\in A\tm C}\bigcup _{x\in B_{\ga }\cap D_{\gb }}G_{\ga ,\gb ,x} \]
so \(U\cap V\in \st \).
We have proved that \(\st \) is a topology on \(X\). Clearly, \(\sb \sub \st \). To show that \(\sb \) is a basis for \(\st \), let \(U\in \st \) and \(x\in U\). There is \(B\in \sb \) with \(x\in B\sub U\). □
Let \(A:=\set {\dfrac {1}{n}:n\in \bn }\). Let \(\sb _{1}\) be the collection of open intervals in \(\br \) and \(\sb _{2}\) be the collection of all subsets of \(\br \) that are of the form \(\cur {a,b}\sem A\) for \(a,b\in \br \) with \(a<b\). Prove that \(\sb :=\sb _{1}\cup \sb _{2}\) is a basis for a topology on \(\br \) and that the sequence \(\cur {\dfrac {1}{n}}_{n\in \bn }\) does not converge to \(0\) in this topology.
Solution. For each \(x\in \br \) we have \(x\in \cur {x-1,x+1}\in \sb _{1}\). Thus \(\bigcup \sb =\br \). Let \(B_{1},B_{2}\in \sb \) with \(x\in B_{1}\cap B_{2}\). We need to find \(B\in \sb \) with \(x\in B\). There are \(a_{1},a_{2},b_{1}\) and \(b_{2}\)such that \(B_{1}=\cur {a_{1},b_{1}}\) or \(B_{1}=\cur {a_{1},b_{1}}\sem A\) and \(B_{2}=\cur {a_{2},b_{2}}\) or \(B_{2}=\cur {a_{2},b_{2}}\sem A\). If \(x\in A\), then \(B_{1}=\cur {a_{1},b_{1}}\) and \(B_{2}=\cur {a_{2},b_{2}}\) so \(B=B_{1}\cap B_{2}\) satisfies the requirements. If \(x\nin A\), then
\[ B=\cur {\cur {a_{1},b_{1}}\cap \cur {a_{2},b_{2}}}\sem A \]
satisfies the requirements. Thus \(\sb \) is a basis for the topology on \(\br \).
Now we show that \(\cur {\dfrac {1}{n}}_{n\in \bn }\) does not converge to \(0\) in this topology. Suppose, for a contradiction, that it does converge to \(0\). Then for every nbhd \(U\) of \(0\) there is \(k\in \bn \) with \(\dfrac {1}{n}\in U\) for every \(n\ge k\). In particular, this holds when \(U=\cur {-1,1}\sem A\). However, for such \(U\) we have \(\dfrac {1}{n}\nin U\) for all \(n\in \bn \) so we have a contradiction. □
Let \(\sb :=\set {\cur {x,\ity }:x\in \br }\). Prove that \(\sb \) is a basis of a topology on \(\br \) and find the closures of \(A:=\set {\dfrac {1}{n}:n\in \bn }\) and \(B:=\bn \) in this topology.
solution. We have \(\bigcap \sb =\br \) since for every \(x\in \br \) we have \(x\in \cur {x-1,\ity }\). If \(B_{1},B_{2}\in \sb \), then there are \(b_{1},b_{2}\in \br \) such that \(B_{1}=\cur {b_{1},\ity }\) and \(B_{2}=\cur {b_{2},\ity }\) and let \(x\in B_{1}\cap B_{2}\). Let \(b=\max \set {b_{1},b_{2}}\). Then \(\cur {b,\ity }\in \sb \) and \(x\in \cur {b,\ity }\sub B_{1}\cap B_{2}\) as required. Thus \(\sb \) is a basis for a topology on \(\br \).
The closure of \(A\) is this topology is \(\ocl {-\ity ,1}\) and the closure of \(B\) is \(\br \). □
Consider \(\br \) with the topology generated by the basis \(\sb :=\set {\clo {a,b}:a,b\in \bq }\). Find the boundary, closure and interior of the subsets \(\cur {0,\sqrt {2}}\) and \(\cur {\sqrt {3},4}\) of \(\br \).
