Triangle inequality for distance to set

[quasar987]

Homework Statement

If X is a metric space and A is a subset of X, show that

$$|d(x,A) - d(y,A)|\leq d(x,y)$$

for any x,y in X.

Homework Equations

Triangle inequality:
$$|d(x,z) - d(y,z)|\leq d(x,y)$$,

$$d(x,y)\leq d(x,z)+d(z,y)$$

The Attempt at a Solution

Fidling around with the triangle inequality.

Edit: This is the third problem in G. Bredon's Topology and Geometry book. The goals of the exercice is to establish the continuity of the map $x\mapsto d(x,A)$, but she gives as a hint, "Use the triangle inequality to show that $|d(x,A) - d(y,A)|\leq d(x,y)$)" from which the result clearly follows in view of the continuity of the map $x\mapsto d(x,x_0)$ for some fixed x_0...

The point of this edit is to bring attention to the exact wording of her hint, from which one can arguably conclude that the result should follow essentially from the triangle inequality.

 

[CompuChip]

As I understand it, $|d(x,A) - d(y,A)|\leq d(x,y)$ should follow from the triangle inequality for d, which is $|d(x,z) - d(y,z)|\leq d(x,y)$ for any x, y and z in X. We know the latter holds because d is a metric. What you want to prove is technically not a triangle inequality, because d(x, A) is not really a metric, in the sense that it doesn't map pairs of points to numbers.

 

[Dick]

Suppose A consists of two points. Label the points x and y so that x is farther from A than y. The only nontrivial case is that x is closer to a different point of A than y. So label the points in A={a,b} so that d(b,x)>d(a,x) and d(a,y)>d(b,y). Now assume the inequality isn't true. |d(x,A)-d(y,A)|>d(x,y) becomes d(x,a)-d(y,b)>d(x,y) -> d(x,a)>d(x,y)+d(y,b)>=d(x,b). That's a contradiction. For a general set A, you should be able to set up some epsilons and find points in A that represent the equivalent of a and b.

 

[mathstew]

Suppose A consists of two points. Label the points x and y so that x is farther from A than y. The only nontrivial case is that x is closer to a different point of A than y. So label the points in A={a,b} so that d(b,x)>d(a,x) and d(a,y)>d(b,y). Now assume the inequality isn't true. |d(x,A)-d(y,A)|>d(x,y) becomes d(x,a)-d(y,b)>d(x,y) -> d(x,a)>d(x,y)+d(y,b)>=d(x,b). That's a contradiction. For a general set A, you should be able to set up some epsilons and find points in A that represent the equivalent of a and b.

You can't assume that there exists points in A such that d(x,A) = d(x,a) and similarly for y. The easiest example is if A is open. Then the inf is most likely not in A.