# How to prove the metric which defined by supremum of all semi-metric?

Define the function $f:X\times X \to R$ by $d(x,y)=\sup\{d_i(x,y):i\in I\}$, when each $d_i$ is a pseudometric; $d_i(x,y)=0$ need not imply $x=y$; for every $i$ in a directed set $(I,\leq)$ and $X$ is any nonempty set. I can't show that $d(x,y)=0 \Longrightarrow x=y$, otherwise i can do by myself. Thank you!

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You mean pseudometric, not semimetric. They’re not the same thing: a pseudometric need not separate points, and a semimetric need not satisfy the triangle inequality. The result that you want isn’t true without some further hypothesis on $X$. What information are you given about $X$? – Brian M. Scott Jan 6 '13 at 15:04
Take $X = \mathbb{R},$ then the set of all pseudometrics include $k|\cdot|$ for any $k \in \mathbb{R}_+.$ It follows $d(x,y),$ as you have defined it, is equal to $0$ if $x=y$ and is infinite otherwise. Certainly this is not a metric. – jspecter Jan 6 '13 at 15:41
I'm sorry! My main purpose is pseudometric. However some book such as functional analysis by Kreyszig P.46 said that it is a same thing. The set $X$ is any nonempty set. I'm sorry for missing. – Nimana Jan 6 '13 at 16:51