Let \(X\) and \(Y\) be topological spaces and
\[ \sb :=\set {U\tm V:U\text { is open in }X\text { and }V\text { is open in }Y}. \]
Then \(\sb \) is a basis for a topology on \(X\tm Y\).
Proof. According to Theorem 1.4.12, need to verify that \(\bigcup \sb =X\tm Y\) and for every \(B_{1},B_{2}\in \sb \) and every \(z\in B_{1}\cap B_{2}\) there is \(B\in \sb \) with \(z\in B\sub B_{1}\cap B_{2}\). Since \(X\tm Y\in \sb ,\) it follows that \(\bigcup \sb =X\tm Y\).
Let \(B_{1},B_{2}\in \sb \) with \(B_{1}=U_{1}\tm V_{1}\) and \(B_{2}=U_{2}\tm V_{2}\). Then \(\ang {x,y}\in B_{1}\cap B_{2}\) if and only if \(x\in U_{1}\cap U_{2}\) and \(y\in V_{1}\cap V_{2}\) so
\[ B_{1}\cap B_{2}=\cur {U_{1}\cap U_{2}}\tm \cur {V_{1}\cap V_{2}}\in \sb . \]
Thus we can take \(B=B_{1}\cap B_{2}\) for any \(z\in B_{1}\cap B_{2}.\) □
Let \(X\) and \(Y\) be topological spaces. The topology induced by the basis
\[ \sb :=\set {U\tm V:U\text { is open in }X\text { and }V\text { is open in }Y} \]
is called the product topology on \(X\tm Y\) and the obtained topological space is called the product space.
The functions \(p_{X}:X\tm Y\to X\) and \(p_{Y}:X\tm Y\to Y\) defined by \(p_{X}\of {x,y}:=x\) and \(p_{Y}\of {x,y}:=y\) are called projections.
Let \(X\) and \(Y\) be topological spaces and \(Z:=X\tm Y\) be the product space. The projections \(p_{X}:Z\to X\) and \(p_{Y}:Z\to Y\) are continuous and open. Moreover, the product topology on \(X\tm Y\) is the smallest topology for which both \(p_{X}\) and \(p_{Y}\) are continuous.
Proof. Let
\[ \sb :=\set {U\tm V:U\text { is open in }X\text { and }V\text { is open in }Y}. \]
Since \(p_{X}^{-1}\bof {U\tm V}=U\) is open in \(X\) for every \(U\tm V\in \sb \), it follows that \(p_{X}\) is continuous. Similarly, \(p_{Y}\) is continuous. Theorem 2.1.16 implies that both \(p_{X}\) and \(p_{Y}\) are open.
Assume that \(\st \) is any topology on \(X\tm Y\) for which both \(p_{X}\) and \(p_{Y}\) are continuous. If \(U\) is open in \(X\) and \(V\) is open in \(Y\), then
\[ U\tm V=\cur {U\tm Y}\cap \cur {X\tm V}=p_{X}^{-1}\bof U\cap p_{Y}^{-1}\bof V \]
belongs to \(\st \). Thus \(\sb \sub \st \), which implies that \(\st \) is finer than the product topology on \(X\tm Y\). □
Let \(X=Y:=\br \) and
\[ C:=\set {\ang {x,y}\in X\tm Y:xy=1}. \]
Then \(C\) is closed in \(X\tm Y\), but \(p_{X}\bof C=X\sem \set 0\) is not closed in \(X\).
If \(X\) and \(Y\) are topological spaces and
\[ \ss :=\set {p_{X}^{-1}\bof U:U\text { is open in }X}\cup \set {p_{Y}^{-1}\bof V:V\text { is open in }Y}, \]
then \(\ss \) is a subbasis for the product topology on \(X\tm Y\).
Let \(\sb \) and \(\sd \) be bases for the topologies on \(X\) and \(Y\), respectively. Then
\[ \se :=\set {B\tm D:B\in \sb ,\,D\in \sd } \]
is a basis for the product topology on \(X\tm Y\).
