Section 1.3 Convergence and Completeness
The idea behind convergence of a sequence is that the distance to the limit approaches \(0\text{.}\) This concept has a natural extension to metric spaces.
Definition 1.29.
Let \((X,d)\) be a metric space and \((x_n)_{n \in \N}\) a sequence in \(X\text{.}\) A point \(x_0 \in X\) is said to be a limit of the sequence if \(d(x_n,x_0) \to 0\) in \(\R\) as \(n \to \infty\text{.}\) If so, we say that the sequence converges to \(x_0\) and we write \(x_0 = \lim_{n \to \infty} x_n\) or \(x_n \to x_0\text{.}\)
Example 1.30.
In \(\R\text{,}\) consider the sequence \((x_n)_{n \in \N}\) with \(x_n =
\frac{1}{n}\text{.}\) Then \(d(x_n,0) = \frac{1}{n}\text{,}\) so \(x_n\) converges to \(0\text{.}\)
Example 1.31.
In \(X = (0,1]\text{,}\) let \(x_n = \frac{1}{n}\text{.}\) Then there is no point \(x_0\) in \(X\) such that \(d(x_n,x_0) \to 0\) as \(n \to
\infty\text{.}\) The sequence \((x_n)_{n \in \N}\) is not convergent in this space.
Convention 1.32. Notation for sequences.
Sequences are one of the main characters in this unit, and so it is important that we agree on clear and consistent notation for them.
- On the board and on problem sheets and exams, we allow ourselves to drop the subscript “\(n \in \N\)” from \(\seq xn\) when it is clear from context.
- It is sometimes convenient to indicate sequences by listing their terms with an ellipsis, \(\seq xn = (x_1,x_2,\ldots)\text{.}\)
- There is unfortunately no standard notation for the phrase “\(\seq xn\) is a sequence in \(X\)”. It is incorrect to write \(\seq xn \in X\text{,}\) since the sequence \(\seq xn\) is not a point in \(X\text{,}\) and it also incorrect to write \(\seq xn \subset X\text{,}\) as \(\seq xn\) is not subset of \(X\) either. It is technically correct to write \(\seq xn \in X^\N\text{,}\) thinking of the sequence as a mapping \(\N \to X\text{,}\) but this is not common. Another uncommon but technically acceptable option is to write \(x_1,x_2,\ldots \in X\text{.}\)
Example 1.33. Uniform convergence.
Given a set \(S\text{,}\) consider the normed space \((B(S),\n\blank_{\sup})\) from Example 1.7. A sequence \((f_n)_{n \in \N}\) converges to \(f_0\) in \(B(S)\) if and only if
\begin{equation*}
\forall \varepsilon \gt 0
\,
\exists N \in \N
\st
\forall n \ge N
\, \forall s \in S
,\;
|f_n(s) - f_0(s)| \lt \varepsilon\text{.}
\end{equation*}
This is also known as uniform convergence.
Theorem 1.34.
In a metric space \((X,d)\text{,}\) let \((x_n)_{n \in \N}\) be a convergent sequence with limit \(x_0 \in X\text{.}\)
- If \(x_n \to y_0\) in \((X,d)\text{,}\) then \(x_0 = y_0\) (i.e., limits are unique).
- For every subsequence \((x_{n_k})_{k \in \N}\text{,}\) the convergence \(x_{n_k} \to x_0\) holds as well.
Proof.
Part i.
By the triangle inequality,
\begin{equation*}
d(x_0,y_0) \le d(x_0,x_n) + d(y_0,x_n) \to 0
\end{equation*}
as \(n \to \infty\text{.}\) Hence \(d(x_0,y_0) = 0\text{,}\) which means that \(x_0
= y_0\text{.}\)
Part ii.
Recall: when we say that \((x_{n_k})_{k \in \N}\) is a subsequence of \((x_n)_{n \in \N}\text{,}\) this means that \((n_k)_{k \in \N}\) is a strictly increasing sequence of natural numbers. Let \(\varepsilon \gt
0\text{.}\) Since \(x_n \to x_0\) as \(n \to \infty\text{,}\) there exists \(N
\in \N\) such that \(d(x_n,x_0) \lt \varepsilon\) for \(n \ge N\text{.}\) Moreover, there exists \(K \in \N\) such that \(n_k \ge N\) for \(k
\ge K\text{.}\) Hence \(d(x_{n_k},x_0) \lt \varepsilon\) for \(k \ge K\text{,}\) which proves the convergence \(x_{n_k} \to x_0\) as \(k \to \infty\text{.}\)
The following criterion is very useful for checking whether or not a given set is closed.
