Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
2  
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
6  
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
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." –  Manny Reyes Nov 2 '12 at 14:47

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.