Proof. It is clear that the members of \(\se \) are open in the product topology on \(X\tm Y\). Let \(W\) be open in the product topology on \(X\tm Y\) and \(\ang {x,y}\in W\). There is an open \(U\) in \(X\) and an open \(V\) in \(Y\) with
\[ \ang {x,y}\in U\tm V\sub W. \]
Then \(x\in B\sub U\) and \(y\in D\sub V\) for some \(B\in \sb \) and \(D\in \sd \) so
\[ \ang {x,y}\in B\tm D\sub U\tm V. \]
Thus \(B\tm D\sub W\) as required. □
The standard topology on \(\br ^{2}\) is the product topology on \(\br \tm \br \).
Proof. Let
\[ \sb :=\set {B\of {\ang {x,y},r}:\ang {x,y}\in \br \tm \br ,\,r>0} \]
and
\[ \sd :=\set {\cur {a,b}\tm \cur {c,d}:a,b,c,d\in \br ,\,a<b,\,c<d}. \]
Then \(\sb \) is a basis of the standard topology on \(\br ^{2}\) and \(\sd \) is a basis for the product topology on \(\br \tm \br \). Let \(B\in \sb \) and \(\ang {x,y}\in B\). There are open intervals \(\cur {a,b}\) and \(\cur {c,d}\) with
\[ \ang {x,y}\in \cur {a,b}\tm \cur {c,d}\sub B. \]
Since \(\cur {a,b}\tm \cur {c,d}\in \sd \), Proposition 1.4.14 implies that the product topology on \(\br \tm \br \) is finer than the standard topology. Similarly, the standard topology is finer than the product topology. □
If \(X,Y\) and \(Z\) are topological spaces then \(X\tm Y\) is homeomorphic with \(Y\tm X\) and \(\cur {X\tm Y}\tm Z\) is homeomorphic with \(X\tm \cur {Y\tm Z}\).
Let \(X\) and \(Y\) be topological spaces with \(A\sub X\) and \(B\sub Y\). Let \(\st \) be the subspace topology on \(A\tm B\) inherited from the product topology on \(X\tm Y\) and \(\st ’\) be the product topology on \(A\tm B\), where \(A\) has the subspace topology inherited from \(X\) and \(B\) has the subspace topology inherited from \(Y\). Then \(\st =\st ’\).
Proof. The family
\[ \sb :=\set {\cur {A\tm B}\cap \cur {U\tm V}:U\text { open in }X\text { and }V\text { open in }Y} \]
is a basis for \(\st \) and
\[ \sb ’:=\set {\cur {A\cap U}\tm \cur {B\cap V}:U\text { open in }X\text { and }V\text { open in }Y} \]
is a basis for \(\st ’\). Since
\[ \cur {A\tm B}\cap \cur {U\tm V}=\cur {A\cap U}\tm \cur {B\cap V} \]
for any \(U\sub X\) and \(V\sub Y\), it follows that \(\sb =\sb ’\) so \(\st =\st ’\). □
Let \(X\), \(Y\) and \(Z\) be topological spaces and \(f:Z\to X\tm Y\). Then \(f\) is continuous if and only if both compositions \(p_{X}\circ f\) and \(p_{Y}\circ f\) are continuous.
Proof. If \(f\) is continuous, then both \(p_{X}\circ f\) and \(p_{Y}\circ f\) are continuous since compositions of continuous functions is continuous.
Assume that both \(p_{X}\circ f\) and \(p_{Y}\circ f\) are continuous. To show that \(f\) is continuous it suffices to prove that \(f^{-1}\bof {U\tm V}\) is open in \(Z\) for any open \(U\sub X\) and any open \(V\sub Y\). Assume that \(U\) is open in \(X\) and \(V\) is open in \(Y\). Note that \(z\in f^{-1}\bof {U\tm V}\) if and only if \(\cur {p_{X}\circ f}\of z\in U\) and \(\cur {p_{Y}\circ f}\of z\in V\) so
\[ f^{-1}\bof {U\tm V}=\cur {p_{X}\circ f}^{-1}\bof U\cap \cur {p_{Y}\circ f}^{-1}\bof V. \]
Since both \(\cur {p_{X}\circ f}^{-1}\bof U\) and \(\cur {p_{Y}\circ f}^{-1}\bof V\) are open in \(Z\), it follows that \(f^{-1}\bof {U\tm V}\) is open in \(Z\), as required. □
Let \(X\), \(X’\), \(Y\) and \(Y’\) be topological spaces and \(f:X\to X’\) and \(g:Y\to Y’\) be continuous. Let
\[ f\tm g:X\tm Y\to X’\tm Y’ \]
be defined by
\[ \cur {f\tm g}\of {x,y}:=\cur {f\of x,g\of y} \]
for any \(x\in X\) and \(y\in Y\). Then \(f\tm g\) is continuous.