Solution. The set \(\cur {0,\sqrt {2}}\) is equal to it’s interior. It’s closure is \(\bra {0,\sqrt {2}}\) and the boundary is \(\set {0,\sqrt {2}}\).
The set \(\cur {\sqrt {3},4}\) is also equal to it’s interior. It’s closure is \(\clo {\sqrt {3},4}\) and the boundary is \(\set {\sqrt {3}}\). □
Let \(\sb \) be a basis of the topological space \(X\) and \(A\sub X\). Prove that \(x\in \ob A\) if and only if \(B\cap A\neq \emp \) for every \(B\in \sb \) such that \(x\in B\).
Solution. Assume that \(x\in \ob A\). Let \(B\in \sb \) with \(x\in B\). Since \(B\) is an open nbhd of \(x\), it follows that \(B\cap A\neq \emp \).
Now assume that \(B\cap A\neq \emp \) for evey \(B\in \sb \) such that \(x\in B\). Let \(U\) be any open nbhd of \(x\). Then there is \(B\in \sb \) with \(x\in B\sub U\). Since \(B\cap A\neq \emp \), it follows that \(U\cap A\neq \emp \). Thus \(x\in \ob A\). □
Let \(\st \) and \(\st ’\) be topologies on the set \(X\) generated by bases \(\sb \) and \(\sb ’\), respectively. Then \(\st ’\) is finer than \(\st \) if and only if for every \(B\in \sb \) and every \(x\in B\) there is \(B’\in \sb ’\) with \(x\in B’\sub B\).
Proof. Assume that \(\st \sub \st ’\). If \(B\in \sb \) and \(x\in \sb \), then \(B\in \st \) so \(B\in \st ’\) and there is \(B’\in \sb ’\) with \(x\in B’\sub B\).
Assume that every \(B\in \sb \) and every \(x\in B\) there is \(B’\in \sb ’\) with \(x\in B’\sub B\). Let \(U\in \st \). For every \(x\in U\), there is \(B_{x}\in \sb \) with \(x\in B_{x}\sub U\). By assumption, there is \(B_{x}’\sub B_{x}\) with \(x\in B_{x}’\). Thus
\[ U=\bigcup _{x\in U}B_{x}’\in \st ’. \]
Therefore \(\st \sub \st ’\). □
Two metrics on the same set \(X\) are equivalent if they induce the same topology on \(X\).
The metrics \(d\) and \(d’\) on a set \(X\) are equivalent if and only if for each \(x\in X\) and each \(\eps >0\) there are \(\gd _{1},\gd _{2}>0\) such that
\[ B_{d}\of {x,\gd _{1}}\sub B_{d’}\of {x,\eps } \]
and
\[ B_{d’}\of {x,\gd _{2}}\sub B_{d}\of {x,\eps }. \]
Proof. Let \(\st \) and \(\st ’\) be the topologies induced by \(d\) and \(d’\), respectively. Assume that \(d\) and \(d’\) are equivalent. Then \(\st =\st ’\). Let \(x\in X\) and \(\eps >0\). Since \(B_{d’}\of {x,\eps }\in \st ’\), it follows that \(B_{d’}\of {x,\eps }\in \st \) so there is \(\gd _{1}>0\) with
\[ B_{d}\of {x,\gd _{1}}\sub B_{d’}\of {x,\eps }. \]
Similarly, there is \(\gd _{2}>0\) with
\[ B_{d’}\of {x,\gd _{2}}\sub B_{d}\of {x,\eps }. \]
Now assume that for each \(x\in X\) and each \(\eps >0\) there are \(\gd _{1},\gd _{2}>0\) such that
\[ B_{d}\of {x,\gd _{1}}\sub B_{d’}\of {x,\eps } \]
and
\[ B_{d’}\of {x,\gd _{2}}\sub B_{d}\of {x,\eps }. \]
Let \(U\in \st \). For each \(x\in U\) there is \(\eps _{x}>0\) with \(B_{d}\of {x,\eps _{x}}\sub U\). For each \(x\in U\), let \(\gd _{x}\) be such that
\[ B_{d’}\of {x,\gd _{x}}\sub B_{d}\of {x,\eps }. \]
Then
\[ U=\bigcup _{x\in U}B_{d’}\of {x,\gd _{x}}\in \st ’. \]
Thus \(\st \sub \st ’\). Similarly \(\st ’\sub \st \). □
Let \(\cur {X,d}\) be a metric space. For each \(\gl >0\), there is a metric \(d_{\gl }\) that is equivalent to \(d\) such that the diameter of \(X\) in \(d_{\gl }\) is at most \(\gl \).
Proof. Define
\[ d_{\gl }\of {x,y}:=\min \set {\gl ,d\of {x,y}} \]
for every \(x,y\in X\). Then \(d_{\gl }\) is a metric on \(X\). Indeed, the positivity and symmetry of \(d_{\gl }\) are clear and the triangle inequality holds since otherwise there are \(x,y,z\in X\) with
\[ d_{\gl }\of {x,y}+d_{\gl }\of {y,z}<d_{\gl }\of {x,z} \]
and, since \(d_{\gl }\of {x,z}\le d\of {x,z}\), this implies that that
\[ d_{\gl }\of {x,y}+d_{\gl }\of {y,z}<d\of {x,z} \]
so at least one of \(d_{\gl }\of {x,y}\), \(d_{\gl }\of {y,z}\) must be equal \(\gl \) and consequently \(d_{\gl }\of {x,z}>\gl \), which is a contradiction. The diameter of \(X\) in \(d_{\gl }\), which is equal to
\[ \sup \set {d_{\gl }\of {x,y}:x,y\in X} \]
is at most \(\gl \) since \(d_{\gl }\of {x,y}\le \gl \) for every \(x,y\in X\).
It remains to show that the metrics \(d\) and \(d_{\gl }\) are equivalent. Let \(x\in X\) and \(\eps >0\). Taking \(\gd _{1}:=\eps \) and \(\gd _{2}=\min \set {\eps ,\gl }\), we get
\[ B_{d}\of {x,\gd _{1}}\sub B_{d_{\gl }}\of {x,\eps }, \]
since \(d\of {x,y}<\gd _{1}=\eps \) implies that \(d_{\gl }\of {x,y}<\eps \), and
\[ B_{d_{\gl }}\of {x,\gd _{2}}\sub B_{d}\of {x,\eps }, \]
since \(d_{\gl }\of {x,y}<\gd _{2}\) implies that \(d_{\gl }\of {x,y}<\gl \) so \(d_{\gl }\of {x,y}=d\of {x,y}\) and so \(d\of {x,y}<\eps \). It follows that \(d\) and \(d_{\gl }\) are equivalent. □
Let \(X\) be a topological space and \(x\in X\). A nbhd basis (local basis) at \(x\) is a collection \(\sb _{x}\) of nbhds of \(x\) such that each nbhd of \(x\) contains a member of \(\sb _{x}\).
The family of all open nbhds of \(x\) is a nbhd basis at \(x\). In a discrete space, the family consisting of the singleton \(\set x\) is a nbhd basis at \(x\). In a metrics space the set \(\set {B\of {x,\eps }:\eps >0}\) is a nbhd basis at \(x\).
Let \(X\) be a topological space and, for each \(x\in X\), let \(\sb _{x}\) be a nbhd basis at \(x\). Then the following conditions hold for every \(x\in X\):
1. \(\sb _{x}\neq \emp \);
2. \(x\in B\) for every \(B\in \sb _{x}\);
3. for every \(B_{1},B_{2}\in \sb _{x}\) there is \(B\in \sb _{x}\) with \(B\sub B_{1}\cap B_{2}\);
4. for each \(B\in \sb _{x}\) there is \(B’\in \sb _{x}\) such that \(B\) contains a member of \(\sb _{y}\) for every \(y\in B’\).
