Notation.
\(a:=b\) mean that \(a\) equals \(b\) by definition.
\(\bn :=\set {1,2,\ds }\) is the set of positive integers.
\(\bz \) is the set of integers.
\(\br \) is the set of real numbers.
\(\bq \) is the set of rational numbers □
A sequence of real numbers \(\cur {a_{n}}_{n=1}^{\ity }\) converges to a real number \(a\) if, for every positive real number \(\eps \), there exists \(k\in \bn \) such that \(\abv {a_{n}-a}<\eps \) whenever \(n\ge k\).
A function \(f:\br \to \br \) is continuous at \(a\in \br \) if for every \(\eps >0\) there is \(\gd >0\) such that \(\abv {f\of x-f\of a}<\eps \) whenever \(\abv {x-a}<\gd \).
A metric space is a set \(X\) together with a function \(d:X\tm X\to \br \), called a metric on \(X\), that satisfies the following conditions:
1. \(d\of {x,y}\ge 0\) with equality if and only if \(x=y\);
2. \(d\of {x,y}=d\of {y,x}\); and
3. \(d\of {x,y}+d\of {y,z}\ge d\of {x,z}\).
Condition 1. is called positivity, 2. is called symmetry and 3. is called triangle inequality.
Note that the function \(d:\br \tm \br \to \br \) defined by \(d\of {x,y}:=\abv {x-y}\) is a metric on \(\br \).
Let \(n\in \bn \) and \(X:=\br ^{n}\). For
\[ x:=\cur {x_{1},x_{2},\ds ,x_{n}}\in X, \]
let
\[ \nm x:=\sqrt {\sum _{i=1}^{n}x_{i}^{2}} \]
and for \(x,y\in X\), let \(d\of {x,y}:=\nm {x-y}.\) Then \(d\) is a metric on \(X\).
Proof. It is clear that \(d\) is positive and symmetric. We prove the triangle inequality. If \(1\le i<j\le n\), then
\[ \cur {x_{i}y_{j}-x_{j}y_{i}}^{2}\ge 0 \]
so
\[ 2x_{i}y_{i}x_{j}y_{j}\le x_{i}^{2}y_{j}^{2}+x_{j}^{2}y_{i}^{2}. \]
Thus
\( \seteqsection {1} \)\begin{align*} \cur {\sum _{i=1}^{n}x_{i}y_{i}}^{2} & =\sum _{i=1}^{n}x_{i}^{2}y_{i}^{2}+\sum _{1\le i<j\le n}2x_{i}y_{i}x_{j}y_{j}\\ & \le \sum _{i=1}^{n}x_{i}^{2}y_{i}^{2}+\sum _{1\le i<j\le n}\cur {x_{i}^{2}y_{j}^{2}+x_{j}^{2}y_{i}^{2}}\\ & =\sum _{1\le i,j\le n}x_{i}^{2}y_{j}^{2}=\cur {\sum _{i=1}^{n}x_{i}^{2}}\cur {\sum _{i=1}^{n}y_{i}^{2}}. \end{align*} Hence
\( \seteqsection {1} \)\begin{align*} \nm {x+y}^{2} & =\sum _{n=1}^{n}\cur {x_{i}+y_{i}}^{2}=\sum _{i=1}^{n}x_{i}^{2}+\sum _{i=1}^{n}y_{i}^{2}+2\sum _{i=1}^{n}x_{i}y_{i}\\ & \le \sum _{i=1}^{n}x_{i}^{2}+\sum _{i=1}^{n}y_{i}^{2}+2\sqrt {\sum _{i=1}^{n}x_{i}^{2}}\sqrt {\sum _{i=1}^{n}y_{i}^{2}}\\ & =\cur {\nm x+\nm y}^{2}, \end{align*} which implies that \(\nm {x+y}\le \nm x+\nm y\). Thus, for \(x,y,z\in X\), we have
\[ d\of {x,y}+d\of {y,z}=\nm {x-y}+\nm {y-z}\ge \nm {x-z}=\left \Vert a\right \Vert d\of {x,z}, \]
so the triangle inequality holds. □
Let \(X:=\ell _{2}\) be the set of all infinite sequences \(\cur {x_{i}}_{i=1}^{\ity }\) of real numbers with
\[ \sum _{i=1}^{\ity }x_{i}^{2}<\ity . \]
For
\[ x:=\cur {x_{1},x_{2},\ds }\in X, \]
let
\[ \nm x:=\sqrt {\sum _{i=1}^{\ity }x_{i}^{2}} \]
and for \(x,y\in X\), let \(d\of {x,y}:=\nm {x-y}.\) Then \(d\) is a metric on \(X\).
