Exercise 1.3

The line $$\mathbb{R}$$ with $$d(x_1,x_2) = |x_1 - x_2|$$ is also a metric space. Check that this defines a metric. Note that if we consider $$\mathbb{R}$$ as $$\{(x,0) : x \in \mathbb{R}\} \subset \mathbb{R}^2$$, then this is the same metric from Example 1.2.

baruch_shahi
That the absolute value satisfies the first two properties is easy to check, and so we do not write it up.

Let $$u$$ and $$v$$ be real numbers. Then by properties of the absolute value, we have

$$-|u|\le u\le |u|$$ and $$-|v|\le v\le |v|$$. Adding these two expressions gives

$$-(|u|+|v|)\le u+v\le |u|+|v|$$. But by definition of absolute value, this means

$$|u+v|\le |u|+|v|$$.

To see how this gives us the triangle inequality for the proposed metric $$d(x,y)=|x-y|$$, set $$u=x-y$$ and $$v=y-z$$ for real numbers $$x,y,z$$. Then the above inequality gives us:

$$d(x,z)=|x-z|=|x-y+y-z|=|u+v|\le |u|+|v|\le |x-y|+|y-z|=d(x,y)+d(y,z)$$.

Thus we have that $$d$$ is a metric on $$\mathbb{R}$$.

baruch_shahi 21:08, February 27, 2011 (UTC -5)

''' Very good! --Steven.clontz 15:55, March 1, 2011 (UTC)'''