Proof. Conditions 1.–3. follow easily from conditions 1.–3. of Proposition 1.2.8. We show that 4. holds. Let \(B\in \sb _{x}\). Since \(B\) is a nbhd of \(x\), condition 5. of Proposition 1.2.8 implies that if
\[ \rg B:=\set {y\in B:B\text { is a nbhd of }y} \]
is a nbhd of \(x\). By the definition of a nbhd basis, there is \(B’\in \sb _{x}\) with \(B’\sub \rg B\). Then for every \(y\in B’\), the set \(B\) is a nbhd of \(y\) so contains a member of \(\sb _{y}\). □
Let \(X\) be a set and, for each \(x\in X\), let \(\sb _{x}\) be a family of subsets of \(X\) satisfying conditions 1.–4. of Proposition 1.4.19. Then there exists a unique topology on \(X\) such that \(\sb _{x}\) is a nbhd basis at \(x\) for every \(x\in X\).
Proof. First note that if \(\st \) is a topology on \(X\) such that for each \(x\in X\), the family\(\sb _{x}\) is a nbhd basis at \(x\) relative to \(\st \), then a subset \(U\) of \(X\) is in \(\st \) if and only if \(U\) contains a member of \(\sb _{x}\) for each \(x\in U\). Thus such a topology \(\st \) is unique provided it exists.
Define
\[ \st :=\set {U\sub X:\text {for every }x\in U\text { there is }B\in \sb _{x}\text { with }B\sub U}. \]
We verify that \(\st \) is a topology on \(X\). We have \(\emp \in \st \) since there are no \(x\in \emp \). We have \(X\in \st \) since for every \(x\in X\) the set \(\sb _{x}\) is nonempty.
Let \(\sa \sub \st \) be arbitrary. We show that \(\bigcup \sa \in \st \). Let \(x\in \bigcup \sa \). Then \(x\in U\) for some \(U\in \sa \) so there is \(B\in \sb _{x}\) with \(B\sub U\) hence \(B\sub \bigcup \sa \).
Now let \(U,V\in \st \). We show that \(U\cap V\in \st \). Let \(x\in U\cap V\). Then there are \(B,D\in \sb _{x}\) with \(B\sub U\) and \(D\sub V\). Let \(G\in \sb _{x}\) with \(G\sub B\cap D\). Then \(G\sub U\cap V\).
It remains to show that \(\sb _{x}\) is a nbhd basis at \(x\) for every \(x\in X\) relative to \(\st \). Let \(x\in X\). First we check that any \(B\in \sb _{x}\) is a nbhd of \(x\).
Given \(B\in \sb _{x}\), let
\[ U:=\set {y\in B:\text {there is }D\in \sb _{y}\text { with }D\sub B}. \]
Since \(x\in U\), it suffices to verify that \(U\in \st \). If \(y\in U\), and \(D\in \sb _{y}\) with \(D\sub B\), then 4. implies that there is \(D’\in \sb _{y}\) such that \(D\) contains a member of \(\sb _{z}\) for every \(z\in D’\). Then \(B\) contains a member of \(\sb _{z}\) for every \(z\in D’\), which implies that \(D’\sub U\). Thus for every \(y\in U\) there is \(D’\in \sb _{y}\) with \(D’\sub U\). Hence \(U\in \st \).
If \(N\) be a nbhd of \(x\), then there is \(U\in \st \) with \(x\in U\sub N\) so there is \(B\in \sb _{x}\) with \(B\sub U\). Thus \(B\sub N\). □
Let \(\ss \) be a subbasis for the topology of a space \(X\) and \(D\sub X\) be such that \(U\cap D\neq \emp \) for each \(U\in \ss \). Does it follow that \(D\) is dense in \(X\)? Give a proof or a counterexample.
