Constructing completion of metric space on closed intervals

Question

Given $$X=\{[a,b] | -\infty < a < b < \infty \}$$ and $|I|=b-a$ for $I=[a,b]$, I have already proved that $d: X\times X \to \mathbb{R}$ where $$d(I,J)=|I|+|J|-2|I\cap J|$$ is a metric. I am now asked to construct a completion $(\tilde{X}, \tilde{d})$ of $(X,d)$ but am lost doing so. I know that $X\subset \tilde{X}$ has to be dense, $\tilde{d}=d$ on $X$ and all cauchy sequences have to converge.

Answer 1

I think the completion just needs one more element, something of size $0$.

Let $(I_n)_n$ be a sequence in $X$. First note that the sequence of intervall lengths $(|I_n|)_n$ needs to be a Cauchy sequence: $$ \begin{align} \big||I_n| - |I_k|\big| &= \Big| \big(|I_n| - |I_n \cap I_k|\big) - \big(|I_k| - |I_n \cap I_k|\big) \Big| \\ &\leq |I_n| - |I_n \cap I_k| + |I_k| - |I_n \cap I_k| \\ &= d(I_n, I_k) \end{align} $$ $(I_n)_n$ is a sequence in $ℝ$ and therefore has a limit.

Case 1: $|I_n|→ 0$ as $n → ∞$. Then $(I_n)_n$ is a Cauchy-Sequence and is equivalent to all other such sequences since $d(I_n, I_k) \leq |I_n| + |I_k| → 0$. Since there is no element in $X$ with size $0$, we should add one. As I commented any does the job, $[1,1]$ is not different to $[0,0]$.

Case 2: $|I_n| → L > 0$. Then drawing pictures leads to he conclusion that the intervalls must agree on a growing portion. My claim is that $$ \lim_{n → ∞} I_n := \overline{\bigcup_{k ∈ ℕ} \underbrace{\bigcap_{n \geq k} I_n}_{=: B_k}} $$ is in $X$ and the limit of the Cauchy sequence $(I_n)_n$.

Let $x_n → x$ be a convergent sequence in $B_k$. Then it is a convergent sequence in all $I_n$ ($n\geq k$) and hence $x ∈ I_n$ for all $n \geq k$, hence $x ∈ B_k$. So $B_k$ is closed. It is an intervall since otherwise there must be an $I_n$ ($n ≥ k$) that has this gap.

Since we are intersecting less and less intervalls while $k → ∞$, the sequence $\left( B_k \right)_k$ is monotonically increasing with respect to $\subset$. Therefore $\bigcup_{k ∈ ℕ} B_k$ is an intervall as well and the closure $\lim_{n→∞} I_n$ is therefore a closed intervall. Since $|I_n|$ converges and is therefore bounded, each of $B_k$ is bounded by the same constant and therefore $\lim_{n→∞} I_n$ is bounded by this constant. Hence $\lim_{n→∞} I_n ∈ X$. (Here bounded means that the size/length/ $|·|$ is bounded.)

Now to $I_n → \lim_{n → ∞} I_n$. Let $ε > 0$. Take $k$ such that for all $m, p \geq k$ we have $d(I_n, I_m) < ε$. Since $B_k \subset I_p$, $d(I_p, B_k) = |I_p \setminus B_k|$. Let us first look at the part of $I_p$ that is left of $B_k$. There must be some $m \geq n$ such that the left (lower) end of $I_m$ is arbitrarily ($ε' \geq 0$) close to $\min B_k$. Then $ε > d(I_p, I_m) = |I_p \setminus B_k| - ε' + d(I_p \setminus B_k, I_m \setminus B_k)$ (the last summand is the part right of $B_k$.) Hence $|I_p \setminus B_k| < ε + ε' - d(\text{parts on the right})$. Since $ε'$ was arbitrary, $|I_p \setminus B_k| < ε$. _{I am somehow suspicious that this argument is fine since I thought I had to do the same thing on the right but somehow I get to my conclusion anyway.}

So $d(I_p, B_k) < ε$ or in other terms $d(I_k, B_k) → 0$ as $k → ∞$.

The last step feels like it shouldn't be too hard but my head is a bit too mushy¹ for that right now.

¹Is that suitable English?