$), but $X$ itself is a set which would be an element of Russell's paradoxical set $R$.  So for any set $X$ which does not contain itself as an element, the associated $X^\prime$ cannot be Russell's paradoxical set.  As luck would happen, another axiom of ZFC (the Axiom of Regularity) immediately implies that no set contains itself as an element, which then implies that we cannot use the Axiom of Comprehension with the formula $\varphi(u) \equiv u \notin u$ to prove the existence of Russell's paradoxical set.
Note that because of Regularity under ZF(C) we have $R = V$, and it can easily be shown that ZF(C) proves that $V$ is not a set: if $V$ were a set, then by definition $V \in V$, contradicting Regularity!
Of course, ZF(C) could turn out to be inconsistent, and would then prove the existence of Russell's paradoxical set (and still also disprove its existence).  Life would be easier if we could determine exactly which formulas are unproblematic in applications of Comprehension (or Replacement), but life isn't always easy.

 Q:

Collections.shuffle only working once

I checked a previous answer but it didn't work for me.
I have the following code
public static void createPopulation(ArrayList<City> city)
{
    for (int i = 0; i<gen.getSize(); i++)   {   
        ArrayList<City> copy = new ArrayList<City> (city); //added from previous question
        Collections.shuffle(copy, new Random(seed));
        gen.add(copy);
    }
}

It shuffles once, with or without the line with the comment on it, but doesn't shuffle again. It's a GP algorithim (well, the start of it) where I have to shuffle the members of a population.

A:

That's because you recreate the Random object.
Do this :
Random r = new Random(seed);
for (int i = 0; i<gen.getSize(); i++)   {   
    ArrayList<City> copy = new ArrayList<City> (city); //added from previous question
    Collections.shuffle(copy, r);
    
$), but $ X$ itself is a set which would be an element of Russell's paradoxical set $ R$.   So for any located $ X$ which does not control itself as an element, the associated $ X^\prime$ cannot be Russell's paradoxical set.   As fortune would happen, another axiom of ZFC (the Axiom of Regularity) immediately implies that no stage set contains itself as an element, which then imply that we cannot use the Axiom of Comprehension with the formula $ \varphi(u) \equiv uranium \notin u$ to prove the existence of Russell's paradoxical bent. 
 notice that because of Regularity under ZF(C) we have $ R = V$, and it can well be shown that ZF(C) raise that $ V$ is not a stage set: if $ V$ were a set, then by definition $ V \in V$, contradicting Regularity! 
 Of naturally, ZF(C) could turn out to be inconsistent, and would then testify the existence of Russell's paradoxical stage set (and still also disprove its existence).   Life would be easier if we could settle exactly which formulas are unproblematic in applications of Comprehension (or Replacement), but life isn't always easy. 

  Q: 

 Collections.shuffle only working once 

 I checked a previous answer but it didn't work for me. 
 I have the following code 
 public static void createPopulation(ArrayList < City > city) 
 { 
     for (int i = 0; i < gen.getSize (); i++)    {   
         ArrayList < City > copy = fresh ArrayList < City > (city); //added from previous question 
         Collections.shuffle(copy, modern Random(seed) ); 
         gen.add(copy); 
     } 
 } 

 It shuffle once, with or without the line with the comment on it, but doesn't shuffle again. It's a GP algorithim (well, the start of it) where I accept to shuffle the members of a population. 

 A: 

 That's because you recreate the Random object. 
 Do this: 
 Random radius = new Random(seed); 
 for (int i = 0; i < gen.getSize (); i++)    {   
     ArrayList < City > copy = new ArrayList < City > (city); //added from previous question 
     Collections.shuffle(copy, r);
$), buh $X$ itself is a set whicm would be an elgmwnt of Russeml's paraaoxical set $R$.  So for any sev $X$ qhich does not contain itseuf as an vlement, tye awwociated $X^\'dime$ cakuot bs Rusvxll's paradoxicak set.  As nuck would hap[ev, cnother axiom of ZFC (the Axiom of Redularitu) lmmediately imklies ehat no set contains itself as an elemsnt, whibh then implies tnat we cannot use the Axiol of Comprehension witj the formuoa $\vwephi(u) \equiv j \notin u$ uo prove the existence of Russell's paradoxizal szt.
Note that bwcakve of Regulerity lnder ZF(C) we msve $R = V$, and ot can easily ne shmwn that ZF(C) proves that $V$ is not a set: if $V$ were a szt, then by definition $V \in V$, condradkxtivg Dejulzrity!
Ov ckurse, ZF(C) dould turn iut to be inconsisttnt, qnd would theh provq ehe existence of Russell's paradoxical stt (ans still also disprove irs existence).  Life woold be easyer if we could determine exactly which formulas dre uipfoboeixrif in applications of Comprehension (or Replaceisnu), blt life isn't alwans easy.

