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I want to prove that the Hausdorff property holds for all $\kappa$-metric spaces.

For $\kappa \neq 1$, $(X,d)$ is a $\kappa$-metric space if $X$ is a set and $d$ is a function $X\times X \rightarrow \mathbb R$ such that for every $x,y,z \in X$

$1$. $d(x,y)\ge 0$

$2$. $d(x,y)=d(y,x)$

$3$. $d(x,y)=0 \iff x=y$

$4$. $d(x,z)\le \kappa [d(x,y)+d(y,z)]$.

We put a topology on $X$ by saying $U\subseteq X$ is open iff for each $x\in U$, there exists an $\epsilon>0$ such that $B_d(x,\epsilon)\subseteq U$.

In my attempted proof, take two distinct points $x,y\in X$ such that $d(x,y)=\epsilon$.

Take the open balls $B_d\left(x,{\epsilon\over {3\kappa}}\right)$ and $B_d\left(y,{\epsilon\over {3\kappa}}\right)$. Then it can be shown that $$B_d\left(x,{\epsilon\over {3\kappa}}\right)\cap B_d\left(y,{\epsilon\over {3\kappa}}\right)=\emptyset.$$ For say there is $z\in B_d\left(x,{\epsilon\over {3\kappa}}\right)\cap B_d\left(y,{\epsilon\over {3\kappa}}\right)$. Then $$d(x,y)\le \kappa \left[{\epsilon\over {3\kappa}}+{\epsilon\over {3\kappa}}\right]\\={2\over 3}{\epsilon}\\\lt \epsilon$$ which gives a contradiction.

So, we have found two disjoint open balls in $X$ that do not intersect.

At this point I was thinking that Hausdorff Property has been proved for these spaces, but then I remembered that open balls in $\kappa$-metric spaces area not necessarily open sets. And for a space to be Hausdorff , we need to find, for any two distinct points, two disjoint open sets each containing one of them.

So the above proof of Hausdorff property in $\kappa$ metric space is wrong.

Please help me prove this.

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    $\begingroup$ How are you defining open sets? $\endgroup$ – User8128 Apr 9 '16 at 16:55
  • $\begingroup$ @User8128 : To be clear, a set is said to be open if it contains an open ball around each of its points. $\endgroup$ – user118494 Apr 9 '16 at 16:57
  • $\begingroup$ @EricWofsey : Yes . $\endgroup$ – user118494 Apr 9 '16 at 16:57
  • $\begingroup$ Is there some reference for $\kappa$-metric spaces? (I have noticed that you had a few question about them recently.) The only papers I was able to find use this name for a different objects, see dx.doi.org/10.1090/S0002-9939-1987-0883422-8 or dx.doi.org/10.1090/S0002-9939-1988-0964884-5 $\endgroup$ – Martin Sleziak Apr 29 '16 at 10:35
  • $\begingroup$ @MartinSleziak : math.stackexchange.com/questions/1733552/… This is the definition of kappa-metric space I have. I was only given the definitions and asked to prove the separation axioms for it. $\endgroup$ – user118494 Apr 29 '16 at 18:34
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You can soup up your idea to inductively build open sets as follows. Fix $x\neq y$; we will define two sequences of sets $U_0\subseteq U_1\subseteq U_2\subseteq\dots$ and $V_0\subseteq V_1\subseteq V_2\subseteq\dots$ by induction. These sets will have the property that for each $n$, $d(U_n,V_n)>0$ (where $d(U_n,V_n)=\inf\{d(p,q):p\in U_n, q\in V_n\}$).

We start with $U_0=\{x\}$ and $V_0=\{y\}$. Given $U_n$ and $V_n$, let $\epsilon=d(U_n,V_n)$, and define $$U_{n+1}=\bigcup_{p\in U_n} B_d(p,\epsilon/3\kappa^2)$$ and $$V_{n+1}=\bigcup_{q\in B_n} B_d(q,\epsilon/3\kappa^2).$$

We must show that $d(U_{n+1},V_{n+1})>0$; let $r\in U_{n+1}$ and $s\in V_{n+1}$. Then there are $p\in U_n$ and $q\in V_n$ such that $d(p,r)<\epsilon/3\kappa^2$ and $d(s,q)<\epsilon/3\kappa^2$. We then have $$d(p,q)\leq \kappa^2(d(p,r)+d(r,s)+d(s,q))<\frac{2\epsilon}{3}+\kappa^2d(r,s).$$

But $d(p,q)\geq d(U_n,V_n)=\epsilon$, so this gives us $d(r,s)>\epsilon/3\kappa^2$. Thus $d(U_{n+1},V_{n+1})\geq \epsilon/3\kappa^2>0$.

Now let $U=\bigcup U_n$ and $V=\bigcup V_n$. Then $U\cap V=\emptyset$, since $U_n\cap V_n=\emptyset$ and the sequences $(U_n)$ and $(V_n)$ are ascending. Also, $x\in U_0\subseteq U$ and $y\in V_0\subseteq V$. Finally, $U$ and $V$ are open, since for any $p\in U$, $p\in U_n$ for some $n$, and then $U_{n+1}$ contains a ball around $p$ (and similarly for $V$). Thus $U$ and $V$ are disjoint open sets containing $x$ and $y$, so $X$ is Hausdorff.

More generally, this argument shows that if $A,B\subseteq X$ and $d(A,B)>0$, then $A$ and $B$ can be separated by open sets (just take $U_0=A$ and $V_0=B$).

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