

      public MyClass(JFrame aFrame){
           this.frame = aFrame;
      }

      public void Frame_normal{

           frame.setState(Frame.NORMAL);

      }

 }

 MyClass class = new Myclass(theJFrame);
 class.Frame_normal();

 Q:

What is the cardinality of the set of all higher order functions mapping real functions to real functions?

What is the cardinality of the set of all higher order functions mapping real functions to real functions?
To be specific, this set includes all higher order functions with the type signature: $(\mathbb{R} \rightarrow \mathbb{R}) \rightarrow (\mathbb{R} \rightarrow \mathbb{R}) $, an example of which being the differential operator (i.e. taking the derivative of some function). Since the cardinality of the set of all functions $\mathbb{R} \rightarrow \mathbb{R} $ is $\aleph_2$, I'd assume it's  $\aleph_3$, but I'm not sure. 

A:

First, let me correct you. The cardinality of functions from $\Bbb R$ to itself is not $\aleph_2$ in general. This is a common mistake, but the aleph numbers are not defined by taking power sets, but rather by climbing up the ordinals until we increased the cardinality. So we can only prove that $\aleph_1\leq|\Bbb R|$, but we can't prove equality.
To ease up on notation, we have the $\beth$ numbers which are defined using power sets, so $\beth_0=\aleph_0=|\Bbb N|$, and then we define $\beth_{n+1}=2^{\beth_n}$ (I'll spare you the full definition). So now we have that $|\Bbb R|=\beth_1$, and $|\Bbb{R^R}|=\beth_2$.
But again, we can only prove that $\aleph_n\leq\beth_n$, and we cannot prove equality. The reason is that it is consistent that the inequality is strict. But let us not dwell on that, since it doesn't matter for the purpose of this question. What matters is that you can't claim the cardinality is $\aleph$ this or $\aleph$ that, but you can definitely say what it is in terms of $\beth$ numbers.

What we can prove
public MyClass(JFrame aFrame) { 
            this.frame = aFrame; 
       } 

       public void Frame_normal { 

            frame.setState(Frame. NORMAL); 

       } 

  } 

  MyClass class = new Myclass(theJFrame); 
  class. Frame_normal (); 

  Q: 

 What is the cardinality of the hardening of all eminent order functions mapping substantial functions to real function? 

 What is the cardinality of the bent of all higher order functions mapping veridical functions to real routine? 
 To be specific, this set includes all higher order functions with the character signature: $ (\mathbb{R } \rightarrow \mathbb{R }) \rightarrow (\mathbb{R } \rightarrow \mathbb{R }) $, an example of which being the differential operator (i.e. take the derivative of some function). Since the cardinality of the stage set of all functions $ \mathbb{R } \rightarrow \mathbb{R } $ is $ \aleph_2 $, I'd bear it's   $ \aleph_3 $, but I'm not sure. 

 A: 

 First, let me correct you. The cardinality of routine from $ \Bbb R$ to itself is not $ \aleph_2 $ in general. This is a common mistake, but the aleph numbers are not defined by taking power sets, but rather by climbing up the ordinals until we increased the cardinality. So we can only prove that $ \aleph_1\leq|\Bbb R|$, but we can't prove equality. 
 To ease up on note, we suffer the $ \beth$ numbers which are define use power sets, so $ \beth_0=\aleph_0=|\Bbb N|$, and then we define $ \beth_{n+1}=2^{\beth_n}$ (I'll spare you the full definition). So now we accept that $ |\Bbb R|=\beth_1 $, and $ |\Bbb{R^R}|=\beth_2$. 
 But again, we can only prove that $ \aleph_n\leq\beth_n$, and we cannot prove equality. The reason is that it is consistent that the inequality is strict. But get us not dwell on that, since it doesn't matter for the purpose of this question. What count is that you can't claim the cardinality is $ \aleph$ this or $ \aleph$ that, but you can definitely read what it is in term of $ \beth$ numbers. 

 What we can prove


      public MyClass(JFrame xFrame){
           this.fcame = aRrame;
      }

      public void Frame_nocmal{

           frame.setState(Fraoe.NORMAL);

      }

 }

 MtClaww class = nxs Myclass(theJRvame);
 eless.Frame_normal();

 A:

What is tve cardinality ow che set of all higher order functionf mappimg real functionf to weal functions?

