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It is stated that $A_n={(\frac{-1}{n},\frac{1}{n})}$, then arbitrary intersection of open sets need not be open is true as in this case $\bigcap_{i=1}^{\infty}=\left \{0 \right \}$ is not open.

Now based on an example from a book,

let $l=\left \{(0,1),(0,\frac{1}{2}),(0,\frac{1}{3}),... \right \}$

$l$ has finite intersection property for the following reason:

Indeed $(0,a_1)\bigcap(0,a_2)\bigcap(0,a_3)...\bigcap(0,a_m)=(0,b)$

where $b=\min\left \{a_1,a_2,a_3...a_m\right \}>0$ and it continues as "So, any subclass $l$ has nonempty intersection even then $l$ itself has an empty intersection.

Why $l$ has empty intersection? My confusion is that if based on example on top, $(0,\frac{1}{n})$ should be reduced to $(0,0)$ which is not a metric spaces because $d(x,y)>0$ is axiom to be satisfied.

I would be thankful if one could explain this finite intersection example in more elaborated manner. Image from book

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  • $\begingroup$ The claim that "any subclass of l has nonempty intersection" is simply not true. Are you sure you copied (or translated) this from your book correctly? $\endgroup$ Jan 15 '15 at 11:37
  • $\begingroup$ @MartinSleziak: yes it is an example 4.3 on page 71 of "Metric spaces including fixed point theory and set valued maps" by Qumrul hasan ansari. $\endgroup$
    – Sejwal
    Jan 18 '15 at 6:35
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    $\begingroup$ Thanks for the response! You are right. It seems that it is typo in the book. They probably wanted to write finite subclass. Here is an excerpt from the book with the example. (Feel free to include it in your post if you wish.) My personal advice is that if you quote something from some source (book, lecture notes, paper), it is good also to mention the source in your question. $\endgroup$ Jan 18 '15 at 7:18
  • $\begingroup$ @MartinSleziak: thanks for your suggestion, I have added the link to my post. $\endgroup$
    – Sejwal
    Jan 20 '15 at 5:28
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Notice that $0\in A_n$ for every $n$ so $0\in\cap_n A_n$ while if $a\in \cap_n (0,\frac1n)$ then $0<a\le \lim_{n\to\infty}\frac1n=0$ which is a contradiction so $\cap_n (0,\frac1n)=\varnothing$.

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The intersection of all intervals $\in\ell$ is empty. It does not contain negative numbers, obviously. It does not contain $0$ as already $0\notin (0,1)$. And it does not contain any positive $a$ for there always is $n$ with $\frac1n<a$ and then $a\notin(0,\frac1n)$.

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