3

I have a specific and a general question. My specific question is this: how would I determine the shape and location of the set of points satisfying $d(x,a) \leq 1$ in the metric space $(\mathbb{R}^2, d)$, where $d(x,y) = |x_{1} - a_{1}| + |x_{2} - a_{2}|$? Since the coordinates of $a$ are constant, the values of $x_{1}$ and $x_{2}$ must range from $a_{i} - 1$ and $a_{i} + 1$, for $i = 1,2$. However, I can't think of a way to determine possible values of the coordinates at these values.

In general, what are some methods that can be used to visualize these sorts of shapes in different metric spaces? I realize that in some exotic spaces things probably can't be visualized, but I'm sure there are tricks for simple ones like this, right?

Wyatt Gregory
  • 423

1 Answers1

1

Think about it this way:

If you want $d(x,a) \leq 1$, i.e., $|x_{1} - a_{1}| + |x_{2} - a_{2}| \leq 1$, then draw out your $XY$ plane, and label some point $a$.

Now put another point somewhere on the $XY$ plane, and connect the two points by a straight line. Now, put your pencil on $a$, and get to $x$ by first going only horizontally (parallel to $x$-axis until you reach either above or below $x$ depending on your drawing, then draw parallel to the $y$-axis (i.e., vertically) up or down until you hit $x$.

The "length" of the horizontal line is represented by the number $|x_{1} - a_{1}|$. The "length" of the vertical line is represented by the number $|x_{2} - a_{2}|$. You want all points such that the sum of these lengths is still $\leq 1$. You can see right away that the further $x_{1}$ is from $a_{1}$, the closer $x_{2}$ must be to $a_{2}$ in order to have the sum of the distances still be $\leq 1$. With all of this in mind, try to figure out what shape you will get.

layman
  • 20,191