Let \(X\) be the set of real-valued functions on the interval \(\bra {0,1}\). Consider the following question. Is there a metric \(d\) on \(X\) such that a sequence \(\cur {f_{n}}_{n=1}^{\ity }\) in \(X\) converges to \(f\in X\) in the metric space \(\cur {X,d}\) if and only if \(\cur {f_{n}\of x}_{n=1}^{\ity }\) converges to \(f\of x\) for every \(x\in X\)?
The answer is no!
Proof. Nested Interval Property (Theorem 1.4.1. in Abbott’s book) says:
For each \(n\in \bn \) let \(I_{n}:=\bra {a_{n},b_{n}}\) be a closed interval such that
\[ I_{1}\sur I_{2}\sur I_{3}\sur \ds . \]
Then
\[ \bigcap _{i=1}^{\ity }I_{i}\neq \emp . \]
Suppose, for a contradiction, that such a metric \(d\) on \(X\) exists. For each \(n\in \bn \), let \(f_{n}\in X\) be defined by
\[ f_{n}\of x:=\begin {cases} 1 & \text {if }x\in \cur {0,\frac {1}{n}}\\ 0 & \text {otherwise. } \end {cases} \]
Then \(\cur {f_{n}}_{n=1}^{\ity }\) converges pointwise to the constant function \(f\) such that \(f\of x:=0\) for each \(x\in \bra {0,1}\). Then there is \(k_{1}\in \bn \) such that \(d\of {f_{k_{1}},f}<1\). Let \(a_{1}:=0\), \(b_{1}:=\dfrac {1}{k_{1}}\) and \(g_{1}:=f_{k_{1}}\).Analogous argument shows that there are \(a_{2},b_{2}\) with \(a_{1}<a_{2}<b_{2}<b_{1}\) and \(d\of {g_{2},f}<\dfrac {1}{2}\), where \(g_{2}\in X\) is defined by:
\[ g_{2}\of x:=\begin {cases} 1 & \text {if }x\in \cur {a_{2},b_{2}}\\ 0 & \text {otherwise,} \end {cases} \]
By induction, for each \(n\ge 2\), we get \(a_{n},b_{n}\) such that \(a_{n-1}<a_{n}<b_{n}<b_{n+1}\) and \(d\of {g_{n},f}<\dfrac {1}{n}\), where \(g_{n}\in X\) is defined by:
\[ g_{n}\of x:=\begin {cases} 1 & \text {if }x\in \cur {a_{n},b_{n}}\\ 0 & \text {otherwise.} \end {cases} \]
Then \(\cur {g_{n}}_{n=1}^{\ity }\) converges to \(f\) in the metric space \(\cur {X,d}\). However, the Nested Interval Property implies that
\[ C:=\bigcap _{n=1}^{\ity }\bra {a_{n},b_{n}}\neq \emp . \]
If \(c\in C\), then \(c\in \cur {a_{n},b_{n}}\) for each \(n\in \bn \) so \(g_{n}\of c=1\) for each \(n\in \bn \). Thus \(\cur {g_{n}\of c}_{n=1}^{\ity }\) does not converge to \(f\of c=0\), which is a contradiction. □
A topological structure (or just topology) on a set \(X\) is a family \(\st \) of subsets (called open sets) of \(X\) such that the following conditions hold:
1. \(X\) is open and \(\emp \) is open.
2. The union of any family of open sets is open.
3. The intersection of any nonempty finite family of open sets is open.
A topological space is a set \(X\) together with a topology on \(X\).
1. The discrete topology on \(X\) is the family of all subsets of \(X\).
2. The trivial topology on \(X\) is the family \(\set {X,\emp }\).
3. The Sierpiński space is the set \(X=\set {1,2}\) with the topology \(\set {X,\emp ,\set 1}\).
4. For any metric space \(X\), the family of open sets is a topology on \(X\).
5. Let \(X\) be an infinite set. The family of cofinite subsets of \(X\) (whose complements are finite) is a topology on \(X\). It is called the cofinite topology.
6. Let \(X\) be an uncountable set. The family of cocountable subsets of \(X\) (with countable complements) is a topology on \(X\). It is called the cocountable topology on \(X\).
Let \(\st \) and \(\st ’\) be topologies on the same set \(X\). If \(\st \sub \st ’\), then we say that \(\st \) is coarser, smaller or weaker than \(\st ’\) and that \(\st ’\) is finer, larger or stronger than \(\st \).
