Complete Metric Spaces

Some Topics Covered Definition of completeness; examples; existence of completion of a metric space

Informally, the completion of a metric space \((X,d)\) can be thought of as an operation which adds some minimal set of new points to \(X\) so that Cauchy spaces always converge (i.e., it fills in all the holes).

Step 1: Defining the completion Let \(\overline{X}\) be the collection of all equivalence classes of Cauchy sequences in \((X,d)\) modulo the following equivalence relation:

\(\{x_n\}_{n=1}^\infty \sim \{y_n\}_{n=1}^\infty\) when \(d(x_n,y_n) \rightarrow 0\) as \(n \rightarrow \infty\).

We also define the distance \(\overline{d}\) between two elements \(\overline{x}, \overline{y}\) of \(\overline{X}\) in the following way:

We need to verify that the limit always exists, that it is independent of the choice of representative, and that it satisfies the properties of a metric.

To see that the limit exists, observe that

where we used the triangle inequality to say that \(d(x_n,y_n) \leq d(x_n,x_m) + d(x_m,y_n)\) and \(d(x_m,y_n) \leq d(y_n,y_m) + d(y_m,x_m)\). By flipping the roles of \(n\) and \(m\), we can conclude that

We can make both terms on the right-hand side less than \(\epsilon/2\) when \(n\) and \(m\) are sufficiently large, which means that \(\{d(x_n,y_n)\}_{n=1}^\infty\) is a Cauchy sequence of real numbers and there fore converges and the limit exists.

To see that the limit is independent of representative, let \(d(x_n,x_n')\) and \(d(y_n,y_n')\) both tend to zero as \(n \rightarrow \infty\). Then

Swapping the roles of primed and unprimed sequences gives that

Since the right-hand side goes to zero, the limit of \(d(x_n,y_n)\) as \(n \rightarrow \infty\) must be the same as the limit of \(d(x_n',y_n')\) as \(n \rightarrow \infty\).

To see that \(\overline{d}\) is a metric is mostly a series of simple observations resting on the fact that \(d\) is a metric. The most delicate part is the triangle inequality. To achieve it, take the limit of the inequality \(d(x_n,z_n) \leq d(x_n,y_n) + d(y_n,z_n)\) as \(n \rightarrow \infty\) to get \(\overline{d}(\overline{x},\overline{z}) \leq \overline{d}(\overline{x},\overline{y}) + \overline{d}(\overline{y},\overline{z})\) whenever \(\{x_n\}_{n=1}^\infty, \{y_n\}_{n=1}^\infty\), and \(\{z_n\}_{n=1}^\infty\) represent \(\overline{x}, \overline{y},\) and \(\overline{z}\), respectively (because we know that the limit of every individual term in the inequality exists).

Step 2: Identifying \(X\) as a subset of \(\overline{X}\) Given any point \(x \in X\), let \(\phi(x)\) be the equivalence class of the constant sequence \(x_n := x\) modulo the equivalence class above. We identify \(x\) with \(\phi(x)\) and note that \(\overline{d}(\phi(x),\phi(y)) = d(x,y)\) since \(d(x_n,y_n) = d(x,y)\) for each \(n\) when \(\{x_n\}_{n=1}^\infty\) and \(\{y_n\}_{n=1}^\infty\) are constant sequences.

Step 3: Showing that \((\overline{X},\overline{d})\) is complete Let \(\{s_n\}_{n=1}^\infty\) be a Cauchy sequence in our metric space \((\overline{X},\overline{d})\). In other words, each \(s_n\) is an equivalence class of Cauchy sequences in \(X\) and \(\overline{d}(s_n,s_m) < \epsilon\) for all \(n,m\) sufficiently large.

For each \(n\), let \(s_{n}^{(1)},s_{n}^{(2)},s_{n}^{(3)},\ldots\) be a Cauchy sequence in \(X\) which represents \(s_n\). In particular, since this sequence is Cauchy, we know that there must be some point \(x_n \in X\) such that \(d(x_n,s_{n}^{(k)}) < \frac{1}{n}\) for all \(k\) sufficiently large (e.g., we can take \(x_n := s_{n}^{(\ell)}\) for any \(\ell\) sufficiently large because we know that \(d(s_{n}^{(\ell)},s_{n}^{(k)}) < \epsilon\) for all large \(\ell\) and \(k\)).

We claim that \(\{x_n\}_{n=1}^\infty\) is itself a Cauchy sequence in \(X\). This is because

for all \(k\) sufficiently large. In particular, taking the limit as \(k \rightarrow \infty\) gives that

If \(n\) and \(m\) are so large that \(n,m > 3\epsilon^{-1}\) and \(\overline{d}(s_n,s_m) < \epsilon/3\), it follows that \(d(x_n,x_m) < \epsilon\) for all such \(n\) and \(m\).

Let \(\overline{x}\) be the equivalence class of the Cauchy sequence \(\{x_n\}_{n=1}^\infty\) just constructed. Then

but

so if both \(n\) and \(k\) are sufficiently large, both terms on the right-hand side are less than \(\epsilon/2\), which means that \(\overline{d}(s_n,\overline{x}) \leq \epsilon\) for all these large \(n\). Therefore we have shown that \(s_n \rightarrow \overline{x}\) in \((\overline{X},\overline{d})\).

Step 4: Showing that \((\overline{X},\overline{d})\) is minimal Suppose that \((Y,d_Y)\) is a complete metric space containing \(X\) such that \(d_Y\) agrees with \(d\) on the set \(X\). Every point \(\overline{x} \in \overline{X}\) is an equivalence class of a Cauchy sequence \(x_1,x_2,\ldots\) in \(X\). Since \(Y\) is complete and contains \(X\), there must be some limit \(\phi(\overline{x})\) of this sequence in the set \(Y\). If \(x_1',x_2',\ldots\) is any other sequence representing \(\overline{x}\), then the fact that \(d(x_n,x_n') \rightarrow 0\) as \(n \rightarrow \infty\) implies that \(d_Y(x_n,x_n') \rightarrow 0\) as \(n \rightarrow \infty\), which means that both sequences converge to the same thing \(\phi(\overline{x})\) in the set \(Y\). We identify the point \(\phi(\overline{x}) \in Y\) with the point \(\overline{x} \in \overline{X}\). Note that by looking at constant sequences, we have to have that \(\phi(x) = x\) for all \(x \in X\).

It suffices now to show that \(d_Y(\phi(\overline{x}),\phi(\overline{z})) = \overline{d}(\overline{x},\overline{z})\). But this is an immediate consequence of the fact that \(d_Y(x_n,z_n) = d(x_n,z_n)\) when \(\{x_n\}_{n=1}^\infty\) and \(\{z_n\}_{n=1}^\infty\) represent \(\overline{x}\) and \(\overline{z}\), respectively. Because \(x_n \rightarrow \phi(\overline{x})\) and \(z_n \rightarrow \phi(\overline{z})\) in \(Y\), \(\lim_{n \rightarrow \infty} d_Y(x_n,z_n) = d_Y(\phi(\overline{x}),\phi(\overline{z}))\). But also \(\lim_{n \rightarrow \infty} d(x_n,z_n) := \overline{d}(\overline{x},\overline{z})\).