I asked a similar question here: Does $d(A,B) := \sum\limits_{n=0}^{\infty} \frac{1}{2^n}\cdot \frac{d_n(A,B)}{1+d_n(A,B)}$ define a metric $d$? and got an answer. I will explain the situation below:
Suppose we have a set $X$ of objects $A,B,C\cdots$, which we wish to compare pairwise. Furthermore we are given a set $M$ and to each object $A \in X$ a map $\phi_A:M \rightarrow \mathbb{N}$. We call the objects $A,B$ equivalent if $\phi_A(x) = \phi_B(x) $ for all $x \in M$. Define a metric $d(A,B) = \sup_{x \in M} |\phi_A(x) - \phi_B(x)|$ on the space of equivalence classes of $X$.
Now I am asking myself if this metric space is complete? I have thought about it, and have come to the conclusion, that it must be complete since every Cauchy sequence must get stationary, that is for every Cauchy sequence $A_1,\cdots,A_n\cdots$ there exists an $N$ such that $A_n \equiv A_N$ for all $n > N$. But I am struggling a bit with the details of the proof. Does somebody know if the metric space defined above is complete or not? If so, how does one prove it. Thanks for your help!
This is what I got so far: Let $A_1,A_2,\cdots$ be a Cauchy - sequence. Then for all $\epsilon>0$ there exists $N$ such that for all $m,n>N$ we have $\sup_{x \in M} |\phi_{A_n}(x) - \phi_{A_m}(x)|< \epsilon$. Choose $\epsilon = 1/2$. Since $\sup_{x \in M} |\phi_{A_n}(x) - \phi_{A_m}(x)|$ is a natural number $<1/2$ it must be equal to $0$. Hence $\sup_{x \in M} |\phi_{A_n}(x) - \phi_{A_m}(x)| = 0$ and it follows that $\forall x \in M: \phi_{A_n}(x) = \phi_{A_m}(x)$ for $m,n>N$. Hence we have for $A_n \equiv A_m$ for $m,n>N_{1/2}$. How does one proceed from here?
