Let $(X,d)$ be a metric space. If $M_i\subset X$ with $M_i \cap M_{i+1}\neq \emptyset$ for $i\in \mathbb{N}$, then:$$\text{diam}\left(\bigcup\limits_{i=1}^{\infty}M_i\right)\leq \sum \limits_{i=1}^{\infty}\text{diam}(M_i)$$ I feel like this proof needs the triangle inequality, but I can't get ahead of this problem - any suggestions on how to prove it?


1 Answer 1


Take $x \in M_i, y \in M_j$ with $i \le j$. Let $x_i=x$ and $x_j = y$

Choose $x_k \in M_k \cap M_{k+1}$ for $k=i,...,j-1$.

Then $d(x,y) = d(x_i,x_j) \le \sum_{k=i}^{j-1} d(x_k,x_{k+1}) \le \sum_{k=i}^{j-1} \operatorname{diam} M_k \le \sum_{k=1}^{\infty} \operatorname{diam} M_k$.

Hence $\operatorname{diam} \cup_k M_k \le \sum_k \operatorname{diam} M_k$.


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