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Prove $B(P,r)\subseteq B(P,r')\implies r\leq r'$

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

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

Proof: Assume $B(P,r)=B\subseteq B'=B(P,r')$. Then the area of $B$ is less or equal than the area of $B'$, so

\begin{align} \pi r^2&\leq \pi r'^2\\ r^2&\leq r'^2\\ r&\leq r'\qquad\blacksquare \end{align}

But I feel like I'm cheating. Is this a valid way or is there another better way to show this?

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    This is valid . – 2017-02-25
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    It seems that you are assuming that the balls live in $\mathbb{R}^2$, in which case you know their volume. Is this assumption valid? A more general argument might be to prove the contrapositive, i.e., if $r > r'$ then there exists a point in $B(P,r)$ that does not belong to $B(P,r')$. – 2017-02-25
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    @K.Miller Could you elaborate that as an answer? – 2017-02-25
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    Where do the closed balls $B(P,\cdot)$ live? Are they in a general metric space or something more specific like $\mathbb{R}^n$. – 2017-02-25

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In certain cases this assertion is true. For example, suppose the points lie in $\mathbb{R}^n$ and $d$ is the usual Euclidean metric. In this case the truth of the claim is easy to prove. For example, suppose on the contrary that $r > r'$ and let $x\in B(P,r)$ such that $d(P,x) = r$ (clearly such a point exists). Then $d(P,x) = r > r'$, which implies that $x\not\in B(P,r')$, a contradiction.

However, I do not believe this claim is true in general. Consider the following counterexample. Let $X$ be a set and consider the metric space $(X,d)$, where $$ d(x,y) = \left\{ \begin{array}{ll} 1, & \text{if}~x \neq y\\ 0, & \text{if}~x = y \end{array}\right. $$ Let $P\in X$ and define the closed ball $B(P,r) = \{x\in X\,|\, d(P,x) \leq r\}$. It is clear that if $r \geq 1$, then $B(P,r) = X$ and if $0 < r < 1$, then $B(P,r) = \{P\}$. Thus it is possible that $B(P,r) = B(P,r') \implies B(P,r) \subset B(P,r')$ with $r > r'$. The only possible issue with this counterexample is the definition of the symbol $\subseteq$. If its intended purpose is to indicate that $B(P,r)$ is a proper subset of $B(P,r')$, then this couterexample fails.

If $\subseteq$ implies proper subset, then the claim is true. Suppose that $B(P,r)$ is a proper subset of $B(P,r')$. Then there exists an $x\in B(P,r')$ that does not belong to $B(P,r)$. Therefore, $r < d(P,x) \leq r'$.

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    How do you know that if $r\geq1$, then $B(P,r)=X$? – 2017-02-26
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    If $r \geq 1$, then by the definition of $d$, $d(P,x) \leq r$ for all $x \in X$. Hence the ball contains all $x\in X$. – 2017-02-26