Theorem 1.35.
Let \((X,d)\) be a metric space. A set \(S \subseteq X\) is closed if and only if, for every sequence in \(S\) that converges in \(X\text{,}\) the limit belongs to \(S\text{.}\)
Proof.
Suppose that \(S\) is closed and \((x_n)_{n \in \N}\) is a sequence in \(S\) with limit \(x_0 \in X\text{.}\) Then for any \(r \gt 0\text{,}\) there exists \(n \in \N\) with \(x_n \in B_r(x_0)\text{.}\) But since \(x_n \in
S\text{,}\) this means that \(B_r(x_0) \cap S \ne \varnothing\text{;}\) so \(x_0 \in
\overline S = S\text{.}\)
Now suppose that \(S\) is not closed. Then \(X \setminus S\) is not open. Hence there exists \(x_0 \in X \setminus S\) such that \(B_r(x_0) \cap S \ne \varnothing\) for any \(r \gt 0\text{.}\) Use this observation for \(r = \frac{1}{n}\) for every \(n \in \N\text{.}\) Thus we find points \(x_n \in B_{1/n}(x_0) \cap S\text{.}\) So \((x_n)_{n \in \N}\) is a sequence in \(S\) that converges to \(x_0\text{.}\) But \(x_0 \not\in
S\text{,}\) so the condition from the theorem is not satisfied.
Thinking back to first and second year analysis, you may remember the Cauchy principle, which states that a sequence in \(\R\) is convergent if and only if it is a Cauchy sequence. The concept of a Cauchy sequence has an obvious generalisation to metric spaces.
Definition 1.36.
A sequence \((x_n)_{n \in \N}\) in a metric space \((X,d)\) is called a Cauchy sequence if
\begin{equation*}
\forall \varepsilon \gt 0
\,
\exists N \in \N
\st
\forall m,n \ge N
,\;
d(x_m,x_n) \lt \varepsilon\text{.}
\end{equation*}
Theorem 1.37.
Any convergent sequence is a Cauchy sequence.
Proof.
Let \((x_n)_{n \in \N}\) be a convergent sequence in the metric space \((X,d)\text{.}\) Denote its limit by \(x_0\text{.}\) Fix \(\varepsilon \gt 0\text{.}\) Then there exists an \(N \in \N\) such that \(d(x_n,x_0) \lt
\frac{\varepsilon}{2}\) for \(n \ge N\text{.}\) For \(m,n \ge N\text{,}\) we then conclude that
\begin{equation*}
d(x_m,x_n) \le d(x_m,x_0) + d(x_n,x_0) \lt \varepsilon\text{.}
\end{equation*}
Is the Cauchy principle true in metric spaces, too? In general, the answer is no. But in many cases, it is true, and we pay these spaces particular attention.
Definition 1.38.
A metric space \((X,d)\) is called complete if every Cauchy sequence in \(X\) is convergent. A complete normed space is called a Banach space and a complete inner product space is called a Hilbert space.
Example 1.39.
The Euclidean spaces \(\R^n\) are complete by results from second year analysis. Hence they are Hilbert spaces.
Theorem 1.40.
Let \(S\) be a set. Then the space \(B(S)\) of bounded functions \(f \maps S \to \R\) with the supremum norm is a Banach space.
Proof.
We already know that \(B(S)\) is a normed space. To verify that the space is complete, let \(\seq fn\) be a Cauchy sequence in \(B(S)\text{.}\) We must show that there exists \(f_0 \in B(S)\) such that \(\n{f_n-f_0}
\to 0\) as \(n \to \infty\text{.}\)
Step 1: Defining \(f_0\).