Note that the trivial topology on a set \(X\) is smaller than any topology on \(X\) and the discrete topology on \(X\) is larger than any topology on \(X\).
A subset \(C\) of a topological space \(X\) is closed if \(X\sem C\) is open.
Let \(X\) be a topological space.
1. The sets \(X\) and \(\emp \) are closed.
2. The intersection of any nonempty family of closed sets is closed.
3. The union of any finite family of closed sets is closed.
Proof. The set \(X\) is closed since \(\emp \) is open and \(\emp \) is closed since \(X\) is open.
Let \(\sc \) be a nonempty family of closed sets. Then
\[ \sa :=\set {X\sem C:C\in \sc } \]
is a family of open sets so \(\bigcup \sa \) is open. Since
\[ \bigcap \sc =X\sem \bigcup \sa , \]
it follows that \(\bigcap \sc \) is closed.
Let \(\sc \) be a finite family of closed sets. If \(\sc =\emp \), then \(\bigcup \sc =\emp \) is closed. If \(\sc \neq \emp \), then
\[ \sa =\set {X\sem C:C\in \sc } \]
is a nonempty family of open sets so \(\bigcap \sa \) is open. Since
\[ \bigcup \sc =X\sem \bigcap \sa , \]
it follows that \(\bigcup \sc \) is closed. □
An infinite union of closed sets does not have to be closed.
The closed interval \(\bra {\dfrac {1}{n},2}\) is closed in \(\br \) for each \(n\in \bn \), but the union
\[ \bigcup _{n=1}^{\ity }\bra {\dfrac {1}{n},2}=\ocl {0,2} \]
is not closed.
In a discrete space every set is both closed and open.
In \(\bz \) with the cofinite topology finite sets are closed, but not open. Cofinite sets are open, but not closed. Then set \(\bn \) is neither closed nor open.
Let \(X:=\set {1,2,3}\) and \(\st _{1}:=\set {\emp ,X,\set 1,\set {1,2}}\) and \(\st _{2}:=\set {\emp ,X,\set 3,\set {2,3}}\) be topologies on \(X\).
• Prove that \(\st _{1}\cup \st _{2}\) is not a topology on \(X\).
• Find the smallest topology on \(X\) containing \(\st _{1}\cup \st _{2}\).
• Find the largest topology on \(X\) contained in \(\st _{1}\cap \st _{2}\).
Solution.
• \(\st _{1}\cup \st _{2}\) is not a topology on \(X\) since \(\set 1,\set 3\in \st _{1}\cup \st _{2}\) but \(\set {1,3}\nin \st _{1}\cup \st _{2}\).
• The smallest topology on \(X\) containing \(\st _{1}\cup \st _{2}\) is the discrete topology.
• The largest topology on \(X\) contained in \(\st _{1}\cap \st _{2}\) is the trivial topology.
□
Let \(X\) be an infinite set and \(x_{0}\in X\). Let
\[ \st :=\set {G\sub X:X\sem G\text { is finite or }x_{0}\nin G}. \]
• Prove that \(\st \) is a topology on \(X\).
• Let \(x\in X\). Prove that \(\set x\) is both open and closed if and only if \(x\neq x_{0}\).
Solution. We have \(\emp \in \st \) since \(x_{0}\nin \emp \) and \(X\in \st \) since \(X\sem X\) is empty hence finite.
Let \(\sa \sub \st \). If for every \(A\in \sa \) we have \(x_{0}\nin A\), then \(x_{0}\nin \bigcup \sa \) so \(\bigcup \sa \in \st \). Otherwise, there is \(A_{0}\in \sa \) such that \(x_{0}\in A_{0}\). Then \(X\sem A_{0}\) is finite. Since \(X\sem \bigcup \sa \sub X\sem A_{0}\), it follows that \(X\sem \bigcup \sa \) is finite so \(\bigcup \sa \in \st \).
Let \(\sa \sub \st \) be finite and nonempty. If there is \(A_{0}\in \sa \) such that \(x_{0}\nin A_{0}\), then \(x_{0}\nin \bigcap \sa \) so \(\bigcap \sa \in \st \). Otherwise, we have \(x_{0}\in A\) for every \(A\in \sa \) so \(X\sem A\) is finite for every \(A\in \sa \). Then
\[ X\sem \bigcap \sa =\bigcup _{A\in \sa }\cur {X\sem A} \]
is finite as a finite union of finite sets. Thus \(\bigcap \sa \in \st \).
