Singleton as a metric space

Question

I am trying to show that I can come up with a metric such that the set $X =\{0\} $ will be a metric space. I was considering the metric $d(x,y)=0$ for all $x,y \in X$ and I showed that this function satisfies the definition of metric. So this metric would turn $(X,d)$ into a metric. Would this argument be okay?

Answer 1

If $X = \{0\}$ and you define $d(x,y) = 0$ for each $x,y \in X$, we have

1. $d(x,x) = 0$ for each $x \in X$ (this is the only case).

2. $d(x,y) > 0$ for each $x,y\in X, x \neq y$ (this is vacuously true).

3. $d(x,y) = d(y,x)$ for each $x,y\in X$ (this is also vacuously true).

4. $d(x,z) \leq d(x,y) + d(y,z)$ for each $x,y,z \in X$ (true because $0 \leq 0 + 0$).

So yes, $d$ is a valid metric on $X$.

Answer 2

Yes, this is the same thing as taking the discrete metric of a space and considering only singleton subsets.

Answer 3

As long as this satisfies the three requisite conditions to be a metric, $(X,d)$ is a metric space.

1) Triangle Inequality: $d(x,y)\le d(x,z)+d(z,y)\:\forall x,y,z\in X$. True because $0$ is the only element of the metric space.

2) Positive Definiteness: $d(x,y)\ge0\:\forall x,y\in X, d(x,y)=0$ if and only if $x=y$. This also holds, as you can verify easily.

3) Symmetry of the Metric: $d(x,y)=d(y,x)$. This is obviously true, because $0=0$.

So, because these three axioms are upheld, your distance metric is a well-defined metric on the set $X=\{0\}.$

Answer 4

For any set $X$, the map $(x,y)\mapsto 0$ is a pseudometric, as both sides of the triangle inequality would always be zero, but is only a metric if $X$ consists of a single point, as otherwise there would be $x\ne y$ with $d(x,y)=0$.

More interesting, though, is the discrete metric: $$ d(x,y) = \begin{cases} 1,& x\ne y\\ 0,& x=y.\end{cases} $$ A good exercise here is to show that $(X,d)$ is compact iff $X$ is finite.