Completing metric spaces

If you’ve read the proof of the “completion of a metric space”, then you surely must have asked yourself “WHY?”! Say we have an incomplete metric space X. Why can’t we just complete X by including the limit points of all its cauchy sequences?!

No. We can’t. The limit points of cauchy sequences may not be determinable.

The new space \overline{X} that we create, is it just X\cup \{\text{limit points of cauchy sequences in X}\}? No. It is a completely different space.

So what exactly is \overline{X}? \overline{X} is a space with a new bunch of points: equivalence classes of cauchy sequences in X such that \{a\}\sim\{b\} iff \lim\limits_{n\to\infty}d(a_n,b_n)=0.

If you read the proof, you’ll realise it does a bunch of random crap to prove \overline{X} is complete. WHY?? Couldn’t it have been simpler with less dense sets and the like?

Let’s create a cauchy sequence of the equivalence classes. How do we know that the limit point of this sequence exists? We’re stuck here. One wouldn’t know how to proceed.

On a more important note, we just have a bunch of equivalence classes whose limits we do not know. We have no idea how they behave with respect to each other. Should we have equivalence classes whose limit points we do know, then we’ll have some perspective on the structure of the space and what the limit point of the cauchy sequence is. We might not even know the terms of some such equivalence classes. How’re we supposed to analyze things we have absolutely no idea about?

Some information is better than no information. If we could find out the limit points of all such equivalence classes (or terms of the cauchy sequence, in this case), we could think of doing something productive. But we can’t determine the limit points. So what now?

Consider all equivalence classes of cauchy sequences which converge to points in the space X. This set is dense in \overline{X} (this is easy to prove).

A fundamental concept is this: Let us take a cauchy sequence \{a_1,a_2,a_3,\dots\}, and another cauchy sequence \{b_1,b_2,b_3,\dots\} which converges to a_N. Then \lim\limits_{n\to\infty}d(a_n,b_n)=\epsilon, where \epsilon is a fixed number. As N increases, the cauchy sequence \{b_i\} converges to \{a_i\}. Hence, we extrapolate from the concept of convergence of points to convergence of converging sequences. Can we think about the convergence of converging sequences in any other way? Something to think about. But this is definitely a useful concept to remember. Note that the limit point of \{a_i\} may not even be known.

So how is this concept relevant to the proof? We’ve associated with the original cauchy sequence \{x_i\} another cauchy sequence \{b_i\} with limit points in the space, as mentioned before. The association is such that \lim b_i=\lim x_i. Now the masterstroke- we map each sequence to the limit in the original space X: we map \{y_i\} converging to l, to the point l in X. Isn’t that a lot of potentially useless mapping? No. This is explained below.

What do we have here? We have a cauchy sequence \{l_1,l_2,l_3,\dots\}. This may or may not have a limit, which is inconsequential to the proof. Now let us take the cauchy sequence \{t^i\} converging to l_i. We know from before that \lim\limits_{n\to\infty}d(l_n,t_n)=0. Now let us take equivalence classes of the sequence \{l_1,l_2,\dots\}, and the sequences \{t_i\}. The cauchy sequence of equivalence classes of \{t_i\} will obviously converge to the equivalence class of l_i. As a result, the original \{x_i\} also converges to the equivalence class of \{l_i\}. We had associated \{x_i\} just so that we could get sequences converging to the terms of \{l_1,_2,\dots\}.

What is the point of creating these equivalence classes? Couldn’t we have formed a complete metric space in some other way? Thinking about cauchy sequences, something that immediately pops into mind is cauchy sequences of cauchy sequences. Cauchy sequences of what else can be formed? Cauchy sequences of squares of points? Will that space really be complete? Maybe there are other possibilites to form a complete metric space as derived from X, but this one is one that easily pops into mind after one gets comfortable with the concept of the cauchy sequence \{l_1,l_2,\dots\} and the sequences \{t_i\} converging to l_i. Whether metric spaces can be completed in other ways is something you and I should think about.

Published by -

Graduate student

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: