All the spaces which are Hausdorff and semi metric spaces, are necesarily metric spaces? My hypothesis: $(X,d)$ is a semi metric space and a Hausdorff space
My thesis: $d$ is metric
My try:
Well $d$ is semi metric so, to prove that $d$ is metric, I only have to prove that: $d(x,y)=0 \Rightarrow x=y \quad \forall x,y \in X$
$X$ is Hausdorff so I know that:
For every distinct $x_1,x_2\in X$, there are two open sets $V$ and $U$ so that $x_1\in V$, $x2\in U$ and $V\cap U =\emptyset$.
Well I know that, $V$ is open iff $\forall v \in V$, there is a $r>0$ so that the $B_r(v)\subset V$.
i.e., $\forall v \in V$, there is a $r>0$, so that $\forall x in X$ for which: $d(v,x) < r \Rightarrow x\in V$.
The same happens for $U$ because it is also open. 
But I think it is impossible to conclude the thesis...I think I have not enough information
Can someone give a hint?
 A: What you’re calling a semimetric is usually called a pseudometric; a semimetric satisfies all of the requirements of a metric except the triangle inequality.
Your conjecture is false. Let $X=\{0,1\}$ with the discrete topology, and define $d(x,y)=0$ for all $x,y\in X$; then $d$ is a pseudometric that is not a metric, and $X$ is Hausdorff.
If you mean that the topology on $X$ is generated by the pseudometric $d$, then yes, $d$ must be a metric, and you’ve started a correct proof. Given distinct $x,y\in X$, there are disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$. Since $d$ generates the topology, there are $r_x,r_y>0$ such that $B(x,r_x)\subseteq U$ and $B(y,r_y)\subseteq V$. Clearly $B(x,r_x)\cap B(y,r_y)\subseteq U\cap V=\varnothing$. If $d(x,y)=0$, then $y\in B(x,r_x)\cap B(y,r_y)$, which is impossible, so $d(x,y)>0$.
A: I assume that what you call a semimetric is what's more commonly known as a pseudometric: so $d(x,x)=0$, $d(x,y)=d(y,x)$, $d(x,z)\le d(x,y)+d(y,z)$ (for all $x,y,z\in X$), lacking the property that $d(x,y)=0$ implies $x=y$. I assume also that the topology on $X$ is the one induced by $d$.
You're almost there. Suppose $d(x,y)=0$, but $x\ne y$. Then there are open sets $U$ and $V$ containing $x$ and $y$, respectively, so that $U\cap V=\emptyset$.
An open set is such that it contains an open ball around each of its points; so there is $r>0$ such that $B(x,r)\subseteq U$ and there is $s>0$ such that $B(y,s)\subseteq V$. The hypothesis implies that $B(x,r)\cap B(y,s)=\emptyset$.
In particular $y\notin B(x,r)$, so $d(x,y)\ge r>0$. A contradiction.
