$\Delta = \{(x,x), x\in M\}\subset M\times M$. Show that if $z\in M\times M - \Delta$ then there is a ball with center $z$, disjoint from $\Delta$

Question

I need to show this:

I need to use the metric $d(z,z') = max\{d_i(x_i, x_i'), i\in \{1,2\}\}$ where $z = (x_1, x_2)$ and $z' = (x_1', x_2')$ for the set $M\times M$. Note that $d_i$ is the metric on $M$.

First of all: WHY would someone ask something like this? I tried to imagine $\Delta$, which could be understood, in the plane, as the set of points $(x,y)$ in which $x=y$. And I Inderstand $z\in M\times M - \Delta$ as the plane without the line $x=y$. I understand the metric of this space as being the greatest distance: the horizontal one or the vertical one.

Now, I can construct an open ball, in my mind, in this metric space. But how do I prove it, and why is this important?

In order to prove, I'd try to construct an open ball with center in $b\in M\times M - \Delta$, to begin with my argument. Then, what argument can I use to prove that no element inside my ball will touch $\Delta$? Why is the metric, as defined that way, useful?

It's related to the condition of being Hausdorff. A space is Hausdorff iff the diagonal is closed in the product. See this question — Michael Burr, Mar 20 '16 at 00:44
This metric is called the $\ell^\infty$ metric. It's a very common metric. See this wikipedia page. — Michael Burr, Mar 20 '16 at 00:45
The reason this is important is that it is an exercise designed to strengthen your mathematical muscles. In particular, this exercise, if successfully completed, will strengthen your triangle inequality muscle. — Lee Mosher, Mar 20 '16 at 16:30

Michael Burr · Accepted Answer · 2016-03-21T10:05:31.357

2

Hint: Let $z=(z_1,z_2)\in M\times M\setminus\Delta$. Let $a=d_M(z_1,z_2)$. Consider the ball of radius $\frac{a}{4}$ centered at $z$. Could this ball intersect the diagonal? Suppose, for contradiction, that it does, then $(w,w)$ is in the ball. Now, use the triangle inequality to derive a contradiction.

More details: Since $(w,w)$ is in the ball, $d_M(z_1,w)<\frac{a}{4}$ and $d_M(z_2,w)<\frac{a}{4}$. But then, $d_M(z_1,z_2)\leq d_M(z_1,w)+d_M(z_2,w)<\frac{a}{2}$, which is a contradiction.

edited Mar 21 '16 at 10:05

answered Mar 20 '16 at 00:51

Michael Burr

32,867

hmmm, what a radius $a/4$ helps in? – Guerlando OCs Mar 21 '16 at 00:37
You need any radius less than $\frac{a}{2}$ for the contradiction to work. It's a constant that "falls out" of the contradiction (it's not obvious when you start, but once you've worked through the contradiction, you'll see where it becomes necessary). – Michael Burr Mar 21 '16 at 00:41
I'm confused. I' ve found arguments not using the metric at all: http://math.stackexchange.com/questions/136922/x-is-hausdorff-if-and-only-if-the-diagonal-of-x-times-x-is-closed but I can' t do this one you're trying to tell me. – Guerlando OCs Mar 21 '16 at 04:01
Sure, you could use the argument that you linked to. First, you need to show that $M$ is Hausdorff which uses essentially the proof that I highlighted above, and then you can follow what's in that link. Or, you can just prove it all at once... – Michael Burr Mar 21 '16 at 10:04
could you give me a geometric interpretation for this? :c – Guerlando OCs Apr 15 '16 at 03:26
The more details section really covers the geometry. Think about the point $(1,2)$. If you have a point $(w,w)$ thatis close to $(1,2)$, then the change from $1$ to $w$ or the change from $2$ to $w$ is at least $1/2$ (by the triangle inequality). Moreover, if you consider the ball of radius $1/4$, then the first coordinate can vary in $(3/4,5/4)$ while the second coordinate can vary in $(7/4,9/4)$. Observe that the two intervals overlap, so there is no point of the form $(w,w)$ in this ball. – Michael Burr Apr 15 '16 at 10:23

$\Delta = \{(x,x), x\in M\}\subset M\times M$. Show that if $z\in M\times M - \Delta$ then there is a ball with center $z$, disjoint from $\Delta$

1 Answers1