Proof. It suffices to show that both \(p_{X’}\circ \cur {f\tm g}\) and \(p_{Y’}\circ \cur {f\tm g}\) are continuous. Since
\[ p_{X’}\circ \cur {f\tm g}=f\circ p_{X}, \]
and since both \(f\) and \(p_{X}\) are continuous, it follows that \(p_{X’}\circ \cur {f\tm g}\) is continuous. Similarly, \(p_{Y’}\circ \cur {f\tm g}\) is continuous. □
Let \(X\), \(Y\) and \(Z\) be topological spaces and \(f:Z\to X\) and \(g:Z\to Y\) be continuous. Let
\[ \cur {f,g}:Z\to X\tm Y \]
be defined by
\[ \cur {f,g}\of z:=\cur {f\of z,g\of z} \]
for any \(z\in Z\). Then \(\cur {f,g}\) is continuous.
Proof. Since \(p_{X}\circ \cur {f,g}=f\) and \(p_{Y}\circ \cur {f,g}=g\) are continuous, it follows that \(\cur {f,g}\) is continuous. □
Let \(X_{\ga }\) be a set for any \(\ga \in A\). The Cartesian product \(X:=\prod _{\ga \in A}X_{\ga }\) is the set of all functions \(f\) with domain \(A\) such that \(f\of {\ga }\in X_{\ga }\) for any \(\ga \in A\). Such a function \(f\) will be denoted by \(\cur {x_{\ga }}_{\ga \in A}\), where \(x_{\ga }\) is the value of \(f\) at \(\ga \).
For each \(\ga \in A\), let \(p_{\ga }:X\to X_{\ga }\) be the projection defined by
\[ p_{\ga }\of {\cur {x_{\gb }}_{\gb \in A}}:=x_{\ga }. \]
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and \(X:=\prod _{\ga \in A}X_{\ga }\). The collection
\[ \sb :=\set {\prod _{\ga \in A}U_{\ga }:U_{\ga }\text { open in }X_{\ga }\text { for every }\ga \in A} \]
can be proved to be a basis for a topology on \(X\). This topology is called the box topology on \(X\).
Let \(X_{n}:=\br \) for every \(n\in \bn \) and \(X:=\prod _{n\in \bn }X_{n}\) with the box topology. Let \(f_{n}:\br \to X_{n}\) be the identity function and \(f:\br \to X\) be defined by
\[ f\of x=\cur {x,x,x,\ds }. \]
Then \(f_{n}=p_{n}\circ f\) is continuous for every \(n\in \bn \). However, \(f\) is not continuous.
Proof. Let \(U_{n}:=\cur {-\dfrac {1}{n},\dfrac {1}{n}}\) for each \(n\in \bn \). Then \(U:=\prod _{n\in \bn }U_{n}\) is open in the box topology on \(X\), but \(f^{-1}\bof U=\set 0\) is not open in \(\br \). □
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and let \(X:=\prod _{\ga \in A}X_{\ga }\). The product topology (or Tychonoff topology) on \(X\) is induced by the subbasis
\[ \ss :=\set {p_{\ga }^{-1}\bof {U_{\ga }}:U_{\ga }\text { is open in }X_{\ga }\text { for each }\ga \in A}. \]
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and let \(X:=\prod _{\ga \in A}X_{\ga }\). The product topology on \(X\) is the smallest topology on \(X\) for which each projection \(p_{\ga }\) is continuous.
Proof. If \(X\) has the product topology, then each \(p_{\ga }\) is continuous since each member of the subbasis
\[ \ss :=\set {p_{\ga }^{-1}\bof {U_{\ga }}:U_{\ga }\text { is open in }X_{\ga }\text { for each }\ga \in A} \]
is open in the product topology on \(X\).
