Metric Spaces II (Completeness) Wednesday, September 11, 2024
In the previous set of notes, we defined cauchy sequences and I commented that they do converge but also they do not converge. I will expand on that more in this set of notes.
Completeness and how it fails
We say a metric space $ (X,d) $ is complete
if every cauchy sequence is convergent.
But isn’t this obvious?
Yes!——but also no!
Cauchy sequences, morally speaking, should converge, but they don’t necessarily.
Where can they fail?
To explore this, consider the metric space $ (X,d) $ with $ X = \left \{ 1/n : n \in \mathbb{Z} \setminus \left \{ 0 \right \} \right \} $ and $ d $ the Euclidean metric. This space is not complete. Why? Well, we know that the sequences $ (a_n), (b_n) \subseteq X $ defined by $ a_n = 1/n $ and $ b_n = -1/n $ are Cauchy and should converge to $ 0 $, but $ 0 \not \in X $. Hence, in some sense, $ a_n ,b_n \to 0 $, but just in a space larger than but still containing $ X $. This is precisely how Cauchy sequences can fail to converge, but it gives us a way to fix this! We (somehow) make our spaces bigger… or we complete the space. B)
Incompleteness and how to fix it
Consider an incomplete metric space $ (X,d) $. How do we fill the holes? By adding in the limits of Cauchy sequences. This is literally the core idea lol.
We proceed with a step-by-step proof.
Given a metric space $ (X,d) $, there exists a complete metric space $ (\mathcal{X},D) $ such that there is an injective map $ \iota \colon X \to \mathcal{X} $ such that $ d(x_1,x_2) = D(\iota(x_1),\iota(x_2)) $.
We’re basically saying, we can put the original metric space inside a bigger, complete metric space.
Solution
Constructing the completion.
Step 1.
(Creating the fillers.)
We proceed by creating equivalence classes of Cauchy sequences.
We say that two Cauchy sequences $ (p_n) \sim (q_n) $ if
$$\begin{equation}
\lim_{n \to \infty} d(p_n,q_n) = 0.
\end{equation}$$
We claim that $ \sim $ defines an equivalence relation as it clearly satisfies reflexivity, symmetry, and transitivity.
Step 2.
(Choosing our space and its metric.)
Now let $ \mathscr{X} $ be the space of $ X $-valued sequences and let $ \mathcal{X} \coloneqq \mathscr{X} / \sim $.
We now claim that $ D \colon \mathcal{X} \times \mathcal{X} \to \mathbb{R} $ defined by $ (\overline{(a_n)},\overline{(b_n)}) \mapsto\lim_{n \to \infty} d(a_n,b_n) $ is a metric on $ \mathcal{X} $ under which $ (\mathcal{X}, D) $ is complete.
(We use the bar-notation to indicate that the object is an equivalence class under $ \sim $.)
Step 3.
(Showing $ D $ is well-defined so $ (\mathcal{X},D) $ is a metric space.)
Suppose $ (a_n) \sim (c_n) $ and $ (b_n) \sim (d_n) $.
Then using $ d $’s triangle inequality:
$$\begin{align}
D \left ( \overline{(a_n)}, \overline{(b_n)} \right ) - D\left ( \overline{(c_n)}, \overline{(d_n)} \right )
&= \lim_{n\to \infty} \left [ d(a_n,b_n) -\lim_{n\to\infty} d(c_n,d_n) \right ] \\
&\leq\lim_{n\to\infty} \left [ d(a_n,b_n) - d(c_n,d_n) \right ] \\
&\leq\lim_{n\to \infty} \left [ d(a_n,b_n) - d(c_n,b_n) \right. \\
&\qquad \left . + d(c_n,b_n) - d(c_n,d_n) \right ] \\
&\leq\lim_{n\to\infty} \left [ d(a_n,c_n) + d(b_n,d_n) \right ] \\
&= 0.
\end{align}$$
Thus $ D $ is well-defined.
