Is X complete with the given metric d. Let $X=\displaystyle\prod_{i=1} ^{\infty}\mathbb{R}$ be the countable product space  of real lines.
Define a metric on $X$ by $$d(f,g)=e^{-k}$$ where $k=\min \left\lbrace i:f(i)\neq g(i)\right\rbrace$.
I want to determine whether $X$ is complete or not. 
My guess is $X$ may be a complete space. 
Any help would be appreciated.
 A: Let $D(f,g) = \{i \in \mathbb N \mid f(i) \ne g(i) \}$. Then you define $d(f,g) = \min D(f,g)$. As pointed out in the comments, this only works for $f \ne g$ and you should additionally define $d(f,f) = 0$. Of course you may use the interpretions $\min \emptyset =  \infty$ and $e^{-\infty} = 0$, but this is not really satisfactory.
Note that $d(f,g) < e^{-k}$ implies $f(i) = g(i)$ for $i \le k$.
Now let $(f_n)_{n\in \mathbb N}$ be a Cauchy sequence. We can find an increasing sequence $(M(k))_{k\in \mathbb N}$ of integers $M(k)$ such that $d(f_n,f_m) < e^{-k}$ for $n,m \ge M(k)$. Hence
$\quad$ $f_n(i) = f_{M(k)}(i)$ for $n \ge M(k)$ and $i \le k$.
Recalling that  $(M(k))$ is increasing and taking $n = M(k')$ and $i = k$ this implies
$\quad$ $f_{M(k')}(k) = f_{M(k)}(k)$ for $k' \ge k$.
In this formula we can replace $(k,k')$ by $(i,k)$ and obtain
$\quad$ $f_{M(i)}(i) = f_{M(k)}(i)$ for $i \le k$.
Define $f \in \prod_{i=1}^\infty \mathbb R$ by
$$f(i) = f_{M(i)}(i) .$$
Then for $n \ge M(k)$ and $i \le k$
$$f(i) = f_{M(i)}(i) = f_{M(k)}(i) = f_n(i)$$
which means $d(f_n,f) \le e^{-(k+1)} < e^{-k}$. Hence $(f_n)$ converges to $f$.