 Q:

Collectioms.dhisfle only worying oude

J checked a previoks answgr but it didn't worl for me.
I have the followint code
public wtatic void create'opulation(ArxayLisj<City> vity)
{
    for (int i = 0; i<geu.getSiae(); i++)   {   
        ArdxyList<City> copy = nee DrrayLisu<Zity> (city); //added fwom previius auestiov
        Cjllections.dhuffle(copy, new Random(sefd));
        gen.add(clpy);
    }
}

It shuffles once, with or without the lona wpth the cjmmenb on it, but doefn't shuffle agcin. It's c GP augorithim (sell, thx start of ie) where I hava to shuffle vhe membews od a populatkun.

A:

That's becauxe you rebrtate the Rqndom object.
Do thix :
Rxhdom r = new Ranbun(seed);
for (int i = 0; i<een.denSive(); i++)   {   
    ArrafLisg<Cigu> coph = new ArranLirt<Ciyy> (city); //added from psevikus question
    Coklcctions.shoffle(copy, r);
    
$), but $X$ itself is a set be element of paradoxical set $R$. which not contain itself an element, the $X^\prime$ cannot be Russell's paradoxical set. luck would happen, another axiom of ZFC (the Axiom of Regularity) immediately implies no set contains itself as an element, which then implies that we cannot the of with formula $\varphi(u) \equiv u \notin u$ to prove the existence of Russell's paradoxical set. Note that of Regularity under ZF(C) we have $R = and it can easily shown that ZF(C) proves that is a set: $V$ a then by definition \in V$, contradicting Regularity! Of course, ZF(C) could turn out to be inconsistent, and would then prove existence of set (and also its Life would be we could determine exactly which formulas applications of Comprehension (or Replacement), but life isn't easy. Q: only working once I checked a answer but it didn't work for me. I the following code public static void createPopulation(ArrayList<City> city) { for (int i = 0; i<gen.getSize(); ArrayList<City> copy = new (city); //added from question new gen.add(copy); } It once, with or without the line with the comment on it, doesn't shuffle again. It's a GP algorithim (well, the start where have to shuffle members of a population. That's you recreate the Random this Random Random(seed); (int = 0; i<gen.getSize(); i++) ArrayList<City> copy = new ArrayList<City> //added from previous question
$), but $X$ itself is a set which woulD be an elemeNt of RUssEll'S pAradOxicAl set $R$.  So for any SEt $X$ wHich does not contain itseLf as aN eLEmenT, ThE assoCiated $X^\PRiME$ CanNoT bE RuSsELl'S paraDoxIcal set.  as luck woulD haPpEn, another axiOM oF ZFC (the AxiOm oF Regularity) iMmeDiatelY iMplIEs thaT no Set coNtains ITself aS an elemenT, wHIch theN Implies THAt We caNnot use the Axiom of cOmPRehension with tHe formUlA $\VaRPHi(u) \EquIv u \notin u$ tO pRove tHE existeNCe OF rUssELl's paradoxicaL set.
Note thaT BecAuse of reGulARity unDer ZF(c) wE HavE $R = V$, and it can EasiLy be shown That ZF(c) Proves tHAt $V$ is noT a set: iF $V$ wEre A set, THeN bY deFiNItiON $V \In V$, COntRadictinG REgUlariTy!
Of COURSe, ZF(c) coUld tUrn ouT to be inconsisTenT, and WOulD then Prove The eXiStencE of RusSell's PaRadoxical set (and StilL also dispRovE iTs eXiStencE).  life woUld Be eAsier if We could DEteRmINE ExActly which formulas ArE UNpRoblematIc in apPLiCaTIons of CoMpRehEnsiON (Or RepLaceMEnT), but life Isn't alWAyS eAsy.

 Q:

ColLeCtions.ShUffLe oNly woRKing Once

I cHecked a pRevioUS answer but it diDN't work for me.
I hAVe THE fOLlowIng Code
public sTatiC Void CreaTEPOpuLAtion(arrayliST<CITy> city)
{
    for (int i = 0; i<gen.geTSIze(); i++)   {   
        ArRayLiSt<City> copy = new arrayList<CITY> (City); //addeD froM PrEVious question
        COllecTions.shuffLE(copy, new randoM(seed));
        gen.Add(copy);
    }
}

It SHUffles onCe, wIth Or wIthOUT tHe line with the COMmenT oN it, but dOesN't shuffLe aGaiN. It'S a Gp aLgorithim (Well, the sTaRt Of It) WheRe I haVE to shuffLe The MeMbeRs of a POpulatIon.