Proof. First, we verify that the values of \(d\) are finite. If \(x,y\in X\), then
\[ \nm {x-y}=\sqrt {\sum _{i=1}^{\ity }\cur {x_{i}-y_{i}}^{2}}. \]
For each \(n\in \bn \) we have
\[ \sqrt {\sum _{i=1}^{n}\cur {x_{i}-y_{i}}^{2}}\le \sqrt {\sum _{i=1}^{n}x_{i}^{2}}+\sqrt {\sum _{i=1}^{n}y_{i}^{2}}\le \nm x+\nm y, \]
so \(\nm {x-y}\le \nm x+\nm y<\ity \).
It is clear that \(d\) is positive and symmetric. It satisfies the triangle inequality since
\( \seteqsection {1} \)\begin{align*} d\of {x,y}+d\of {y,z} & =\nm {x-y}+\nm {y-z}\\ & =\nm {x-y}+\nm {z-y}\\ & \ge \nm {\cur {x-y}-\cur {z-y}}\\ & =\nm {x-z}=d\of {x,z}. \end{align*} □
Let \(X:=\sc \of I\) be the set of all continuous real-valued functions on the interval \(I:=\bra {0,1}\). Given \(f\in X\), let
\[ \nm f:=\int _{0}^{1}\abv f\,dx \]
and for \(f,g\in X\), let \(d\of {f,g}:=\nm {f-g}\). Then \(d\) is a metric on \(X\).
Let \(Y\) be any set and let \(X:=\sb \of Y\) be the set of all real-valued bounded functions on \(Y\). Given \(f\in X\), let
\[ \nm f:=\sup \set {\abv {f\of y}:y\in Y} \]
and for \(f,g\in X\) let \(d\of {f,g}:=\nm {f-g}.\) Then \(d\) is a metric on \(X\). It is called the supremum metric.
Let \(\cur {X,d}\) and \(\cur {Y,d’}\) be metric spaces and \(f:X\to Y\). Then \(f\) is continuous at \(x\in X\) provided for each \(\eps >0\) there exists \(\gd >0\) such that for every \(x’\in X\) with \(d\of {x,x’}<\gd \) we have
\[ d’\of {f\of x,f\of {x’}}<\eps . \]
The function \(f\) is continuous if it is continuous at each \(x\in X\).
Let \(\cur {x_{n}}_{n=1}^{\ity }\) be a sequence in a metric space \(\cur {X,d}\). The sequence converges to \(x\in X\) provided for each \(\eps >0\) there exists \(k\in \bn \) such that \(d\of {x_{n},x}<\eps \) whenever \(n\ge k\).
Remark. We are going to describe continuity of functions and convergence of sequences using the concept of an open set. □
Let \(\cur {X,d}\) be a metric space, \(x\in X\) and \(r>0\). An open ball with center \(x\) and radius \(r\) is the set
\[ B\of {x,r}:=\set {y\in X:d\of {x,y}<r}. \]
Let \(\cur {X,d}\) be a metric space and \(U\sub X\). We say that \(U\) is open if for each \(x\in U\) there is \(r>0\) such that \(B\of {x,r}\sub U\).
Each open ball is an open set.
Proof. Let \(U:=B\of {x,r}\) be an open ball in \(X\) and let \(y\in U\). Then \(d\of {x,y}<r\). Let \(r’:=r-d\of {x,y}\). If \(z\in B\of {y,r’}\), then \(d\of {y,z}<r’\) so
\[ d\of {x,z}\le d\of {x,y}+d\of {y,z}<\cur {r-r’}+r’=r, \]
which implies that \(z\in U\). Thus \(B\of {y,r’}\sub U\). □
Let \(\cur {X,d}\) be a metric space. 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.
Proof. \(X\) is open since for any \(x\in X\) we have \(B\of {x,1}\sub X\). The empty set \(\emp \) is open since there are no \(x\in \emp .\)
Assume that \(\sa \) is any family of open sets. Let \(x\in \bigcup \sa .\) Then there is \(U\in \sa \) with \(x\in U\) so there is \(r>0\) with \(B\of {x,r}\sub U\). Then \(B\of {x,r}\sub \bigcup \sa \). It follows that \(\bigcup \sa \) is open.
