Why does symmetry allow the assumption that $D(a,c) \ge D(b,c)$?

Question

In Kaplansky, Set Theory and Metric Spaces (pg. 69) there is the following theorem:

Theorem: For any points $a,b,c$ in a metric space, we have:

$$ |D(a,c) -D(b,c)| \le D(a,b)$$

The following proof is provided:

Proof: Because of symmetry in [the above] between $a$ and $b$, we can assume $D(a,c) \ge D(b,c)$. Then $D(a,c) - D(b,c) \ge 0$, so that we can remove the absolute values in [the above]. We then find that [this equation] corresponds with the triangle inequality [in the definition of a metric space].

What I do not understand is how symmetry allows the assumption that $D(a,c) \ge D(b,c)$. An explanatin would be appreciated.

Answer 1

You can understand $|D(a,c) -D(b,c)| \le D(a,b)$ in this manner,

$D(a,c)\le D(b,c)+D(a,b)\Rightarrow D(a,c) -D(b,c)\le D(a,b)$

Similarly $D(b,c)\le D(a,c)+D(a,b)\Rightarrow D(b,c)- D(a,c)\le D(a,b)$

The above two inequalities imply $|D(a,c) -D(b,c)| \le D(a,b)$

Here symmetry is used to emphasize on the fact that we can get both the inequalities $D(a,c) -D(b,c)\le D(a,b)$ and $D(b,c)- D(a,c)\le D(a,b)$ . So if $D(a,c) -D(b,c)\ge0$ then we will use the first inequality else we will use the 2nd inequality so we can easily assume one to be $\ge $ the other and apply any one of the inequality to get the desired result..

Answer 2

Since $d(a,c) \le d(a,b)+d(b,c)$, we have $d(a,c)-d(b,c) \le d(a,b)$. If you switch $a,b$ you get $d(b,c)-d(a,c) \le d(b,a) = d(a,b)$. It follows that $|d(a,c)-d(b,c)| \le | d(a,b) |$.