Fix \(s \in S\text{.}\) For any given \(\varepsilon \gt 0\text{,}\) there exists \(N \in \N\) such that \(d(f_m,f_n) \lt \varepsilon\) when \(m,n \ge
N\text{.}\) But then
\begin{equation*}
|f_m(s) - f_n(s)| \le \|f_m - f_n\|_{\sup} \lt \varepsilon\text{.}
\end{equation*}
Hence \((f_n(s))_{n \in \N}\) is a Cauchy sequence in \(\R\text{.}\) Since \(\R\) is complete, there exists a limit. Call the limit \(f_0(s)\text{.}\) Thus for every \(s \in S\) we obtain a number \(f_0(s)\text{,}\) and this gives rise to a function \(f_0 \maps S \to \R\text{.}\)
Step 2: \(f_0 \in B(S)\).
Since \(\seq fn\) is a Cauchy sequence, it is bounded by Exercise 1.3.3. Since \(B(S)\) is a normed space, this is equivalent (by Exercise 1.1.11) to
\begin{gather*}
M = \sup_{n \in \N} {\n{f_n}_{\sup}} \lt \infty \text{.}
\end{gather*}
Therefore, for any \(s \in S\text{,}\)
\begin{equation*}
|f_0(s)| = \lim_{n \to \infty} |f_n(s)| \le M\text{.}
\end{equation*}
This proves that \(f_0\) is bounded, i.e., that \(f_0 \in B(S)\text{.}\)
Step 3: \(f_n \to f_0\).
Fix \(\varepsilon \gt 0\text{,}\) and choose \(N \in
\N\) such that \(\|f_m - f_n\|_{\sup}\lt \frac{\varepsilon}{2}\) when \(m,n \ge N\text{.}\) Then for any \(s \in S\text{,}\)
\begin{align*}
|f_n(s) - f_0(s)|
\amp= \left|f_n(s) - \lim_{m \to \infty} f_m(s)\right| \\
\amp= \lim_{m \to \infty} |f_n(s) - f_m(s)| \le \frac{\varepsilon}{2}
\end{align*}
whenever \(n \ge N\text{.}\) Hence
\begin{equation*}
\|f_n - f_0\|_{\sup}
= \sup_{s \in S} |f_n(s) - f_0(s)|
\le \frac{\varepsilon}{2}
\lt \varepsilon
\end{equation*}
for all \(n \ge N\text{.}\)
Theorem 1.41.
Let \((X,d)\) be a complete metric space and let \((Y,d')\) be a metric subspace. Then \((Y,d')\) is complete if and only if \(Y\) is closed as a subset of \(X\text{.}\)
Proof.
Suppose that \(Y\) is closed and let \((x_n)_{n \in \N}\) be a Cauchy sequence in \(Y\text{.}\) Then it is also a Cauchy sequence in \(X\text{,}\) and it follows that it has a limit \(x_0\) in \(X\text{.}\) By Theorem 1.35, we know that \(x_0 \in Y\text{.}\) Since
\begin{equation*}
d'(x_n,x_0) = d(x_n,x_0) \to 0\text{,}
\end{equation*}
this means that \(x_n \to x_0\) in \(Y\text{.}\) So \(Y\) is complete.
Conversely, suppose that \(Y\) is not closed. Then by Theorem 1.35 again, there exists a sequence \((x_n)_{n \in
\N}\) in \(Y\) that is convergent in \(X\) with a limit \(x_0 \in X
\setminus Y\text{.}\) According to Theorem 1.37, this is a Cauchy sequence in \(X\) and therefore a Cauchy sequence in \(Y\) as well. Then by the uniqueness of limits (Theorem 1.34), it cannot be convergent in \(Y\text{.}\) So \(Y\) is not complete.
Exercises Exercises
1. (PS3) Some basic limits.
Let \((X,d)\) be a metric space, and suppose that \(\seq xn\) and \(\seq
yn\) are convergent sequences in \(X\) with \(x_n \to x_0 \in X\) and \(y_n \to y_0 \in X\text{.}\)
(a)
Show that \(d(x_n,y_n) \to d(x_0,y_0)\text{.}\)
Hint.
