Negative exponential distance Let $X := \left\{(a_k)_{k \in \mathbb N}, a_k \in \mathbb C\right\}$. Let $d\left( (a_k)_{k \in \mathbb N}, (b_k)_{k \in \mathbb N} \right) := e^{-u}$ with $u$ the smallest integer $k$ such that $a_k \ne b_k$ be a distance on $X$. Is $(X, d)$ compact/complete/connected?
Here's my not very rigorous take : $X$ is not complete since it's not even a normed space (I don't think there exists a norm that induces $d$). It is not compact since $\left((1, \dots ), (2, \dots), (3, \dots), \dots\right)$ is a sequence of elements of $X$ that does not have a convergent subsequence. I can't find an argument for connectedness, any ideas?
 A: $X$ does not have to be a normed space in order to be complete. In fact, I think $X$ is complete indeed, you just need to work through the definitions.
In short: let $a^{(n)}$ be a Cauchy sequence. This means $a^{(n)}$ is very close to $a^{(m)}$ provided $n$ and $m$ are large enough. In other words, $a^{(n)}$ and $a^{(m)}$ agree up to an arbitrarily big rank $N$ provided that $n$ and $m$ are both large enough. Define $a_k = a^{(n)}_k = a^{(m)}_k$ for $k \leqslant N$. If you think about it, this is well defined and that $a^{(n)}$ converges to $a$.
I agree with your argument that shows that $X$ is not compact.
As for connectedness, what about this: Let $\nu : X \rightarrow \mathbb{N}$ defined by $\nu(a) = \min \{n\in \mathbb{N}, a_n \neq 0\}$ (valuation). Check that $\nu$ is continuous. Conclude that $X$ is not connected.
A: You are correct that $X$ is not compact, by exactly the example you mention.
On the other hand, $X$ is in fact complete.  Observe that if $d((a_k), (b_k)) < e^{-n}$, then $a_i = b_i$ for all $i \leq n$.  It follows that if $(\mathbf{a}_n)_{n\in \mathbb{N}} = ((a_{n,k})_{k\in\mathbb{N}})_{n\in \mathbb{N}}$ is a Cauchy sequence, then for each $k$, the sequence $(a_{n,k})_{n\in\mathbb{N}}$ is eventually constant (call its eventual value $b_k$).  One can then show that the sequence $(b_k)_{k\in \mathbb{N}}$ is the limit of $(\mathbf{a}_n)_{n\in \mathbb{N}}$.
Finally, $X$ is not connected.  Indeed, for any $\mathbf{a} = (a_k)_{k\in \mathbb{N}}$, the $\varepsilon$-ball $B$ centered at $\mathbf{a}$ is disjoint from the union of the $\varepsilon$-balls centered at at every $\mathbf{b} \not \in B$.  These are open, so it follows that $X$ is not connected.