Assume that \(\sa \) is a nonempty finite family of open sets. Let \(x\in \bigcap \sa .\) Then \(x\in U\) for every \(U\in \sa \), so for every \(U\in \sa \) there is \(r_{U}>0\) with \(B\of {x,r_{U}}\sub U\). Since \(\sa \) is finite,
\[ r:=\inf \set {r_{U}:U\in \sa }=\min \set {r_{U}:U\in \sa }>0. \]
Then \(B\of {x,r}\sub U\) for every \(U\in \sa \) so \(B\of {x,r}\sub \bigcap \sa \). Thus \(\bigcap \sa \) is open. □
Let \(\cur {X,d}\) and \(\cur {Y,d’}\) be metric space. A function \(f:X\to Y\) is continuous if and only if \(f^{-1}\bof U\) is open in \(X\) whenever \(U\) is open in \(Y\).
Proof. Assume that \(f\) is continuous. Let \(U\sub Y\) be open and \(x\in f^{-1}\bof U\). Then \(f\of x\in U\) so there is \(\eps >0\) such that \(B\of {f\of x,\eps }\sub U.\) Since \(f\) is continuous, there exists \(\gd >0\) such that
\[ d’\of {f\of x,f\of {x’}}<\eps \]
whenever \(d\of {x,x’}<\gd .\) If \(x’\in B\of {x,\gd }\), then \(d\of {x,x’}<\gd \) so \(d’\of {f\of x,f\of {x’}}<\eps \) and
\[ f\of {x’}\in B\of {f\of x,\eps }\sub U, \]
implying that \(x’\in f^{-1}\bof U\). Thus \(B\of {x,\gd }\sub f^{-1}\bof U\) so \(f^{-1}\bof U\) is open.
Now assume that \(f^{-1}\bof U\) is open in \(X\) whenever \(U\) is open in \(Y\). Let \(x\in X\). We will show that \(f\) is continuous at \(x\). Let \(\eps >0\) and \(U=B\of {f\of x,\eps }\). Then \(U\) is open in \(Y\) so \(f^{-1}\bof U\) is open in \(X\). Since \(x\in f^{-1}\bof U\), there is \(\gd >0\) such that \(B\of {x,\gd }\sub f^{-1}\bof U\). If \(x’\in X\) with \(d\of {x,x’}<\gd \), then \(x’\in f^{-1}\bof U\) so \(f\of {x’}\in U\) and \(d’\of {f\of x,f\of {x’}}<\eps \) as required. □
Given a set \(X\), define \(d\of {x,y}:=0\) if \(x=y\) and \(d\of {x,y}:=1\) if \(x\neq y\). Prove that \(d\) is a metric.
Solution. We have \(d\of {x,y}\ge 0\) for each \(x,y\in X\) with equality only when \(x=y\) so positivity holds. Since \(d\of {x,y}=d\of {y,x}\) for every \(x,y\in X\), symmetry holds. It remains to verify the triangle inequality. Suppose, for a contradiction, that the triangle inequality fails so there are \(x,y,z\in X\) with
\[ d\of {x,y}+d\of {y,z}<d\of {x,z}. \]
Then \(d\of {x,z}=1\) and \(d\of {x,y}=d\of {y,z}=0\) so \(x=y\) and \(y=z\). Thus \(x=z\) and we have a contradiction. □
Let \(\cur {X,d}\) be a metric space. Define
\[ d_{1}\of {x,y}:=\dfrac {d\of {x,y}}{1+d\of {x,y}} \]
and
\[ d_{2}\of {x,y}:=\min \cur {1,d\of {x,y}}. \]
Prove that \(d_{1}\) and \(d_{2}\) are metrics on \(X\).
Solution. Since \(d\of {x,y}\ge 0\), it follows that \(d_{1}\of {x,y}\ge 0\) for every \(x,y\in X\). If \(d_{1}\of {x,y}=0\), then \(d\of {x,y}=0\) so \(d_{1}\) satisfies positivity. Since \(d\) satisfies symmetry, it follows that \(d_{1}\) satisfies symmetry.
Now, we verify the triangle inequality for \(d_{1}\). Let \(x,y,z\in X\). We have
\( \seteqsection {1} \)\begin{align*} d_{1}\of {x,y}+d_{1}\of {y,z} & =\dfrac {d\of {x,y}}{1+d\of {x,y}}+\dfrac {d\of {y,z}}{1+d\of {y,z}}\\ & =2-\frac {1}{1+d\of {x,y}}-\frac {1}{1+d\of {y,z}}. \end{align*} Moreover,
\( \seteqsection {1} \)\begin{align*} \frac {1}{1+d\of {x,y}}+\frac {1}{1+d\of {y,z}} & \le \frac {2+d\of {x,y}+d\of {y,z}}{1+d\of {x,y}+d\of {y,z}}\\ & =1+\frac {1}{1+d\of {x,y}+d\of {y,z}}. \end{align*} Since \(d\) satisfies the triangle inequality, it follows that \(d\of {x,y}+d\of {y,z}\ge d\of {x,z}\) so
\[ \frac {1}{1+d\of {x,y}}+\frac {1}{1+d\of {y,z}}\le 1+\frac {1}{1+d\of {x,z}}. \]
Thus
\( \seteqsection {1} \)\begin{align*} d_{1}\of {x,y}+d_{1}\of {y,z} & \ge 2-\cur {1+\frac {1}{1+d\of {x,z}}}\\ & =1-\frac {1}{1+d\of {x,z}}=\frac {d\of {x,z}}{1+d\of {x,z}}=d_{1}\of {x,z}. \end{align*} Thus \(d_{1}\) is a metric.
