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.

  • 1
    $\begingroup$ 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

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 $V_0=B$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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