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

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 ?

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By {$\phi$}, do you mean the empty set? –  Chris Eagle Nov 2 '12 at 14:19
yes !! Sir..... –  ram Nov 2 '12 at 14:20
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 $\{\varnothing\}\neq\varnothing$. –  Asaf Karagila Nov 2 '12 at 14:23
You can specify whatever metric you want as it is vacuous. –  copper.hat Nov 2 '12 at 14:24
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." –  Manny Reyes Nov 2 '12 at 14:47

Yes. A singleton is a metric space.

To see that, note that a subspace of a metric space is a metric space, and $\{x\}$ is a subset of $\mathbb R$ whenever $x\in\mathbb R$.

Since all singletons are "essentially" the same, this means that $\{\varnothing\}$ can also be thought as a metric space.

On the other hand whether or not $\varnothing$ itself, the empty set, is a metric space is up to definition, whether or not you are allowing empty structures in your universe, or does the empty set carries no structure.

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Obviusly $(\{\emptyset\},d)$ defines a metric space, the trivial one. Or you can say vacuously. You can check that the rules of a metric space are satisfied.

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