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Can somebody please help me to show that a pseudo metric space is a metric space if and only if it is a $T_0$-space?

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    You can deduce it from my answer to [this question](http://math.stackexchange.com/questions/234610/pseudo-metric-spaces-are-not-hausdorff/234658#234658): if a pseudo-metric space $\langle X,d\rangle$ is not metric, there are distinct points $x,y\in X$ such that every open set containing one also contains the other, so by definition it is not $T_0$.2012-11-11

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A pseudometric space is a set $X$ with a function (called a pseudometric on $X$) $d:X\times X\to\Bbb R$ such that the following hold for all $x,y,z\in X$:

(i) $d(x,x)=0$,

(ii) $d(x,y)\geq 0$,

(iii) $d(x,y)=d(y,x)$,

(iv) $d(x,z)\leq d(x,y)+d(y,z)$.

We drop the "pseudo" from the above definitions precisely when we change the first of these conditions to

(i$'$) $d(x,y)=0$ if and only if $x=y$.

In other words, a function $d:X\times X\to\Bbb R$ is a metric on $X$ if and only if it is a pseudometric on $X$ and has the property that $x\neq y$ implies $d(x,y)>0$.


Proof Outline:

In a metric space (which is necessarily a pseudometric space), any points $x,y$ with $x\neq y$ have disjoint neighborhoods, so metric spaces are certainly $T_0$.

On the other hand, if a pseudometric space is $T_0$, then for any distinct points $x,y$, (at least) one of the two has a neighborhood not containing the other. Thus, one of them is contained in an open ball not containing the other, and so they're a positive distance apart.

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    You're quite welcome.2012-11-13