Use the triangle inequality to show \(d(x_0,y_0) \le d(x_0,x_n) +
d(x_n,y_n) + d(y_n,y_0)\text{,}\) as well as the same inequality with the roles of \(0\) and \(n\) reversed. Then use this to estimate the difference \(\abs{d(x_n,y_n) - d(x_0,y_0)}\text{.}\)
(b)
Suppose that \((X,\n\blank)\) is in fact a normed space, and let \(\seq \alpha n\) be a convergent sequence in \(\R\) with \(\alpha_n \to \alpha_0 \in
\R\text{.}\) Show that
- \(\n{x_n} \to \n{x_0}\text{,}\)
- \(x_n + y_n \to x_0 + y_0\) and
- \(\alpha_n x_n \to \alpha_0 x_0\text{.}\)
Hint.
For the first part, find a way to express \(\n{x_n}\) in terms of the metric \(d\text{.}\) For the other two parts you will want to use the triangle inequality, perhaps after adding and subtracting an appropriate ‘cross term’.
(c)
Suppose that \((X,\scp\blank\blank)\) is an inner product space. Show that \(\scp{x_n}{y_n} \to \scp {x_0}{y_0}\text{.}\)
Hint.
Again, experiment with adding and subtracting an appropriate ‘cross term’. The Cauchy–Schwarz inequality is also useful.
2. (PS3) Convergence in product spaces.
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces, and consider the product metric space \((X \times Y, d_{X \times Y})\) defined in Definition 1.14. Let \((x_n)_{n \in \N}\) be a sequence in \(X\) and \((y_n)_{n \in \N}\) a sequence in \(Y\text{.}\) Let \(x_0 \in X\) and \(y_0 \in
Y\text{.}\) Show that \((x_n, y_n) \to (x_0, y_0)\) in \(X \times Y\) as \(n \to
\infty\) if and only if \(x_n \to x_0\) in \(X\) and \(y_n \to y_0\) in \(Y\) as \(n \to \infty\text{.}\)
3. (PS4) Cauchy implies bounded.
A sequence \((x_n)_{n \in \N}\) in a metric space \((X,d)\) is called bounded if the set \(\set{x_n}{n \in \N}\) is bounded. Show that any Cauchy sequence in a metric space is bounded.
Hint.
Apply the definition of \(\seq xn\) being Cauchy with \(\varepsilon=1\) to find an appropriate \(N\in\N\text{.}\) Then estimate \(d(x_n,x_m)\) in different ways depending on whether \(n\ge N\text{,}\) \(m \ge N\text{,}\) neither, or both.
4. (PS4) Pairs of Cauchy sequences.
Let \(\seq xn\) and \(\seq yn\) be Cauchy sequences in a metric space \((X,d)\text{.}\) Show that the sequence \((d(x_n,y_n))_{n \in \N}\) is also Cauchy, and conclude that the limit \(\lim_{n\to \infty} d(x_n,y_n)\) exists.
Hint.
The inequality \(\abs{d(x_n,y_n) - d(x_m,y_m)} \le d(y_n,y_m) +
d(x_n,x_m)\) is quite useful here, and has already been established in Exercise 1.3.1.
5. Cauchy sequences and double limits.
Let \(\seq xn\) be a Cauchy sequence in a metric space \((X,d)\text{.}\)
(a)
For any \(n \in \N\text{,}\) show that the limit
\begin{equation*}
\lim_{k \to \infty} d(x_n,x_k)
\end{equation*}
exists.
Hint.
One method is to show that the sequence \((d(x_n,x_k))_{k\in \N}\) is Cauchy. This is related to Exercise 1.3.4.
(b)
Show that
\begin{equation*}
\lim_{n\to \infty} \left(
\lim_{k \to \infty} d(x_n,x_k) \right) = 0 \text{.}
\end{equation*}
6. Limit points.
Suppose that \((x_n)_{n \in \N}\) is a sequence in a metric space \((X,
d)\) and \(x_0 \in X\) is a point such that \(B_r(x_0) \cap
\set{x_n}{n \in \N} \not\subseteq \{x_0\}\) for all \(r \gt 0\text{.}\) Show that there exists a subsequence \((x_{n_k})_{k \in \N}\) with \(x_0 =
\lim_{k \to \infty} x_{n_k}\text{.}\)
