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Let $X$ be a non empty set. Let $M$ the set of all sequences $(x_{n})$ of elements of $X$. For $x=(x_{n})$ and $y=(y_{n})$ in $M$, let $k(x,y)$ the smallest integer $n$ such that $x_{n}\neq y_{n}$. Let $d:M\times M\to \mathbb{R}$ $d(x,y)=\dfrac{1}{k(x,y)}$ if $x\neq y$ and $d(x,x)=0$. Show that $d$ is metric.

My approach: We have $d(x,y)=0\iff y=x$, by hypothesis, and $d(x,y)=\dfrac{1}{k(x,y)}$, where $k(x,y)$ is the smallest integer n, then $d(x,y)>0$. And $d(x,y)=\dfrac{1}{k(x,y)}=\dfrac{1}{k(y,x)}=d(y,x)$, because $k(x,y)$ is a integer. But How I prove triangle inequality...

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First, you cannot say "by hypothesis of metric d", as you need to prove that $d$ is a metric. So you need to prove that $d(x,y) = 0 \Leftrightarrow x=y$

About the triangle inequality :

Let $x,y,z$, and suppose that $k(x,y) = a$ and $k(y,z)=b$. Then $\forall n \leq \min (a,b), x_n=y_n=z_n$, this imply that $k(x,z)\geq \min (k(x,y),k(y,z))$ and it follow immediatly that $d(x,z)\leq d(x,y)+d(y,z)$

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