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Consider the set of all sequences of real numbers.For $x={(x_n)_n}$ and $y={(y_n)_n}$ we define $N(x,y)=\inf \{n\in\Bbb N :x_n\ne y_n,\text{if $x\ne y$} \}$. Now, $$d(x,y)= \begin{cases} 0, & \text{if $x=y$} \\ 1/N(x,y), & \text{otherwise} \end{cases}$$

Prove that $d(x,y)$ defines a metric.

I have difficulties with the triangular inequality i.e, if $z={z_n}$ then $d(x,y)\le d(x,z)+d(z,y)$. Help.

Martin Sleziak
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Rupsa
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1 Answers1

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If the sequence of z differs from the sequence of x before it it differs from the sequence of y (i.e $N(x,z) \leq N (x,y)$) one has $d(x,y) \leq d(x,z)$ so the triangle inequality will be fullfiled trivially.

Therefore it remains to look at the case $N(x,y) < N(x,z)$ this means, that the digits $x_n,y_n,z_n$ are the same for $n < N(x,y)$. But by definition $x_N=z_N$ for $N=N(x,y)$, therefore one has $z_N=x_N \neq y_N$ which implies N(z,y) = N(x,y).

This gives d(z,y) = d(x,y) and again the triangle inequality is fullfilled trivially.

asterisk
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