# Is this proof that all metric spaces are Hausdorff spaces correct?

Let $x$ and $y$ be distinct points of a metric space $M$. Prove that there exist in $M$ disjoint open sets $U$ and $V$ with $x \in U$ and $y \in V$.

Let $U$ and $V$ be open balls centered at $a$ and $b$, respectively. Also, $a, b \in M, x \in U, y \in V$. Since $M$ is a metric space, $M$ has a real defined distance function $D(x,y) < \epsilon$. Since $x\ne y$, $y$ is a limit point of $U$ for $D(x,y)<\epsilon.$ That is, $B_{r}(x)\subset U.$ Thus, an open set is contained in $U$, since open balls are open sets.

Similarly, $x$ is a limit point of $V$. Applying the same definitions as we did to $U$, we find an open set in $V$. These open sets are disjoint. QED.

• You say "$y$ is a limit point of $U$". What is this $U$? Where does it come from? Aug 24, 2013 at 22:55
• You don't define $U$, or $r$, or $\epsilon$. I also don't see how the $V$ (which also isn't defined) will be disjoint from $U$.
– user61527
Aug 24, 2013 at 22:56
• @Jossie You're trying to prove the existence of $U$ and $V$; you can't assume it exists, since that's circular.
– user61527
Aug 24, 2013 at 23:15
• @Jossie, try to draw a picture illustrating the case in Euclidean plane. It is easy to find the radius for the open balls. Aug 24, 2013 at 23:22
• @Jossie $U$ and $V$ can't be arbitrary; at a minimum, $U$ must contain $x$ for this to have any hope of working. The statement "y is a limit point of $U$ for $D(x, y) < \epsilon$" also doesn't make sense.
– user61527
Aug 24, 2013 at 23:33

I think you've made it too complicated. There's no reason to mention limit points.

You want to show that for points $x\ne y$, there are disjoint open sets containing them.

Just use the open neighborhoods of radius $r=D(x,y)/2$ centered at $x$ and $y$.

Since $D(x,y)>0$, then points $x$ and $y$ are members of those respective open neighborhoods.

To show that they are disjoint, suppose $z$ is a point in their intersection. Then $D(x,z)<r$ and $D(y,z)<r$. So by the triangle inequality, $D(x,y)\le D(x,z)+D(z,y)<r+r = D(x,y)$, and we have a contradiction.

• Why $r = {D(x, y) \over 2}$? Aug 24, 2013 at 23:42
• He defined $r$ in that way and that is the trick to show the needed result Aug 24, 2013 at 23:54
• If you draw the picture, you'll see why: Draw two dots on a page, then draw non-overlapping circles around them, of equal sizes. If the radius is more than half the distance between the two points, then the circles overlap, and that's exactly what we're trying to avoid. But also, if $r$ were bigger than half the distance, the we wouldn't be able to say $r+r=D(x,y)$. I could have said: choose $r$ so that $0<r\le D(x,y)/2$, and then at the end we'd say $r+r\le D(x,y)$. Our contradiction is $D(x,y)<D(x,y)$. Those inequalities need to be there in order to get that result. Aug 24, 2013 at 23:54
• I understand now why open sets need be disjoint. Also, then, you have basically shown that closed sets that intersect cannot exist. Correct @MichaelHardy? Aug 25, 2013 at 0:09
• It is not generally true that open sets are disjoint. But these open sets are disjoint. But I have no idea why you would say I showed that closed sets that intersect don't exist. Aug 25, 2013 at 2:18

As T. Bongers points out, you can't assume that $U$ and $V$ exist in your proof. You can ask yourself what properties $U$ and $V$ must have if they are to satisfy the requirements of the problem statement, and then try to work backwards. But after you figure out a proof strategy, you should start over from the beginning and prove for yourself that $U$ and $V$ exist.

The easiest way to show that $U$ and $V$ exist is to explicitly define them! To produce a definition, it is not enough just to say that they are sets. You have to define them unambiguously, in such a way that anyone can read the definition and understand which points are contained in the sets.

Hint: You can make $U$ an open ball centered and $x$ and $V$ and open ball centered at $y$. To complete the definition, you'll have to define the radius of these balls. What radius should you choose? It will depend somehow on $x$ and $y$. As Sigur suggests, try drawing a picture...

Once you've chosen a definition for $U$ and $V$, it remains only to prove that they are disjoint. This will follow from certain basic facts about metric spaces...