$liminf E_{n_0+n} = liminflimits E_{n}$ and $limsup E_{n_0+n} = limsup E_{n}$ where $E_n$ is a decreasing...












1














I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.



In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)




My attempt at a proof is as follows:



Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$



Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$



But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.



I however couldn't prove the reverse inclusion in my attempt below:



Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$



Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$



I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.



I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.



Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...




Original Text:



The main part of my question comes from here:



enter image description here



Theorem 1.26:



enter image description here



And finally, Lemma 1.7:



enter image description here




Hopefully someone can shed some light!






real-analysis measure-theory elementary-set-theory limsup-and-liminf







share|cite|improve this question




















  • 2




    If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
    – Mike Earnest
    Jan 4 at 1:01










  • Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
    – Darius
    Jan 4 at 1:56










  • $I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
    – Kavi Rama Murthy
    Jan 4 at 6:20












  • @MikeEarnest, I just realized I was being a dummy! Thanks for your help!
    – Darius
    2 days ago
















1














I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.



In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)




My attempt at a proof is as follows:



Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$



Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$



But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.



I however couldn't prove the reverse inclusion in my attempt below:



Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$



Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$



I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.



I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.



Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...




Original Text:



The main part of my question comes from here:



enter image description here



Theorem 1.26:



enter image description here



And finally, Lemma 1.7:



enter image description here




Hopefully someone can shed some light!






real-analysis measure-theory elementary-set-theory limsup-and-liminf







share|cite|improve this question




















  • 2




    If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
    – Mike Earnest
    Jan 4 at 1:01










  • Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
    – Darius
    Jan 4 at 1:56










  • $I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
    – Kavi Rama Murthy
    Jan 4 at 6:20












  • @MikeEarnest, I just realized I was being a dummy! Thanks for your help!
    – Darius
    2 days ago














1












1








1







I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.



In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)




My attempt at a proof is as follows:



Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$



Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$



But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.



I however couldn't prove the reverse inclusion in my attempt below:



Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$



Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$



I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.



I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.



Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...




Original Text:



The main part of my question comes from here:



enter image description here



Theorem 1.26:



enter image description here



And finally, Lemma 1.7:



enter image description here




Hopefully someone can shed some light!






real-analysis measure-theory elementary-set-theory limsup-and-liminf







share|cite|improve this question















I dropped the $nrightarrowinfty$ in the title as it was exceeding the character limit.



In the book I'm currently reading, the author claims that $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ and $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ where $mu(E_{n_{0}}) < infty$ and $E_n$ is a decreasing sequence? ($mu$ is an arbitrary measure!)




My attempt at a proof is as follows:



Let $x in liminflimits_{nrightarrowinfty} E_{n_0+n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$



Then $x in bigcap_{kgeq n}E_{n_0+n_{0}'}$ for some $n_{0}' in mathbb{N}$



But $bigcap_{kgeq n}E_{n_0+n_{0}'} subset bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n} = liminflimits_{nrightarrowinfty} E_{n}$ and so $liminflimits_{nrightarrowinfty} E_{n_0+n} subset liminflimits_{nrightarrowinfty} E_{n}$.



I however couldn't prove the reverse inclusion in my attempt below:



Let $x in liminflimits_{nrightarrowinfty} E_{n} = bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$



Then $x in bigcap_{kgeq n}E_{n_{0}''}$ for some $n_{0}'' in mathbb{N}$



I then realized it's not guaranteed that $n_{0}'' = n_{0}$ and so we may be "missing" some elements in $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n_0+n}$ from $bigcup_{ninmathbb{N}}bigcap_{kgeq n}E_{n}$.



I suspect the proof for $limsuplimits_{nrightarrowinfty} E_{n_0+n} = limsuplimits_{nrightarrowinfty} E_{n}$ is similar and would also fall apart at the same point as my proof for $liminflimits_{nrightarrowinfty} E_{n_0+n} = liminflimits_{nrightarrowinfty} E_{n}$ ($n_{0} neq n_{0}''$) and so I didn't attempt it.



