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Let $X$ be a Banach space, and $B_1\supseteq B_2 \supseteq \cdots $ a sequence of closed balls with radius $r_i$ and center $x_i$. Show that $$\bigcap_{i=1}^\infty B_i\neq\emptyset$$

I proved that $r_i\leq r_j$ for $B_i\subseteq B_j$ and $r_i\rightarrow\alpha\geq0$ so, for $\varepsilon>0\ \exists N\in\mathbb{N}$ such that $|r_i-r_j|<\varepsilon$ for $i,j>N$ and just left to show that $\|x_i-x_j\|<\varepsilon$ so the centers is a chauchy sequence and it converges, so my question is how can I prove that $\|x_i-x_j\|<\varepsilon$?

José
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  • What exactly is your question here? If you're asking how to show that $|| x_i-x_j||<\epsilon$ then consider that those centres must lie in balls of radius $r_i$ and $r_j$ respectively, and you've already proven something about how far apart such points can be. – postmortes Oct 13 '15 at 16:08
  • The fact is that I have no idea how to do that – José Oct 13 '15 at 16:10
  • See here: http://math.stackexchange.com/a/22484/260918 – Joey Zou Oct 13 '15 at 16:13
  • If $T:X \to \mathbb R$ is a continuous linear functional, can one show that the image of $B_i$ under $T$ is compact and then that the intersection of all those images is non-empty? ${}\qquad{}$ – Michael Hardy Oct 13 '15 at 16:13

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