Assume that \(\st \) is a topology on \(X\) such that \(p_{\ga }:X\to X_{\ga }\) is continuous for each \(\ga \in A\). Then \(\ss \sub \st \) so \(\st \) is finer than the product topology on \(X\). □
The box topology on \(X\) is finer than the product topology and when \(A\) is finite, these topologies are identical.
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and let \(X:=\prod _{\ga \in A}X_{\ga }\). If \(\ss _{\ga }\) is a subbasis for the topology on \(X_{\ga }\) for every \(\ga \in A\), then
\[ \ss :=\set {p_{\ga }^{-1}\bof {S_{\ga }}:S_{\ga }\in \ss _{\ga }\text { for every }\ga \in A} \]
is a subbasis for the product topology on \(X\).
Proof. Let
\[ \ss ’:=\set {p_{\ga }^{-1}\bof {U_{\ga }}:U_{\ga }\text { is open in }X_{\ga }\text { for each }\ga \in A}, \]
with \(\st \) being the topology induced by \(\ss \) and \(\st ’\) being the product topology (induced by \(\ss ’\)). Since \(\ss \sub \ss ’\), it follows that \(\st \sub \st ’\). To show that \(\st ’\sub \st \), it suffices to show that \(\ss ’\sub \st \).
Assume that \(S\in \ss ’\). Then \(S=p_{\ga }^{-1}\of {U_{\ga }}\) for some \(\ga \in A\) and \(U_{\ga }\) open in \(X_{\ga }\). If \(U_{\ga }=X_{\ga }\), then \(S=X\in \st \). If \(U_{\ga }=\emp \), then \(S=\emp \in \st \). Otherwise, \(U_{\ga }\) is a union of a family \(\sa \) of nonempty finite intersections of members of \(\ss _{\ga }\) so
\[ S=p_{\ga }^{-1}\bof {\bigcup \sa }=\bigcup _{A\in \sa }p_{\ga }^{-1}\bof A. \]
For \(A\in \sa \), if
\[ A=S_{1}\cap S_{2}\cap \ds \cap S_{n}, \]
where \(S_{1},\ds ,S_{n}\in \ss \), then
\[ p_{\ga }^{-1}\bof A=p_{\ga }^{-1}\bof {S_{1}}\cap p_{\ga }^{-1}\bof {S_{2}}\cap \ds \cap p_{\ga }^{-1}\bof {S_{n}}\in \st . \]
Since \(p_{\ga }^{-1}\bof A\in \st \) for every \(A\in \sa \), it follows that \(S\in \st \), as required. □
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and let \(X:=\prod _{\ga \in A}X_{\ga }\). Assume that \(\sb _{\ga }\) is a subbasis for the topology on \(X_{\ga }\) for every \(\ga \in A\), and let \(\sb \) consist of products \(\prod _{\ga \in A}B_{\ga }\) such that there is a finite \(A’\sub A\) with \(B_{\ga }\in \sb _{\ga }\) for \(\ga \in A’\) and \(B_{\ga }=X_{\ga }\) for \(\ga \in A\sem A’\). Then \(\sb \) is a basis for the product topology on \(X\).
Proof. Since \(\sb _{\ga }\) is a subbasis for the topology on \(X_{\ga }\) for every \(\ga \in A\), Proposition 2.2.12 implies that the family
\[ \ss :=\set {p_{\ga }^{-1}\bof {B_{\ga }}:B_{\ga }\in \sb _{\ga }\text { for every }\ga \in A} \]
is a subbasis for the product topology on \(X\). Since \(\sb \) consists of \(X\) and all intersections of finite nonempty subfamilies of \(\ss \), it follows that \(\sb \) is a basis for the topology on \(X\). □
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and let \(X:=\prod _{\ga \in A}X_{\ga }\). For each \(\ga \in A\) let \(Y_{\ga }\) be a subspace of \(X_{\ga }\) and let \(Y:=\prod _{\ga \in A}Y_{\ga }\). Then the product topology on \(Y\) coincides with the subspace topology inherited from \(X\).
