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Let $ X $ be the set of $ n$-letters word, that is $ X = \{(x_{1}, x_{2}, \dots, x_{n}) \} $ where $ x_{i} $ is an alphabetical character. Define $ d(x, y) $ between two words $ x = (x_{1}, \; x_{2}, \; \dots, \; x_{n}) $ and $ y = (y_{1}, \; y_{2}, \; \dots, \; y_{n}) $ to be the number of places in which the $ 2 $ words have difference letters. Prove that $ (X, d) $ is a metric space.

I am having trouble proving the triangle inequality $ d(x, y) \le d(x, z) + d(z, y). $ Also, if one says that $ (\mathbb{C}, |.|) $ is a metric space where $ \mathbb{C} $ is the set of complex numbers, what does the symbol $ |.| $ mean? Does it mean the absolute value of the product of $ 2 $ elements in $ \mathbb{C}? $

user351703
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2 Answers2

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HINT: If $x_k\ne y_k$, then either $x_k=z_k\ne y_k$, $x_k\ne z_k=y_k$, or $x_k\ne z_k\ne y_k\ne x_k$. Each $k$ of the first type also contributes to $d(x,y)$ $d(z,y)$ but not to $d(x,z)$. Each $k$ of the second type also contributes to $d(x,z)$ but not to $d(z,y)$. And each $k$ of the third type contributes to both $d(x,z)$ and $d(z,y)$. If there are $a$ indices $k$ of the first type, $b$ of the second, and $c$ of the third, then $d(x,y)=a+b+c$. Show that $d(x,z)+d(z,y)$ must be at least this big.

One does not correctly say that $\langle\Bbb C,|\cdot|\rangle$ is a metric space: $|\cdot|$ is the absolute value function on $\Bbb C$, so that $|a+bi|=\sqrt{a^2+b^2}$, and this is not a metric. (The dot stands for the missing argument of the function.) What is meant is presumably that the metric is given by $d(x,y)=|x-y|$.

Brian M. Scott
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Hint: If $x_i\neq y_i$ then either $x_i\neq z_i$ or $y_i\neq z_i$. Thus $i$ is a place counted to calculate $d(x,z)$ or $d(z,y)$.

$| x|=\sqrt{a^2+b^2}$ $x=a+bi$.

Tsemo Aristide
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