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Prove $s\leq r-d(Q,P)\implies B(Q,s)\subseteq B(P,r)$

where $B(P,r)$ is the set of points inside a circle with radius $r$ centered at $P$ and $d(P,Q)$ is the distance between $P$ and $Q$. So

$$B(P,r)=\{x\mid d(P,x)\leq r\}$$

Assuming $s\leq r-d(Q,P)$:

$$d(Q,P)\leq r-s$$

I started by proving

$$B(Q,s)\subseteq B(Q,r)$$ $$B(P,s)\subseteq B(P,r)$$

and

$$Q\in B(P,r-s)\subseteq B(P,r)$$ $$P\in B(Q,r-s)\subseteq B(Q,r)$$

but I don't know if that's the right path.

1 Answers 1

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You can do it in a simpler way, which is the standard way to prove inclusion. I'll show you here:

Suppose $s\le r-d(Q,P)$. Then let $y\in B(Q,s)$.
We are gonna show that $y\in B(P,r)$. To do this, let's see if $d(y,P)\le r$.

$$y\in B(Q,s)\implies d(y,Q)\le s\le r-d(Q,P)\implies d(y,Q)+d(Q,P)\le r$$ But from the triangular inequality property of $d$ we know that $d(y,P)\le d(y,Q)+d(Q,P)$, and using the inequality showed before we get $d(y,P)\le d(y,Q)+d(Q,P)\le r\implies y\in B(P,r)$

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    I don't understand how you knew $d(y,P)\leq d(y,Q)+d(Q,P)$. Edit: $d(a,c)\leq d(a,b)+d(b,c)$ – 2017-02-25
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    That's the triangular inequality of every distance function $d$ – 2017-02-25
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    I finally noticed, thanks. – 2017-02-25