Perhaps I'm missing something about how $n_0$ was chosen such that $mu(E_{n_{0}}) <infty$...




Original Text:



The main part of my question comes from here:



enter image description here



Theorem 1.26:



enter image description here



And finally, Lemma 1.7:



enter image description here




Hopefully someone can shed some light!






real-analysis measure-theory elementary-set-theory limsup-and-liminf




real-analysis measure-theory elementary-set-theory limsup-and-liminf






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 0:38

























asked Jan 4 at 0:18









Darius

878




878








  • 2




    If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
    – Mike Earnest
    Jan 4 at 1:01










  • Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
    – Darius
    Jan 4 at 1:56










  • $I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
    – Kavi Rama Murthy
    Jan 4 at 6:20












  • @MikeEarnest, I just realized I was being a dummy! Thanks for your help!
    – Darius
    2 days ago














  • 2




    If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
    – Mike Earnest
    Jan 4 at 1:01










  • Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
    – Darius
    Jan 4 at 1:56










  • $I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
    – Kavi Rama Murthy
    Jan 4 at 6:20












  • @MikeEarnest, I just realized I was being a dummy! Thanks for your help!
    – Darius
    2 days ago








2




2




If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01




If $x$is in all but finitely many $E_{n+n_0}$, then it is also in all but finitely many $E_n$: the only sets $x$ is not in are the finitely many $E_i$ for $i ge n_0$, plus at most finitely many $E_j$ for j< n_0$. The converse is easier. Can you try to apply similar logic to the limsup part?
– Mike Earnest
Jan 4 at 1:01












Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56




Hey @MikeEarnest, I agree that $xinliminflimits_{nrightarrowinfty}E_{n+n_0}$ means that $x$ is not in finitely many $E_i$ for $igeq n_0$ and not in at most finitely many $E_j$ for $j < n_0$. But my problem is with the converse as $xinliminflimits_{nrightarrowinfty}E_{n}$ means that $x$ is not in finitely many $E_i$ for $ileq n_{0}' in mathbb{N}$. But $n_{0}'$ could be less than $n_0$ which means that $liminflimits_{nrightarrowinfty}E_{n}$ may include elements in sets $E_j$ for $n_{0}' leq j leq n_{0}$ .
– Darius
Jan 4 at 1:56












$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20






$I_{lim sup A_n}=lim sup I_{A_n}$ and $I_{lim inf A_n}=lim inf I_{A_n}$ so this reduces to elementary facts about limsup and liminf of sequences of real numbers. [Def: $I_A (x)=1$ if $x in A$, $0$ otherwise].
– Kavi Rama Murthy
Jan 4 at 6:20














@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago




@MikeEarnest, I just realized I was being a dummy! Thanks for your help!
– Darius
2 days ago










1 Answer
1






active

oldest

votes


















2














Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.



Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.






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    Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.



    Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.






    share|cite|improve this answer




























      2














      Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.



      Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.






      share|cite|improve this answer


























        2












        2








        2






        Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.



        Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.






        share|cite|improve this answer














        Let $F_n = E_{n+n_0}$ for some fixed $n_0$ as in your text.



        Then $liminf_{n to infty} F_n$ is the set of all $x$ that are in all but finitely many $F_n$. But then $x$ is also in all but finitely many $E_n$ as $x$ can only miss the sets $E_0, ldots E_{n_0-1}$, i.e. finitely many. And if $x$ is in all but finitely many $E_n$ the same holds for the $F_n$ as we only throw some sets away. So the lemma indeed immediately applies that the liminfs of these shifted sequences of sets are the same. Similar reasoning applies to the limsups. Note that the $E_n$ need not be decreasing, they can be any sets.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered Jan 4 at 5:54









        Henno Brandsma

        105k347114




        105k347114






























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            An IMO inspired problem

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            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
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            Management

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            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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            Has there ever been an instance of an active nuclear power plant within or near a war zone?

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            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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