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Prove that

$$\sup_{i \in I} \operatorname{dist}(x,S_i) \leq \operatorname{dist}(x, \bigcap_{i \in I}S_i)$$

where $(X,d)$ is a metric space, $S_i \subseteq X : i \in I$, and $x \in X$.

The question had two parts: one part to prove the above inequality and another to provide an example where the inequality was strict. I was able to make the inequality strict by letting the intersection be empty, so that the distance from $x$ to $\emptyset$ would be $\infty$ and I could say that any real number is less than infinity.

I figured that I could use this technique to prove the other end (where the inequality was in fact equality), but I'm finding it fruitless to apply the same type of logic (letting the intersection be non-empty, for example). Is there another technique or nugget of gnosis that I'm missing?

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Fix $i$.

Then $\bigcap_{j}S_j\subseteq S_i$ so $$ d(x,S_i)\leq d(x,\bigcap_{j}S_j). $$

Now take the sup over $i$ to get the inequality.

For an example where the inequality is strict, but the intersection non-empty, consider $$ S_1=\{(x,0)\in\mathbb{R}^2\;;\; x\geq 0\}\quad\mbox{and}\quad S_2=\{(0,y)\in\mathbb{R}^2\;;\; y\geq 0\}. $$ Then take the point $X=(1,1)$. You'll find $$ 1<\sqrt{2}. $$

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alright, but I don't follow how the form $\bigcap_jS_j$ goes to $\bigcap_iS_i$ (even with the definition of $S_j$)... – Sean Allred Feb 17 '13 at 16:31
@vermiculus I am not sure I understand your question. Obviously, $\bigcap_iS_i=\bigcap_jS_j$. Can you see it? And it is also equal to $\bigcap_kS_k$, etc... – 1015 Feb 17 '13 at 16:33
sorry -- I understood your definition to be $(\bigcap_jS_j) \subseteq S_i$ :P – Sean Allred Feb 17 '13 at 16:49

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