Proof. Let \(\st \) be the product topology on \(Y\) and \(\st ’\) be the subspace topology. Let \(\ss _{\ga }\) be a subbasis for the topology on \(X_{\ga }\) for all \(\ga \in A\). Then
\[ \ss :=\set {p_{\ga }^{-1}\bof {S_{\ga }}:S_{\ga }\in \sb _{\ga }\text { for every }\ga \in A} \]
is a subbasis for the product topology on \(X\) so
\[ \ss ’:=\set {p_{\ga }^{-1}\bof {S_{\ga }}\cap Y:S_{\ga }\in \sb _{\ga }\text { for every }\ga \in A} \]
is a subbasis for \(\st \). Since
\[ p_{\ga }^{-1}\bof {S_{\ga }}\cap Y=\cur {p_{\ga }\res Y}^{-1}\bof {S_{\ga }\cap Y_{\ga }} \]
and since
\[ \ss _{\ga }’:=\set {S_{\ga }\cap Y_{\ga }:\ga \in A} \]
is a subbasis for the topology on \(X_{\ga }\) for each \(\ga \in A\), it follows the \(\ss ’\) is a subbasis for \(\st ’\). □
Let \(X_{\ga }\) be a topological space for each \(\ga \in A\) and let \(X:=\prod _{\ga \in A}X_{\ga }\). Let \(f:Y\to X\) for some topological space \(Y\). Then \(f\) is continuous if and only if \(p_{\ga }\circ f:Y\to X_{\ga }\) is continuous is continuous for each \(\ga \in A\). Moreover, the product topology on \(X\) is the unique topology with such a property.
Proof. If \(f\) is continuous, then it is clear that \(p_{\ga }\circ f\) is continuous for each \(\ga \in A\). Assume that \(p_{\ga }\circ f\) is continuous for each \(\ga \in A\). Since
\[ \ss :=\set {p_{\ga }^{-1}\bof {U_{\ga }}:U_{\ga }\text { is open in }X_{\ga }\text { for each }\ga \in A} \]
is a subbasis for the product topology on \(X\), and since
\[ f^{-1}\bof {p_{\ga }^{-1}\bof {U_{\ga }}}=\cur {p_{\ga }\circ f}^{-1}\bof {U_{\ga }} \]
is open in \(Y\) for each \(\ga \in A\), it follows that \(f\) is continuous.
Suppose that \(\st \) is any topology on \(X\) such that for any topological space \(Y\) and any \(f:Y\to X\), the continuity of \(p_{\ga }\circ f\) for each \(\ga \in A\) is equivalent to the continuity of \(f\). Taking \(Y:=X\) with the same topology \(\st \) and \(f\) to be the identity function (which is continuous), we conclude that \(p_{\ga }\) is continuous for each \(\ga \in A\). This implies that \(\st \) is finer than the product topology on \(X\) by Proposition 2.2.11.
Taking \(Y:=X\) with the product topology (and \(X\) with topology \(\st \)) and \(f:Y\to X\) to be the identity function, we have \(p_{\ga }\circ f\) continuous for each \(\ga \in A\) so \(f\) is continuous. It follows that any member of \(\st \) is open in \(Y\), thus \(\st \) is coarser than the product topology on \(X\). □
Let \(\cur {X_{n},d_{n}}\) be a metric space for each \(n\in \bn \) and let \(X:=\prod _{n\in \bn }X_{n}\). Consider \(X_{n}\) to be the topological space with the topology induced by \(d_{n}\). Then there exists a metric \(d\) on \(X\) that induces the product topology on \(X\).
Proof. For each \(n\in \bn \), let \(\gl _{n}>0\) with \(\lim _{n\to \ity }\gl _{n}=0\) and let \(d_{n}’\) be a metric on \(X_{n}\) such that the diameter of \(X_{n}\) in \(d_{n}’\) is at most \(\gl _{n}\). Such a metric \(d_{n}’\) exists by Corollary 1.4.17. For \(x:=\cur {x_{n}}_{n\in \bn }\) and \(y:=\cur {y_{n}}_{n\in \bn }\) in \(X\), define
\[ d\of {x,y}=\sup \set {d_{n}’\of {x_{n},y_{n}}:n\in \bn }. \]
It is clear that \(d\) is positive and symmetric. We verify the triangle inequality. Let \(x:=\cur {x_{n}}_{n\in \bn }\) , \(y:=\cur {y_{n}}_{n\in \bn }\) and \(z:=\cur {z_{n}}_{n\in \bn }\) be in \(X\). Then
\[ d_{n}’\of {x_{n},y_{n}}+d_{n}’\of {y_{n},z_{n}}\ge d_{n}’\of {x_{n},z_{n}} \]
for each \(n\in \bn \) so
\[ d\of {x,y}+d\of {y,z}\ge d_{n}’\of {x_{n},z_{n}} \]
for each \(n\in \bn \). Thus
\[ d\of {x,y}+d\of {y,z}\ge d\of {x,z}, \]
as required. Thus \(d\) is a metric on \(X\).