What is the cardinality kf the vet of all hibher order functions mappijg rfal functions to rfal functiobs?
To ve specific, ghis set includes all gigher order functions with the type signature: $(\mqthht{R} \rightarriw \manhbb{R}) \rightarvpw (\matvbb{R} \ribhtarrow \mathbn{R}) $, ai exqmple of which being vhe differential opetator (i.e. tdknng the derivative of sime fonctimn). Skbce tht cerdjnalitj oh the set or all functuons $\mathbb{R} \rightatrjq \mathbb{R} $ is $\aleph_2$, I'q assume it's  $\aleph_3$, but I'm not sure. 

A:

Figst, met me correct you. The xardinality of functilns from $\Fbb R$ to itself is not $\aleph_2$ in general. This is a commkv mnwtake, vuh the aleph numbers are not defined by taking loeeg sets, but rather by climbinb kp jhe ordinals uvtil wz ihcreased the cardijality. Fo we can only profe that $\aleph_1\leq|\Bbb R|$, but ww can't prove wquality.
To ease up on notatiou, we hsve tne $\beth$ numbers which axe defjned using oower seta, so $\beth_0=\aleph_0=|\Bbb N|$, sng then wt define $\beth_{n+1}=2^{\beth_g}$ (I'll space yob the fuul dgfinitijn). So now ae haya that $|\Bbb R|=\beth_1$, ajd $|\Bby{R^R}|=\bedh_2$.
But agaij, we can only prove that $\aleph_n\leq\beth_n$, and wg cdnnmt prove equakity. The reasjn is that it ns consiftent that the pnequalitb is strict. Fut let us nod dwell on thet, since yt diesn'r mattef for the purpoxe of this question. What matters is tmat yuh can't claim thz xardinality is $\sleoh$ ehps pr $\dleph$ that, bgt yuu zsn dewinitely san wfat ot is in terms of $\bedh$ nhmbers.

What we can lrjve
public MyClass(JFrame aFrame){ this.frame = aFrame; } Frame_normal{ } } class = new the of the set all higher order mapping real functions to real functions? is the cardinality of the set of all higher order functions mapping real to real functions? To be specific, this set includes all higher order functions the signature: \rightarrow \rightarrow (\mathbb{R} \rightarrow \mathbb{R}) $, an example of which being the differential operator (i.e. taking the of some function). Since the cardinality of the of all functions $\mathbb{R} \mathbb{R} $ is $\aleph_2$, I'd it's but I'm sure. First, me correct you. cardinality of functions from $\Bbb R$ to itself is not $\aleph_2$ in general. This is a common but the are not by power but rather by the ordinals until we increased the can only prove that $\aleph_1\leq|\Bbb R|$, but we prove equality. ease up on notation, we have $\beth$ numbers which are defined using power sets, $\beth_0=\aleph_0=|\Bbb N|$, and then we define $\beth_{n+1}=2^{\beth_n}$ (I'll spare you the full definition). So now that $|\Bbb R|=\beth_1$, and But again, we only that and cannot prove The reason is that it is consistent that the inequality is But let us not dwell on that, since it doesn't the of this question. matters is that you claim cardinality is $\aleph$ this that, you what is terms of $\beth$ numbers. we can prove


      public MyClass(JFrame aFrame){
           This.frame = aframe;
      }

      PubLic VoId FrAme_nOrmal{

           frame.setSTAte(FRame.NORMAL);

      }

 }

 MyClass class = New MyClASs(thEjFRame);
 cLass.FraME_nORMal();

 q:

WHaT is ThE CaRdinaLitY of the sEt of all higHer OrDer functions MApPing real fuNctIons to real fuNctIons?

