Examples of Incomplete Spaces

Question

A metric space is complete if every cauchy sequence is convergent. To make space incomplete either i can change the metric or the ambient space. For example if I change real numbers into rational number with usual metric ( absolute value ) it would be incomplete. On the other hand if have a some kind of metric on some space it would be incomplete though.

My question is: Can someone give examples of incomplete spaces such that either they have unusual metric or unusual ambient space other than rational numbers etc ?

Answer 1

Let $$m_0 := \{t \in \mathbb{R}^\mathbb{N} : \{t_1, t_2, \ldots \} \text{ is finite} \}$$ be the sequences of real numbers that only take on finitely many values. Equip it with the sup-norm, i.e. $$\| t \|_\infty := \sup_{i \in \mathbb{N}} |t_i|.$$ This is an incomplete normed vector space (so it's also a metric space). To see this, consider the sequences $$t^{(1)} = (1, 0, 0, \ldots),$$ $$t^{(2)} = (1, \frac{1}{2}, 0, 0, \ldots ),$$ $$t^{(3)} = (1, \frac{1}{2}, \frac{1}{3}, 0, 0, \ldots )$$ and so on. Assume $i > j$, then we have $$\|t^{(i)} - t^{(j)} \|_\infty = \| (0, \ldots, 0, \frac{1}{j+1}, \ldots, \frac{1}{i}, 0, 0, \ldots ) \|_\infty = \frac{1}{j+1},$$ this converges to $0$ for $i,j \rightarrow \infty.$ Therefore $(t^{(i)})_{i\in \mathbb{N}}$ is a cauchy sequence in $m_0$. But it's limit (in the bigger space $\mathbb{R}^\mathbb{N}$ of sequences of real numbers) is $$(1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{i}, \ldots ),$$ which is not an element of $m_0$. Hence, $m_0$ cannot be complete.

Answer 2

Here is an example of a metric $d$ on $\mathbf R$ such that $(\mathbf R,d)$ is not complete.

Set $d(x,y)=\bigl\lvert\mathrm e^{-x}-\mathrm e^{-y}\bigr\rvert$. It trivially satisfies the axioms of a metric.

Now the sequence of natural numbers is a Cauchy sequence. Indeed choose $\varepsilon> 0$ and let $N$ be an integer such that $\;\mathrm e^{-N}<\dfrac\varepsilon2$. If $p,q>N$, we have: $$d(p,q)=\bigl\lvert\mathrm e^{-p}-\mathrm e^{-q}\bigr\rvert\le\mathrm e^{-p}+\mathrm e^{-q}<2\cdot\mathrm e^{-N}<\varepsilon. $$

This sequence does not converge. If it did converge to some $u\in \mathbf R$, we would have $\;d(n,u)=\bigl\lvert\mathrm e^{-n}-\mathrm e^{-u}\bigr\rvert<\varepsilon$ if $n>N_1$ for some $N_1$. If $n>\max(N,N_1)$, this implies $$\mathrm e^{-u}<\mathrm e^{-n}+\varepsilon<\frac{3\varepsilon}2.$$ As $\varepsilon$ was chosen arbitrarily, this means $\mathrm e^{-u}=0$, which cannot be.