We have proved that \(\st \) is a topology on \(X\).
Let \(x\in X\). Assume that \(\set x\) is both open and closed. Since \(\set x\) is open, it follows the either \(X\sem \set x\) is finite or \(x_{0}\nin \set x\). Since \(X\) is infinite, it follows that \(X\sem \set x\) is infinite so \(x_{0}\nin \set x\) and \(x\neq x_{0}\).
Now assume that \(x\neq x_{0}\). Then \(x_{0}\nin \set x\) so \(\set x\) is open. Let \(A:=X\sem \set x\). Then \(X\sem A=\set x\) is finite so \(A\) is open. It follows that \(\set x\) is closed. □
Let \(X:=\sb \of {\bra {0,1}}\) with the supremum metric (see Example 1.1.7). Show that
\[ C:=\set {f\in X:f\text { is continuous}} \]
is closed in \(X\).
Solution. We will show that \(X\sem C\) is open in \(X\). Let \(g\in X\sem C\). We need \(r>0\) such that the ball \(B\of {g,r}\sub X\sem C\).
Since \(g\) is not continuous, there is \(x\in \bra {0,1}\) such that \(g\) is not continuous at \(x\). Thus there is \(\eps >0\) such that for every \(\gd >0\) there is \(y\in \bra {0,1}\) with \(\abv {x-y}<\gd \) and \(\abv {g\of x-g\of y}\ge \eps \). Let \(r:=\eps /3\). If \(f\in B\of {g,r}\), then \(\abv {f\of z-g\of z}<\eps /3\) for every \(z\in \bra {0,1}\). We show that \(f\in X\sem C\) by showing that \(f\) is not continuous at \(x\). Actually, we will show that for every \(\gd >0\) there is \(y\in \bra {0,1}\) such that \(\abv {x-y}<\gd \) and \(\abv {f\of x-f\of y}>\eps /3\).
Let \(\gd >0\). There is \(y\in \bra {0,1}\) with \(\abv {x-y}<\gd \) and \(\abv {g\of x-g\of y}\ge \eps \). Then
\( \seteqsection {1} \)\begin{align*} \eps & \le \abv {g\of x-g\of y}\le \abv {g\of x-f\of x}+\abv {f\of x-f\of y}+\abv {f\of y-g\of y}\\ & <\eps /3+\abv {f\of x-f\of y}+\eps /3=2\eps /3+\abv {f\of x-f\of y}, \end{align*} which implies that \(\abv {f\of x-f\of y}>\eps /3\). □
Consider \(X:=\br ^{2}\) with the standard Euclidean metric \(d\). Give an example of nonempty disjoint closed subsets \(A,B\sub X\) such that
\[ \inf \set {d\of {x,y}:x\in A,\,y\in B}=0. \]
Solution. Let
\[ A:=\set {\ang {a,b}:a,b\in \br \text { with }ab=1} \]
and \(B:=\set {\ang {0,b}:b\in \br }\). Then \(A\) and \(B\) are nonempty disjoint closed subsets of \(X\). For any \(\eps >0\) we have \(x:=\ang {\eps ,1/\eps }\in A\) and \(b:=\ang {0,1/\eps }\in B\) with \(d\of {x,y}=\eps \), which implies that
\[ \inf \set {d\of {x,y}:x\in A,\,y\in B}=0. \]
□
Let \(X\) be a topological space and \(x\in X\). A set \(N\sub X\) is a neighborhood of \(x\) (nbhd for short) if there exists an open set \(U\) such that \(x\in U\sub N\).
The interval \(\clo {0,2}\) is a nbhd of \(1\) in \(\br \) that is not open.
If \(U\) is an open set, then it is a nbhd of each \(x\in U\). In particular, \(X\) is a nbhd of each \(x\in X\).
Let \(X\) be a topological space and, for each \(x\in X\), let \(\sn _{x}\) be the family of all nbhds of \(x\). Then the following conditions hold for each \(x\in X\):
1. \(\sn _{x}\neq \emp \);
2. \(x\in N\) for every \(N\in \sn _{x}\);
3. if \(N_{1},N_{2}\in \cal N_{x}\), then \(N_{1}\cap N_{2}\in \sn _{x}\);
4. if \(N\in \sn _{x}\) and \(N\sub M\sub X\), then \(M\in \sn _{x}\);
5. if \(N\in \sn _{x}\), then \(\set {y\in N:N\in \sn _{y}}\in \sn _{x}\).