Now, we verify that \(d_{2}\) is a metric. It is clear that \(d_{2}\) satisfies positivity and symmetry. We prove that \(d_{2}\) satisfies the triangle inequality. Let \(x,y,z\in X\). Suppose, for a contradiction, that
\[ d_{2}\of {x,y}+d_{2}\of {y,z}<d_{2}\of {x,z}. \]
Since \(d_{2}\of {x,z}\le 1\), it follows that
\[ d_{2}\of {x,y}+d_{2}\of {y,z}<1, \]
so \(d_{2}\of {x,y}<1\), which implies that \(d_{2}\of {x,y}=d\of {x,y}\). Similarly \(d_{2}\of {y,z}=d\of {y,z}\). Since \(d_{2}\of {x,y}<d\of {x,y}\), we conclude that
\[ d\of {x,y}+d\of {y,z}<d\of {x,z}, \]
which is a contradiction. □
Prove that what we defined as the “supremum metric” in Example 1.1.7 is a metric.
Solution. If \(f,g\in X\)
\[ d\of {f,g}:=\sup \set {\abv {f\of y-g\of y}:y\in Y}. \]
Since the absolute value is never negative, we have \(d\of {f,g}\ge 0\) for any \(f,g\in X\). If \(d\of {f,g}=0\), then \(\abv {f\of y-g\of y}=0\) for every \(y\in Y\) so \(f=g\). Thus positivity holds. Since
\[ \abv {f\of y-g\of y}=\abv {g\of y-f\of y}, \]
the symmetry holds. It remains to verify the triangle inequality. Let \(f,g,h\in X\). Then
\[ \abv {f\of z-g\of z}+\abv {g\of z-h\of z}\ge \abv {f\of z-h\of z} \]
for any \(z\in Y\) so
\( \seteqsection {1} \)\begin{align*} d\of {f,g}+d\of {g,h} & =\sup \set {\abv {f\of y-g\of y}:y\in Y}\\ & +\sup \set {\abv {g\of y-h\of y}:y\in Y}\ge \abv {f\of z-h\of z} \end{align*} for any \(z\in Y\). Thus
\[ d\of {f,g}+d\of {g,h}\ge \sup \set {\abv {g\of y-h\of y}:y\in Y}=d\of {f,h}. \]
□
Let \(\cur {X,d}\) be a metric space and \(\cur {x_{n}}_{n=1}^{\ity }\) be a sequence in \(X\). Prove that the sequence \(\cur {x_{n}}_{n=1}^{\ity }\) converges to \(x\in X\) if and only if for every open \(U\sub X\) with \(x\in U\) there exists \(k\in \bn \) such that \(x_{n}\in U\) for every \(n\ge k\).
Solution. Let \(x\in X\). Assume that \(\cur {x_{n}}_{n=1}^{\ity }\) converges to \(x\). Let \(U\) be open in \(X\) with \(x\in U\). There is \(\eps >0\) such that \(B\of {x,\eps }\sub U\). Since \(\cur {x_{n}}_{n=1}^{\ity }\) converges to \(x\), there is \(k\in \bn \) such that \(d\of {x_{n},x}<\eps \) for every \(n\ge k\). Then \(x_{n}\in B\of {x,\eps }\) so \(x_{n}\in U\) for every \(n\ge k\).
Now assume that for every open \(U\sub X\) with \(x\in U\) there exists \(k\in \bn \) such that \(x_{n}\in U\) for every \(n\ge k\). Let \(\eps >0\) be arbitrary. Let \(U:=B\of {x,\eps }\). Then \(U\) is open and \(x\in U\) so there is \(k\in \bn \) with \(x_{n}\in U\) for every \(n\ge k.\) Thus \(d\of {x_{n},x}<\eps \) for every \(n\ge k\), which implies that \(\cur {x_{n}}_{n=1}^{\ity }\) converges to \(x\). □