# Show R^2 is locally compact with non-standard metric: I need help

1. Nov 30, 2005

### benorin

EDIT: I posted this in the wrong forum, will repost in textbook questions. Please delte this (or move it).

The Q: Define the distance between points $\left( x_1 , y_1\right)$ and $\left( x_2 , y_2\right)$ in the plane to be

$$\left| y_1 -y_2\right| \mbox{ if }x_1 = x_2 \mbox{ and } 1+ \left| y_1 -y_2\right| \mbox{ if }x_1 \neq x_2 .$$

Show that this is indeed a metric, and that the resulting metric space is locally compact. I need help with the second part.

My A: Write

$$d\left( \left( x_1 , y_1\right) , \left( x_2 , y_2\right) \right) = \delta_{x_1}^{x_2} + \left| y_1 -y_2\right| ,$$

where

$$\delta_{x_1}^{x_2}=\left\{\begin{array}{cc}0,&\mbox{ if } x_{1} = x_{2}\\1, & \mbox{ if } x_{1} \neq x_{2}\end{array}\right.$$

is the Kronecker delta function. Then $d:\mathbb{R} ^2 \times \mathbb{R} ^2 \rightarrow \mathbb{R}$ is a metiric on $\mathbb{R} ^2$ since the following hold:

i. d is positive definite since d is obviously positive and

$$\delta_{x_1}^{x_2}=0 \Leftrightarrow x_{1} = x_{2} \mbox{ and } \left| y_1 -y_2\right| = 0 \Leftrightarrow y_{1} = y_{2}$$

ii. d is symmetric in its variables, that is

$$d\left( \left( x_1 , y_1\right) , \left( x_2 , y_2\right) \right) = \delta_{x_1}^{x_2} + \left| y_1 -y_2\right| = \delta_{x_2}^{x_1} + \left| y_2 -y_1\right| = d\left( \left( x_2 , y_2\right) , \left( x_1 , y_1\right) \right)$$

iii. d the triangle inequality, that is: if $\left( x_j , y_j\right) \in \mathbb{R} ^2, \mbox{ for } j=1,2,3,$ then

$$d\left( \left( x_1 , y_1\right) , \left( x_2 , y_2\right) \right) \leq d\left( \left( x_1 , y_1\right) , \left( x_3 , y_3\right) \right) + d\left( \left( x_3 , y_3\right) , \left( x_2 , y_2\right) \right) ,$$

which can be reasoned thus: the triangle inequality in R^2 with the Euclidian metric gives

$$\left| y_1 -y_2\right| \leq \left| y_1 -y_3\right| + \left| y_3 -y_2\right| , \forall y_{1},y_{2},y_{3}\in\mathbb{R}$$

and

$$\delta_{x_1}^{x_2} \leq \delta_{x_1}^{x_3} + \delta_{x_3}^{x_2} \mbox{ holds } \forall x_{1},x_{2},x_{3}\in\mathbb{R}$$

for suppose not: then

$$\exists x_{1},x_{2},x_{3}\in\mathbb{R} \mbox{ such that }\delta_{x_1}^{x_2} > \delta_{x_1}^{x_3} + \delta_{x_3}^{x_2} \Leftrightarrow x_1 \neq x_2 \mbox{ and } x_1 = x_3 = x_2 ,$$

which is a contradiction; adding these inequalities yields the required result, viz. the triangle inequality.

By i,ii, and iii, d is a metric on $\mathbb{R} ^2$.

The locally compact part I don't get: a metric space is locally compact iff every point of has a neighborhood with compact closure.

An open neighborhood of a point, say $\left( x_0 , y_0\right)$, is given by: for some k>0, put

$$\left\{ \left( x , y\right) : d\left( \left( x , y\right) , \left( x_0 , y_0\right) \right) < k \right\}$$

but what does that look like? How do I grasp what compact means in this metric?

The delta function above is the discrete metric on R^1 and the absolute value is the Euclidian metric on R^1, and their sum is indeed a metric on the product space R^2. Do I get to keep Heine-Borel? Does Heine-Borel even hold for R^1 with the discrete metric? I don't get the idea of compact sets with H-B, I can tell you "A set is compact if every open cover has a finite subcover," but that topology stuff is so abstract. What does it mean for a set to be compact in terms of a given metric? Is that given by sequential compactness?

Thanks,
-Ben

PS: Please don't answer all the questions in the last paragraph, just the ones that help.

Last edited: Nov 30, 2005
2. Nov 30, 2005

### AKG

Draw yourself a picture of the open ball about (x0, y0) of radius k. I know you asked what it looks like, but you should be able to easily draw this for yourself. Note that there's a drastic change in what these sets look like when k < ? vs. when k > ?. What is the "?"? Well you should be able to figure that out easily as well. Once you do it, there should be an obvious choice (or range of choices) for k. Then I think you might use something like Heine-Borel.

3. Nov 30, 2005

### mathwonk

in any metric space, compact is equivalent to complete and totally bounded. so just show every point ahs a complete totally bounded nbhd.

i am a little schmooked right now but i think totally bounded means every bounded set has a finite open cover by sets of diameter less than e, for any e>0.