Now we show that \(d\) induces the product topology on \(X\). Let \(n\in \bn \) and \(U_{n}\) be open in \(X_{n}\). For each
\[ x=\cur {x_{k}}_{k\in \bn }\in p_{n}^{-1}\bof {U_{n}} \]
there is \(r_{x}>0\) such that
\[ x\in B_{d_{n}’}\of {x_{n},r_{x}}\sub U_{n}. \]
If
\[ y=\cur {y_{k}}_{k\in \bn }\in B_{d}\of {x,r_{x}}, \]
then \(d\of {x,y}<r_{x}\) so \(d\of {x_{k},y_{k}}<r_{x}\) for each \(k\in \bn \) and, in particular, \(d\of {x_{n},y_{n}}<r_{x}\) so \(y_{n}\in U_{n}\) and consequently \(y\in p_{n}^{-1}\bof {U_{n}}\). Thus
\[ B_{d}\of {x,r_{x}}\sub p_{n}^{-1}\bof {U_{n}}, \]
which implies that \(p_{n}^{-1}\bof {U_{n}}\) is open in the topology induced by \(d\). Since
\[ \ss :=\set {p_{n}^{-1}\bof {U_{n}}:U_{n}\text { is open in }X_{n}\text { for each }n\in \bn } \]
is a subbasis for the product topology on \(X\), it follows that the topology induced by \(d\) is finer than the product topology.
Let \(U\) be open in the topology induced by \(d\). We will show that \(U\) is open in the product topology. Let \(x:=\cur {x_{k}}_{k\in \bn }\in U\). There is \(r>0\) such that \(B_{d}\of {x,r}\sub U\). Let \(n\in \bn \) be such that \(\gl _{k}<r/2\) for each \(k>n\). For each \(k=1,2,\ds ,n\) let
\[ U_{k}:=B_{d_{k}’}\of {x_{k},r} \]
and for \(k>n\), let \(U_{k}:=X_{k}\). Then
\[ U’:=\prod _{k=1}^{\ity }U_{k} \]
is open in the product topology and
\[ x\in U’\sub B_{d}\of {x,r}\sub U, \]
which implies that \(U\) is open in the product topology. □
A topological space \(X\) is metrizable if there exists a metric \(d\) on \(X\) that induces the given topology.
We have proved that the product of countably many metrizable spaces is metrizable.
Let \(A\) be an uncountable set and \(X_{\ga }:=\br \) for each \(\ga \in A\). Then \(X:=\prod _{\ga \in A}X_{\ga }\) is not metrizable.
Proof. Suppose, for a contradiction, that \(d\) is a metric on \(X\) that induces the product topology. For each \(n\in \bn \) let \(B_{n}\) be the open ball \(B_{d}\of {0,1/n}\), where \(0\in X\) is the constant function with value \(0\), and let \(A_{n}\sub A\) be finite such that
\[ 0\in U_{n}:=\prod _{\ga \in A}U_{n,\ga }\sub B_{n}, \]
where \(U_{n,\ga }\) is an open interval in \(\br \) for all \(\ga \in A_{n}\) and \(U_{n,\ga }=\br \) for all \(\ga \in A\sem A_{n}\). Since \(\bigcup _{n\in \bn }A_{n}\) is countable and \(A\) is not, there is
\[ \gb \in A\sem \bigcup _{n\in \bn }A_{n}. \]
Let
\[ x:=\cur {x_{\ga }}_{\ga \in A}\in X, \]
with \(x_{\gb }:=1\) and \(x_{\ga }:=0\) for \(\ga \in A\sem \set {\gb }\). Then \(x\in U_{n}\) for each \(n\in \bn \). Since \(x\neq 0\) and since
\[ \bigcap _{n\in \bn }U_{n}\sub \bigcap _{n\in \bn }B_{n}=\set 0, \]
we have a contradiction. □