0

$d_1(x,y)=d(x,y)+|f(x)-f(y)|$

Suppose $f(x)=1$ if $x\in\mathbb{Q}$ and 0 otherwise for a metric space = $\mathbb{R}$.

With this function I am shocked that $d_1$ is a metric. I do not doubt my proof though.

It's still very strange, for example consider $d_1(0,p)\le1$, what p satisfy this? p=1 does (assuming $d$ is nice), but $p=\frac{1}{\sqrt{2}}$ does not yet is less than 1. I am most muddled.

What if rather than 0 it is some irrational number?

Alec Teal
  • 5,515
  • 3
  • 36
  • 59

2 Answers2

3

One way to visualize the space is this. Suppose you have two parallel roads, distance 1 mile apart. One road represents the rational numbers, the other represents the irrational numbers. You can cross between the two roads, but only perpendicular. (This stretches the road analogy, I'll have to think of something a bit better. Maybe grocery store aisles?) Then given this constraint, a ball is what you would expect. So for example, suppose you are at zero. (This is on Rational Rd.) The open ball of radius 2 includes all rational numbers within 2 miles of zero, since you can just walk down Rational Road to them. (Never mind that Rational Road has uncountably many potholes, i.e. missing points.) But you could also cross over to Irrational Road, walking 1 mile, and then walk at most 1 mile further. (Walking on Irrational Road is nicer since while its potholes are dense, they are at least countable.) So you can get any irrational number within 1 mile of the point (which is a pothole) on Irrational Road across from zero, namely any irrational number of absolute value at most 1.

Other radii greater than 1 are similar. For smaller radii, you can't even make it to Irrational Road. For radius exactly 1, you can just make it, but you land in a pothole.

Rob Silversmith
  • 827
  • 4
  • 11
1

The metric $d_1$ gives an isometry between $(X,d_1)$ and the graph of $f$,

$$\Gamma(f) = \left\lbrace (x,f(x)) : x \in X\right\rbrace \subset X\times\mathbb{R},$$

when the product $X\times\mathbb{R}$ is endowed with the metric

$$\delta((x,s),(y,t)) = d(x,y) + \lvert s-t\rvert,$$

which induces the product topology.

So a ball with respect to $d_1$ is the projection of the intersection of a $\delta$-ball with $\Gamma(f)$ to $X$.

For the special case $f = \chi_\mathbb{Q}$, view the space as

$$\mathbb{Q}\times \{1\} \cup (\mathbb{R}\setminus\mathbb{Q})\times \{0\}.$$

The balls are

$$B^{d_1}_r(x) = (B^d_r(x) \cap \mathbb{Q}) \cup (B^d_{r-1}(x)\cap(\mathbb{R}\setminus\mathbb{Q}))$$

for $x\in\mathbb{Q}$ (where balls of zero or negative radius are empty), and analogously with the radii swapped if $x\notin\mathbb{Q}$.

Daniel Fischer
  • 206,697