Solution. No. Here is a counterexample. Let \(X:=\br \),
\[ \ss :=\set {\cur {-\ity ,a}:a\in \br }\cup \set {\cur {a,\ity }:a\in \br } \]
and \(D:=\bz \). Then \(\ss \) is a subbasis for the standard topology on \(X\) and \(U\cap D\neq \emp \) for each \(U\in \ss \). However, \(D\) is not dense in \(X\). □
Let \(\cur {X,d}\) be a metric space. Show that the metric \(d’\), defined by
\[ d’\of {x,y}:=\dfrac {d\of {x,y}}{1+d\of {x,y}} \]
is equivalent to \(d\).
Solution. Let \(x\in X\) and \(\eps >0\). We find \(\gd _{1},\gd _{2}>0\) such that
\[ B_{d}\cur {x,\gd _{1}}\sub B_{d’}\of {x,\eps }\qand B_{d’}\cur {x,\gd _{2}}\sub B_{d}\cur {x,\eps }. \]
Define \(\gd _{1}:=\eps \). If \(y\in B_{d}\cur {x,\gd _{1}}\), then \(d\of {x,y}<\gd _{1}=\eps \) so \(d’\cur {x,y}\le d\cur {x,y}<\eps \). Thus \(y\in B_{d’}\cur {x,\eps }\) as required.
Define \(\gd _{2}:=\dfrac {\eps }{1+\eps }\). If \(y\in B_{d’}\cur {x,\gd _{2}}\), then \(d’\of {x,y}<\gd _{2}\) so
\[ \dfrac {d\of {x,y}}{1+d\of {x,y}}<\frac {\eps }{1+\eps } \]
\[ \cur {1+\eps }d\of {x,y}<\eps \cur {1+d\of {x,y}} \]
\[ d\of {x,y}<\eps \]
and \(y\in B_{d}\cur {x,\eps }\) as required. □
Let \(d\) and \(d’\) be metrics defined on the set \(\sc \of I\) of all continuous function \(f:\bra {0,1}\to \br \) defined by
\( \seteqsection {1} \)\begin{align*} d\of {f,g} & :=\int _{0}^{1}\abv {f\of t-g\of t}\,dt\\ d’\of {f,g} & :=\sup \set {\abv {f\of t-g\of t}:t\in \bra {0,1}}. \end{align*} Prove that \(d\) and \(d’\) are not equivalent.
Solution. For each \(n\in \bn \), let \(f_{n}:\bra {0,1}\to \br \) be defined by
\[ f_{n}\of x:=\begin {cases} 1-nx & x\in \bra {0,\dfrac {1}{n}};\\ 0 & x\in \bra {\dfrac {1}{n},1}; \end {cases} \]
and let \(A=\set {f_{n}:n\in \bn }\sub \sc \of I.\) Let \(f:\bra {0,1}\to \br \) be the constant function with \(f\of x:=0\) for every \(x\in \bra {0,1}\). Then \(f\) is in the closure of \(A\) when \(\sc \of I\) has the topology induced by the metric \(d\), but \(f\) is not in the closure of \(A\) when \(\sc \of I\) has the topology induced by the metric \(d’\). Thus these two topologies are not the same, which means that \(d\) and \(d’\) are not equivalent. □
Consider \(\br \) with the standard topology. Prove that for each \(x\in \br \) the collection
\[ \sb _{x}:=\set {\cur {x-r,x+r}:r\in \bq ,\,r>0} \]
is a nbhd basis at \(x\).
Solution. Each member of \(\sb _{x}\) is open and contains \(x\) so it is a nbhd of \(x\). Let \(N\) be any nbhd of \(x\). There is open \(U\) with \(x\in U\sub N\). There is an open interval \(\cur {a,b}\sub U\) with \(x\in \cur {a,b}\). Let \(r\in \bq \) be such that
\[ 0<r<\min \set {x-a,b-x}. \]
Then \(x\in \cur {x-r,x+r}\sub U\) and \(\cur {x-r,x+r}\in \sb _{x}\). Therefore, \(\sb _{x}\) is a nbhd basis at \(x\). □