Proof. 1. holds since \(X\in \sn _{x}\).
2. holds since if \(N\in \sn _{x}\), then there is open \(U\) with \(x\in U\sub N\), which implies that \(x\in N\).
3. holds since if \(N_{1},N_{2}\in \sn _{x}\), then there are open \(U_{1},U_{2}\) with \(x\in U_{1}\sub N_{1}\) and \(x\in U_{2}\sub N_{2}\). Let \(U:=U_{1}\cap U_{2}\). Then \(U\) is open and
\[ x\in U\sub N_{1}\cap N_{2}. \]
4. holds since if \(N,M\) are as assumed, then there is open \(U\) with \(x\in U\sub N\). Since \(N\sub M\), this implies that \(U\sub M\) so \(M\in \sn _{x}\).
5. holds since \(N\in \sn _{x}\) implies that there is open \(U\) with \(x\in U\sub N\). Since \(U\in \cal N_{y}\) for every \(y\in U\), we have \(N\in \sn _{y}\) for every \(y\in U\). Thus
\[ U\sub \rg N:=\set {y\in N:N\in \sn _{y}}. \]
Since \(U\in \sn _{x}\), it follows that \(\rg N\in \sn _{x}\). □
Let \(X:=\set {1,2,3}\) with \(\sn _{1}:=\set {\set {1,2},X}\) and \(\sn _{2}:=\sn _{3}:=\set X\). Then conditions 1.–4. of Proposition 1.2.8 hold, but 5. fails since \(N:=\set {1,2}\in \sn _{1}\), but \(\rg N=\set 1\nin \sn _{1}\).
If
\[ \st :=\set {U\sub X:U\in \sn _{x}\text { for every }x\in U}, \]
then \(\st =\set {X,\emp }\) is the trivial topology on \(X\). Then \(N\) is not a nbhd of \(1\) relative to \(\st \).
Let \(X\) be a set and for each \(x\in X\) let \(\sn _{x}\) be a family of subsets of \(X\) such that the conditions 1.–4. of Proposition 1.2.8 hold. Then
\[ \st :=\set {U\sub X:U\in \sn _{x}\text { for every }x\in U} \]
is a topology on \(X\) such that each nbhd of \(x\in X\) relative to \(\st \) belongs to \(\sn _{x}\). If 5. is also satisfied, then \(\sn _{x}\) is equal to the family of all nbhds of \(x\) relative to \(\st \) and \(\st \) is the unique topology having that property.
Proof. It is clear that \(\emp \in \st \). Now we show that \(X\in \st \). Given \(x\in X\) we have \(\sn _{x}\neq \emp \) by 1. so there is \(N\in \sn _{x}\) which implies that \(X\in \sn _{x}\) by 4. Thus \(X\in \sn _{x}\) for every \(x\in X\), which implies that \(X\in \st \).
Let \(\sa \) be a family of members of \(\st \). To show that \(\bigcup \sa \in \st \) we need to show that \(\bigcup \sa \in \sn _{x}\) for every \(x\in \bigcup \sa \). Let \(x\in \bigcup \sa \). Then there is \(U\in \sa \) with \(x\in U\). Since \(U\in \st \), we have \(U\in \sn _{y}\) for each \(y\in U\) so, in particular, \(U\in \sn _{x}\). Then 4. implies that \(\bigcup \sa \in \sn _{x}\).
Let \(\sa \) be a nonempty finite family of members of \(\st \). To show that \(\bigcap \sa \in \st \) we need to show that \(\bigcap \sa \in \sn _{x}\) for every \(x\in \bigcap \sa \). Let \(x\in \bigcap \sa \). Then \(x\in U\) for every \(U\in \sa \). Since \(\sa \sub \st \), we have \(U\in \sn _{x}\) for every \(U\in \sa \). Since \(\sa \) is finite, applying 3. and induction we conclude that \(\bigcap \sa \in \sn _{x}\).
Let \(N\) be a nbhd of \(x\in X\) relative to \(\st \). Then there is \(U\in \st \) with \(x\in U\sub N\). The definition of \(\st \) implies that \(U\in \sn _{x}\). Thus \(N\in \sn _{x}\) by \(4.\)
Now assume 5. as well. Let \(x\in X\) and \(N\in \sn _{x}\). To show that \(N\) is a nbhd of \(x\) relative to \(\st \), it suffices to show that
\[ \rg N:=\set {y\in N:N\in \sn _{y}}\in \st . \]
If \(z\in \rg N\), then 5. implies that \(\rg N\in \sn _{z}\) so \(\rg N\in \sn _{z}\) for every \(z\in \rg N\) and consequently \(\rg N\in \st \) as required.
