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Suppose $(X,d)$ is a metric space and that $F$ is a nest of non empty subsets of $X$ for which $\inf\{diam ~(A)~|~A \in F\} =0.$ Show that $\bigcap F = \emptyset$ or $\bigcap F$ is a singleton set.

Attempt:

Let $A_i \in F$ where $F$ is a nest of non empty subsets of $X$ such that

$$\cdots \subseteq A_0 \subseteq A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n \subseteq \cdots .$$

Then, $~~ \cdots \le diam(A_0) \le diam(A_1) \le diam(A_2) \le\cdots \le diam(A_n) \le \cdots $

$$\inf\{diam ~(A)~|~A \in F\} =0 \implies ~\forall~\epsilon_i >0,~\exists~A_i~|~diam(A_i)<\epsilon_i$$

$\sup ~d(a,b) <\epsilon_i$ where $~a,b \in A_i.$

Since, this is valid for all $\epsilon_i >0$, this means, $d(a,b)=0~\forall~a,b \in A_i~~\implies a=b.$

Since, $F$ is a nest of non empty subset of $X\implies \bigcap F = \{a\},$ a singleton set.

However, I am not able to prove the possibility for $\bigcap F = \emptyset$.

If $\bigcap F = \emptyset$, then, will the condition $\inf\{diam ~(A)~|~A \in F\} =0$ be satisfied?

Could someone please check if my above attempt is correct . Also please tell me how to prove the possibility for $\bigcap F = \emptyset$.

Thank you very much for your help in this regard.

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The thing you want to prove is that there cannot be two distinct points in the intersection $\bigcap F$; this will imply that the intersection has zero or one points. Suppose you had two points $x,y\in\bigcap F$. Then for all $A\in F$ you have $x,y\in A$, so $d(x,y)\leq\text{diam}(A)$. But this holds for all $A\in F$ and the infimum of the diameters is zero, so you must in fact have $d(x,y)=0$, a contradiction. (Note that there is no need to assume that $F$ is a nest of sets. It can be any collection of sets so that the infimum of the diameters is zero.)

About your proof: The roles of $a$ and $b$ are not clear; where are they chosen from? It would be clearer to pick both of them from $\bigcap F$, because you want to show that any two points there have to be the same. You also have equations or identities one after another without words connecting them, and it's hard to follow what is it exactly that you want to say.

About the empty possibility: If $F$ is a nested sequence of sets in a metric space $X$, the intersection can well be empty. Consider the case $X=[0,1]$ with $F=\{(0,1/n);n\in\mathbb N\}$.

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  • $\begingroup$ It's specified in the problem that F is a nest of non empty subsets of X $\endgroup$ – MathMan Jul 15 '15 at 8:17
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    $\begingroup$ @Wanderer, sorry, you are right. But that assumption is actually not needed. The argument works in more generality and it's not that complicated. If you want to use the sequence of sets $A_i$, you can use $d(x,y)\leq\text{diam}(A_i)$ and let $i\to-\infty$ to obtain the same conclusion. I will edit my answer a bit. $\endgroup$ – Joonas Ilmavirta Jul 15 '15 at 8:23
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I would prove it by contradiction instead.

Let $A_0 = \bigcap_{A\in F} A$

Suppose that $A_0$ has two distinct elements, $a$ and $b$.

Then $\forall A \in F, a \in A$ and $b\in A$, so $\text{diam}A \geq d(a,b)$

This imply that $\forall A \in F, \text{diam}A \geq d(a,b)$

Hence $\inf_{A\in F} (\text{diam}A ) \geq d(a,b) > 0$ : contradiction

So $A_0$ can't have 2 distinct elements : $A_0 = \{a\}$ or $\emptyset$

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  • $\begingroup$ @Tryss I understandthat $A_o$ can't have 2 distinct elements. But, how do we prove that $A_o$ can't be non empty? $\endgroup$ – MathMan Jul 15 '15 at 8:20
  • $\begingroup$ @postmortes Thanks for pointing this typo ;) $\endgroup$ – Tryss Jul 15 '15 at 8:20
  • $\begingroup$ @Wanderer : But $A_0$ can be empty ! Take The nest composed of the sets $A_n = ]0, \frac{1}{n}]$, the intersection is empty, but each is non empty $\endgroup$ – Tryss Jul 15 '15 at 8:22

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