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Is $\phi$ set forms a metric space or not ?

I think, it does not form a metric space, because, we can't specify a metric on $\phi$.

But, In many text book, it is not mention that, the set on which, we define metric should be non empty.

If I may suppose, that d is a function define on $\phi$ $ \times$ $\phi$ such that

d is constant function with range set { $0$ }. Then it must be metric on {$\phi$}.

Plz help... what is the right thingh ?

  • 2
    By {$\phi$}, do you mean the empty set?2012-11-02
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    First, $\phi$ is **not** the letter for the empty set. $\varnothing,\emptyset$ are the notation for the empty set. Secondly note that $\{\varnothing\}$ is the set whose only element is the empty set. In particular $\{\varnothing\}$ is **not empty**, and therefore $\{\varnothing\}\neq\varnothing$.2012-11-02
  • 0
    You can specify whatever metric you want as it is vacuous.2012-11-02
  • 4
    Well, you can't specify *whatever* metric you want. The metric would have to be a function $d \colon \varnothing \times \varnothing \to \mathbb{R}$. But $\varnothing \times \varnothing = \varnothing$, and there is a unique function from the empty set to any other set. Thus $d \colon \varnothing \to \mathbb{R}$ must be the "empty function."2012-11-02
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    With all due respect, why do you worry about this actually?2012-11-02
  • 0
    [Why are metric spaces non-empty?](http://math.stackexchange.com/q/45145/) might be related (you still haven't specified whether you mean the empty space or the space containing the empty set as only point).2013-01-03

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