Suppose that \(\st ’\) is any topology having the property that \(\sn _{x}\) is the family of nbhds of \(x\) relative to \(\st ’\). To show that \(\st ’=\st \) it suffices to show that \(U\in \st ’\) if and only if \(U\in \sn _{x}\) for every \(x\in U\). Assume that \(U\in \st ’\). Then \(U\) is a nbhd of every \(x\in U\) by the definition of a nbhd. Assume that \(U\in \sn _{x}\) for every \(x\in U\). Then for each \(x\in U\), there is \(V_{x}\in \st ’\) such that \(x\in V_{x}\sub U\). Since
\[ U=\bigcup _{x\in U}V_{x}, \]
it follows that \(U\in \st ’\). □
In a metric space \(\cur {X,d}\), for any real number \(r\ge 0\), the closed r-ball at \(x\in X\) is the set \(\set {y\in X:d\of {x,y}\le r}\). Show that a closed ball is always closed in the metric topology.
Solution. Let \(r\ge 0\) and
\[ C:=\set {y\in X:d\of {x,y}\le r}. \]
We will show that \(X\sem C\) is open in \(X\). Let \(z\in X\sem C\). Then \(d\of {x,z}>r\) so \(\eps :=d\of {x,z}-r>0\). To show that \(X\sem C\) is open in \(X\) it suffices to show that \(B\of {z,\eps }\sub X\sem C\).
Let \(y\in B\of {z,\eps }\). Then \(d\of {y,z}<\eps \) so
\[ r+\eps =d\of {x,z}\le d\of {x,y}+d\of {y,z}<d\of {x,y}+\eps , \]
which implies that \(d\of {x,y}>r\) so \(y\in X\sem C\) as required. □
What is the topology determined by the metric on \(X\) given by \(d\of {x,y}:=1\) if \(x\neq y\) and \(d\of {x,y}=0\) if \(x=y\)?
Solution. The resulting topology is the discrete topology. For any \(x\in X\), the set \(\set x\) is open in \(X\) since \(B\of {x,1}\sub \set x\). Since any subset of \(X\) is a union of singletons, any subset of \(X\) is open so the obtained topology is discrete. □
Let \(X\) be a set with at least two elements. Prove that there are no metric on \(X\) that induces the trivial topology on \(X\).
Solution. Let \(x,y\in X\) with \(x\neq y\). Suppose, for a contradiction, that \(d\) is a metric on \(X\) that induces the trivial topology. Let \(d\of {x,y}=r>0\). Then \(U:=B\of {x,r}\) is open in \(X\) with \(x\in U\) and \(y\nin U\). Thus \(U\neq \emp \) and \(U\neq X\). This is a contradiction since \(\emp \) and \(X\) are the only open sets in the trivial topology. □
Let \(\cur {X,d}\) be a metric space and \(C\sub X\) be closed. Prove that there is a sequence \(\cur {U_{n}}_{n\in \bn }\) of open subsets of \(X\) such that \(C=\bigcap _{n\in \bn }U_{n}\).
Solution. For each \(n\in \bn \), let
\[ U_{n}:=\bigcup _{x\in C}B\of {x,\frac {1}{n}}. \]
Then \(U_{n}\) is open in \(X\) as a union of a family of open sets. It remains to show that
\[ C=\bigcap _{n\in \bn }U_{n}. \]
Since \(x\in U_{n}\) for every \(x\in C\) and every \(n\in \bn \), it follows that \(C\sub U_{n}\) for every \(n\in \bn \) so
\[ C\sub \bigcap _{n\in \bn }U_{n}. \]
Now let \(y\in \bigcap _{n\in \bn }U_{n}\). We aim at showing that \(y\in C\). Suppose, for a contradiction, that \(y\in X\sem C\). Since \(X\sem C\) is open there is \(\eps >0\) such that \(B\of {y,\eps }\sub X\sem C\). Let \(n\in \bn \) be such that \(\dfrac {1}{n}<\eps \). Since \(y\in U_{n}\), there is \(x\in C\) with \(d\of {x,y}<\dfrac {1}{n}\). Then \(d\of {x,y}<\eps \) so \(x\in B\of {y,\eps }\), implying that \(B\of {y,\eps }\cap C\neq \emp \), which is a contradiction. □