I further claim that $ D $ is indeed a metric.
(It is clear that $ D $ is symmetric, reflexive, and satisfies the triangle inequality.)
Step 4.
(Showing $ (\mathcal{X},D) $ is complete.)
This will take a few steps.
Step 4.1. (Figuring out notation! :skull:)
Now, suppose $ (F_n) \subseteq \mathcal{X} $ is a Cauchy sequence with each $ F_n $ being an equivalence class of Cauchy sequences in $ (X,d) $.
For each $ n $, choose a representative and call it $ (f_n(j))_{j=1}^\infty $.
Step 4.2. (Choosing a limit candidate.)
By definition of $ (F_n) $ being Cauchy in $ (\mathcal{X},D) $, we know that, by our aforementioned notation the quantity $\lim_{j\to\infty} d(f_n(j),f_m(j)) $ can be arbitrarily small.
Since each $ (f_k(j))_{j=1}^\infty $ is Cauchy, for each $ k \in \mathbb{Z}^+ $, there exists $ N_k $ such that $ i,j \geq N_k $ implies that $ d(f_k(i),f_k(j)) < 1/k $.
From this, we are now prepared to choose our candidate: consider the sequence $ (\ell_j) $ defined by
$$\begin{equation}
\ell_j = f_j(N_j),
\end{equation}$$
and we claim that $ F_n \to [(\ell_j)] $ (where $ [(\ell_j)] $ is the $ \sim $-equivalence class on our space of $ d $-Cauchy $ X $-sequences).
But now we have to show that $ (\ell_j) $ is actually a point in our space, i.e., we have to…
Step 4.3. (Showing $ (\ell_j) $ is Cauchy in $ (X,d) $.) Let $ \varepsilon > 0 $. Then, can employ the triangle inequality to get that: $$\begin{align} d(\ell_i,\ell_j) &\leq d(f_i(N_i), f_j(N_j)) \\ &\leq \underbrace{d(f_i(N_i), f_i(r))}_A + \underbrace{d(f_i(r), f_j(r))}_B + \underbrace{d(f_j(r),f_j(N_j))}_C \end{align}$$ with $ r $ is large enough that:
- $ A,C < \varepsilon/3 $ (which exists by our choice of $ N_i, N_j$)
- $ B < \varepsilon/3 $ (which exists as $ (F_n) $ is $ D $-Cauchy so $\lim_{s\to\infty} d(f_i(s),f_j(s)) = 0 $).
Thus we have that
$$\begin{equation}
d(\ell_i,\ell_j) \leq \frac{ \varepsilon }{ 3 } + \frac{ \varepsilon }{ 3 } + \frac{ \varepsilon }{ 3 } = \varepsilon
\end{equation}$$
as desired.
Thus $ (\ell_j) $ is $ d $-Cauchy as desired.
Step 4.4. (Proving convergence to the candidate.)
Let $ \varepsilon > 0 $.
Then, for sufficiently large $ n $ (by our construction of $ (\ell_j) $ and $ (\ell_j) $’s Cauchy-ness):
$$\begin{align}
D(F_n, [(\ell_j)]) &= \lim_{j\to\infty} d(f_n(j), \ell_j) \\
&\leq \lim_{j\to\infty} \left [ d(f_n(j),\ell_n) + d(\ell_n,\ell_j) \right ] \\
&<\lim_{j\to\infty} \left [ \frac{ 1 }{ n } + \frac{ \varepsilon }{ 2 } \right ] \\
&< \frac{ \varepsilon }{ 2 } + \frac{ \varepsilon }{ 2 } = \varepsilon
\end{align}$$
as desired.
Constructing the injective map.
This part is actually pretty straight-forward. Define the map $ \iota \colon X \to \mathcal{X} $ by $ x \mapsto [(a_n)] $ where $ a_n = x $ for all $ n $ (i.e., the constant sequence). I leave the verification of its properties up to you.