3 Complete Metric Spaces
A Cauchy sequence is one whose terms eventually crowd together. In a complete space every such sequence actually converges, and the point of this section is that completeness is what lets a limit be constructed before it is known.
After establishing that \(\mathbb {R}\) and \(l_p\) are complete, and that a metric space can always be completed, the section reaches the contraction mapping theorem: a map that shrinks distances by a fixed factor has exactly one fixed point, and the point is found by iterating from anywhere at all.
Picard’s theorem is the application, and it is the reason the integral equation on the title page is worth writing down. Recasting an initial value problem \(y'=f(t,y)\), \(y(x_0)=y_0\) as \[g(x)=y_0+\int _{x_0}^{x}f\big (t,g(t)\big )\,dt\] turns ”solve the differential equation” into ”find a fixed point”, and the contraction mapping theorem then supplies existence and uniqueness at once.
Definition 3.1. A metric space \((X,d)\) is said to be a complete metric space if every Cauchy sequence of points in \(X\) converges to some point \(x\in X\).
Solution. Let \((x_n)\) be Cauchy in \(\mathbb {R}\).
It is bounded
Taking \(\varepsilon =1\) there is \(N\) with \(|x_n-x_N|<1\) for \(n\geq N\), so every term lies within \(\max \{|x_1|,\dots ,|x_{N-1}|,|x_N|+1\}\) of the origin.
It has a convergent subsequence
By the Bolzano–Weierstrass theorem a bounded real sequence has a convergent subsequence, say \(x_{n_k}\rightarrow l\).
The whole sequence converges to the same limit
Given \(\varepsilon >0\) choose \(N\) with \(|x_n-x_m|<\frac {\varepsilon }{2}\) for \(n,m\geq N\), and then \(k\) with \(n_k\geq N\) and \(|x_{n_k}-l|<\frac {\varepsilon }{2}\). For \(n\geq N\), \[|x_n-l|\leq |x_n-x_{n_k}|+|x_{n_k}-l|<\varepsilon .\]
So every Cauchy sequence in \(\mathbb {R}\) converges in \(\mathbb {R}\). This rests on the completeness axiom for \(\mathbb {R}\), by way of Bolzano–Weierstrass; it is not a fact that can be proved from the ordered-field axioms alone, as \(\mathbb {Q}\) demonstrates.
Example 3.3. \(\mathbb {R}^2\) is complete.
Let \(x_n=(x^n_1,x^n_2)\) be a Cauchy sequence in \(\mathbb {R}^2\), then given \(\varepsilon >0,\exists n_0\in \mathbb {N}\),
\(d(x_m,x_n)<\varepsilon \hspace {0.3cm} \forall m,n>n_0\). i.e
\[\sqrt {(x^n_1-x_1^m)^2+(x^n_2-x^m_2)^2}<\varepsilon \]
i.e \(\hspace {0.2cm}(x_1^n-x_1^m)^2+(x_2^n-x_2^m)^2<\varepsilon ^2\hspace {0.3cm} m,n>n_0\) \(\implies (x_1^n-x_1^m)^2<\varepsilon ^2\) and \((x_2^n-x_2^m)^2<\varepsilon ^2\).
Hence we have that \(x^n_1\) and \(x_2^n\) are Cauchy sequences in \(\mathbb {R}\). Now, every Cauchy sequence in \(\mathbb {R}\) converges to some limit in \(\mathbb {R}\). It means \[x_1^n\rightarrow x_1\in \mathbb {R}\] \[x^n_2\rightarrow x_2\in \mathbb {R}\] i.e \[|x^n_1-x_1|<\varepsilon \hspace {0.5cm} \forall n\geq n_0\] \[|x^n_2-x_2|<\varepsilon \hspace {0.5cm} \forall n\geq n_0.\] \[\sqrt {(x^n_1-x_1)^2+(x^n_2-x_2)^2}<\varepsilon \implies d(x_n,x)<\varepsilon \hspace {0.5cm} n\geq n_0\] Thus \(x_n\) converges to a limit \(x=(x_1,x_2)\in \mathbb {R}^2\).
Example 3.4. Show that the metric space \(\mathbb {R}^n\), where for \(x,y\in \mathbb {R}^n\), \(\hspace {0.3cm}\displaystyle { d(x,y)=\Bigg [\sum ^n_{i=1}(x_i-y_i)^2\Bigg ]^{\frac {1}{2}}}\) is complete.
Solution. Let \(\{x^{(k)}\}\) be Cauchy in \(\mathbb {R}^n\), writing \(x^{(k)}=\big (x^{(k)}_1,\dots ,x^{(k)}_n\big )\).
Each coordinate converges
For a fixed \(i\), every term of the sum is at most the whole sum, so \[\big |x^{(k)}_i-x^{(l)}_i\big |\leq \left [\sum _{j=1}^{n}\big (x^{(k)}_j-x^{(l)}_j\big )^2\right ]^{\frac 12} =d\big (x^{(k)},x^{(l)}\big ).\] Hence \(\{x^{(k)}_i\}_k\) is Cauchy in \(\mathbb {R}\), which is complete, so \(x^{(k)}_i\rightarrow x_i\) for some \(x_i\in \mathbb {R}\).
The vector of limits is the limit
Put \(x=(x_1,\dots ,x_n)\). Given \(\varepsilon >0\), choose for each \(i\) an index \(N_i\) beyond which \(\big |x^{(k)}_i-x_i\big |<\varepsilon /\sqrt {n}\), and let \(N=\max \{N_1,\dots ,N_n\}\) — a maximum of finitely many numbers, so finite. For \(k\geq N\), \[d\big (x^{(k)},x\big )=\left [\sum _{i=1}^{n}\big (x^{(k)}_i-x_i\big )^2\right ]^{\frac 12} <\left [n\cdot \frac {\varepsilon ^2}{n}\right ]^{\frac 12}=\varepsilon .\] So \(x^{(k)}\rightarrow x\) in \(\mathbb {R}^n\), and \(\mathbb {R}^n\) is complete.
The case \(n=2\) above is this argument written out; what makes it work for every \(n\) is that the maximum is over finitely many indices. For sequence spaces that step fails and a different argument is needed.
Theorem 3.5. Let \((X,d)\) be a complete metric space and let \(A\subseteq X\) with some metric \(d\) and \(A\) is closed. Then \((A,d)\) is a complete metric space.
Proof. Let \((x_n)\) be a Cauchy sequence in \((A,d)\). We need to show that \(x_n\rightarrow x\in A\).
Now, since \(A\subset X\), then \(x_n\) is a Cauchy sequence in \(X\). Thus, since \(X\) is complete, \((x_n)\) must converge to a point \(x\in X\). Hence \(x\) is a limit point of \(A\). Since \(A\) is closed, we get \(x\in A\). So every Cauchy sequence in \(A\) converges to a point \(x\) of \(A\). Hence \(A\) is complete. □
Example 3.6. \([0,1]\) is a closed subspace of \(\mathbb {R}\). Now, since \(\mathbb {R}\) is complete so is \([0,1]\).
3.2 Separable Metric Spaces
3.3 Closed Subspaces and the Nested Set Theorem
3.4 Completion of Metric Spaces
3.5 Fixed Point Theorems and Some Applications
3.6 Elementary Theory of Complete Metric Spaces
3.7 Practice Problems
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