metrics: Let d(x,y) = 0 for x=y, 1 for x not eq. to y. Prove

woolley

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Sep 9, 2007
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For x, y ∈ R^n, consider the distance function given by

d(x,y) = 1 if x does not equal y

0 if x = y

Prove that d(·) is a metric on R^n.

sorry for all the posts. please help! I am mainly having problems with exactly how to write proofs. Thank you
 
All you need to do is to show:
\(\displaystyle \begin{array}{l}
1)\;d(x,y) \ge 0 \\
2)\;d(x,y) = 0\;\mbox{iff}\;x = y \\
3)\;d(x,y) \le d(x,z) + d(z,y) \\
\end{array}\).

Using the definition it is easy to do.
 
what does the dot mean?

I get the metric definition, I am unclear about the d-dot notation. my teacher briefly explained in class, but I did not really get it. does d( ·) simply mean d(x,y)? i.e. is d( ·) = d(x,y)?
 
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