Let $X$ and $Y$ be points contained in the disk of radius $r$ around the point P. Explain why $d(X, Y) \leq 2r$.

Question

Working on the book: Lang, Serge & Murrow, Gene. "Geometry - Second Edition" (p. 23)

8. Let $X$ and $Y$ be points contained in the disk of radius $r$ around the point P. Explain why $d(X, Y) \leq 2r$. Use the Triangle Inequality.

The solution given by the author is:

Since $X$ and $Y$ are contained in the disc, $d(P,X) \leq r$ and $d(P,Y) \leq r$.

$d(X,Y) \leq d(P,X) + d(P,Y)$ by the Triangle inequality.

Therefore, $d(X,Y) < r + r = 2r$.

I know this property of Real numbers:

If $a < b$ and $c$ is any real number, then $$a + c < b + c$$ $$a - c < b - c$$

My question is: what's the justification for the conclusion $d(X,Y) < r + r = 2r$ ?

How is that property instantiated in order to reach it (in case this is the appropiate property)?

Answer 1

Since $d(P,X)<r$ and $d(P,Y)<r$ and since you know that $d(X,Y)\leqslant d(P,X)+d(P,Y)$, you have\begin{align}d(X,Y)&\leqslant d(P,X)+d(P,Y)\\&<r+d(P,Y)\\&<r+r\\&=2r.\end{align}