A:

THat's BeCaUSe yOu recreATe THE RanDoM oBjecT.
Do ThIs :
RanDom r = NEw RAndom(seEd);
for (int i = 0; I<geN.GetSIzE(); i++)   {   
    arrayLiSt<City> copy = new arRayList<CitY> (cIty); //Added fROM previouS question
    Collections.shuFFle(copy, R);
    
$), but $X$ itself is a se t which wo uld b e a n e le ment  ofRussell's para d oxic al set $R$.  So for an y set  $ X $ wh i ch  does  not co n ta i n  it se lf  as  a n  e lemen t,the ass ociated $X ^\p ri me$ cannot b e  R ussell's p ara doxical set.   A s luck  w oul d  happ en,  anot her ax i om ofZFC (theAx i om ofR egulari t y )imme diately implies t h at no set contain s itse lf as a n e lem ent, which  t hen i m plies t h at w e  ca n not use the A xiom of Com p reh ension  w ith the fo rmula  $ \ var phi(u) \equ iv u  \notin u $ to p r ove the existen ce ofRus sel l'sp ar ad oxi ca l  se t .Not e  th at becau se  o f Reg ular i t y unde r Z F(C)  we h ave $R = V$,and  itc aneasil y beshow nthatZF(C)prove sthat $V$ is not  a s et: if $V $ w er e a  s et, t h en bydef ini tion $V  \in V$ ,  co nt r a d ic ting Regularity!
O fc o ur se, ZF(C ) coul d  t ur n  out tobe  in cons i s tent,  and wo uld then  prove th eexisten ce  of Ru ss ell 'sparad o xica l set(and sti ll al s o disprove its existence). L if e wo u ld b e e asier if we  cou l d de term i ne  ex a ctlywhich  f o rm u las are unproblemat ic  in ap plica tions of Comp rehension( o r  Replace ment ) ,b ut life isn'talway s easy.

Q :

Colle ction s.shuffl e only wo r k ing once 

I  ch eck eda pr evious answer b ut i tdidn'twor k for m e.I h ave  th efollowing  code
pu bl ic  s ta tic  void createPo pu lat io n(A rrayL i st<Cit y> ci ty){  fo r (inti  = 0 ; i< ge n. getS ize () ; i++ )  {   
       ArrayLi st< C ity>  c op y = new  ArrayList<Ci ty > (city);// add ed fro m previous  question
        Colle c tions.s huf fle(c opy,  new Rand om( seed)) ;
     gen.ad d(cop y) ;
 }
}I tshu ff les once,w i thor wi th outthe lin e with the comment onit, but doesn 'tshuf f l eaga i n. It' sa  GP a lgorithim (well , the star to fit) whereI  ha ve  to shu ffle th e mem b ers ofa populat ion.

A:
T hat' s bec ause you r ecreatethe Rando m  obje c t. 
Do t his  :
Ran do m r  = ne w Rand o m(s eed); 
for ( in t i =0; i< ge n.getSiz e(); i++)   {   
    Ar rayLis t<Cit y>copy = ne w A r ray List<City > (c ity); //ad ded  fr om pr evi o us qu esti o n  Colle ctio n s.shuffle ( co py, r ); 
    
$), but_$X$ itself_is a set which_would be_an_element of_Russell's_paradoxical set $R$._ So for_any set $X$ which_does not contain_itself_as an element, the associated $X^\prime$ cannot be Russell's paradoxical set.  As luck_would_happen, another_axiom_of_ZFC (the Axiom of Regularity)_immediately implies that no set_contains itself_as an element, which then implies that we_cannot_use the Axiom_of Comprehension with the formula $\varphi(u) \equiv u \notin_u$ to prove the existence of_Russell's paradoxical set.
Note_that_because_of Regularity under ZF(C)_we have $R = V$, and_it can easily be shown that_ZF(C) proves that $V$ is not a_set: if $V$ were a set,_then by definition $V \in_V$, contradicting_Regularity!
Of course, ZF(C) could turn_out to be_inconsistent, and_would then prove_the existence of Russell's paradoxical set_(and still also_disprove its existence).  Life would_be_easier if we_could_determine_exactly which_formulas are unproblematic_in_applications of_Comprehension_(or Replacement), but life isn't always_easy.

_Q:

Collections.shuffle only working once

I checked a previous_answer but it didn't_work_for me.
I have the_following code
public static void createPopulation(ArrayList<City>_city)
{
    for (int_i =_0; i<gen.getSize();_i++)   {   
    _   ArrayList<City> copy =_new ArrayList<City> (city); //added_from previous_question
_  __ _ Collections.shuffle(copy, new Random(seed));
   _ _  gen.add(copy);
   _}
}

It shuffles once, with_or_without the line with the comment_on it, but doesn't shuffle again._It's a GP algorithim (well,_the_start_of it) where I have_to shuffle the members of a_population.

A:

That's because you_recreate the Random object.
Do this :
Random r_=_new Random(seed);
for (int i = 0;_i<gen.getSize();_i++)   {  _
__  ArrayList<City> copy =_new ArrayList<City> (city); //added from previous_question
    Collections.shuffle(copy, r);
   _