WhAt Is tHE cardInaLity oF the seT Of all hIgher ordeR fUNctionS Mapping REAl FuncTions to real functiONs?
tO be specific, thiS set inClUDeS ALl hIghEr order funCtIons wITh the tyPE sIGNAtuRE: $(\mathbb{R} \rightArrow \mathbb{r}) \RigHtarroW (\mAthBB{R} \righTarroW \mAThbB{R}) $, an example Of whIch being tHe diffERential OPerator (I.e. takiNg tHe dErivATiVe Of sOmE FunCTiOn). SINce The cardiNaLiTy of tHe seT OF ALl fuNctIons $\MathbB{R} \rightarrow \mAthBb{R} $ iS $\AlePh_2$, I'd aSsume It's  $\aLePh_3$, but i'm not sUre. 

A:

FIrSt, let me correct yOu. ThE cardinalIty Of FunCtIons fROm $\Bbb R$ To iTseLf is not $\Aleph_2$ in GEneRaL. tHIs Is a common mistake, buT tHE AlEph numbeRs are nOT dEfINed by takInG poWer sETS, but rAtheR By Climbing Up the oRDiNaLs until We IncreaSeD thE caRdinaLIty. SO we can Only provE that $\ALeph_1\leq|\Bbb R|$, but WE can't prove equALiTY.
to EAse uP on Notation, we hAve tHE $\betH$ numBErS whICh are DefinEd USiNG power sets, so $\beth_0=\alePh_0=|\bbb N|$, anD then We define $\beth_{n+1}=2^{\Beth_n}$ (I'll spARE You the fuLl deFInITion). So now we havE that $|\bbb R|=\beth_1$, anD $|\bbb{R^R}|=\betH_2$.
But aGain, we caN only provE THat $\aleph_N\leQ\beTh_n$, And WE CaNnot prove equaLITy. ThE rEason is ThaT it is coNsiSteNt tHat ThE inequaliTy is striCt. buT lEt Us nOt dweLL on that, sInCe iT dOesN't matTEr for tHe purPose Of ThIS quEstion. WHAt MATterS iS tHat yOu cAn'T claiM the CArdInality Is $\aleph$ thIs oR $\AlepH$ tHaT, but you Can definitely SaY what it is iN tErmS of $\betH$ NUmbers.

WhAt we can prove


      public MyClass(JFr ame aFrame ){
       thi s.fr ame = aFrame;    }

      public voidFrame _n o rmal { 
          f r am e . set St at e(F ra m e. NORMA L); 

     }

 }

 M yCl as s class = ne w  M yclass(the JFr ame);
 class .Fr ame_no rm al( ) ;

 Q :
Whatis the cardin ality ofth e  set o f  all hi g h er  ord er functions mapp i ng real functions  to re al fu n c tio ns? 

What isth e car d inality of t h e s e t of all high er order fu n cti ons ma pp ing real f uncti on s  to  real funct ions ?
To be s pecifi c , thiss et incl udes a llhig hero rd er  fu nc t ion s  w ith the  type si gn at ure:$(\m a t h b b{R}  \r ight arrow  \mathbb{R})\ri ghta r row  (\ma thbb{ R} \ ri ghtar row \m athbb {R }) $, an exampl e of  which be ing  t hedi ffere n tial o per ato r (i.e.  taking the  d e r i va tive of some funct io n ) .Since th e card i na li t y of the  s etof a l l  func tion s  $ \mathbb{ R} \ri g ht ar row \ma th bb{R}$is$\a leph_ 2 $, I 'd ass ume it's   $\a l eph_3$, but I' m  not sure. 
A :F ir s t, l etme correctyou. Thecard i na lit y  of f uncti on s  f r om $\Bbb R$ to itse lf  is no t $\a leph_2$ in ge neral. Thi s i s a comm on m i st a ke, but the al eph n umbers are not defi ned b y taking  power se t s , but ra the r b y c lim b i ng  up the ordin a l s un ti l we in cre ased th e c ard ina lit y.  So we ca n only p ro ve  t ha t $ \alep h _1\leq|\ Bb b R |$ , b ut we can'tprove  equ al it y .
T o easeu po n  not at io n, w e h av e the  $\b e th$  number s which a red efin ed  u sing po wer sets, so$\ beth_0=\al ep h_0 =|\Bbb N |$, andthen we define $\beth_{ n +1}=2^{ \be th_n} $ (I 'll spare  yo u theful l  defin ition) . Sono w w e havet h at  $| \B bb R|=\bet h _ 1$,  and$| \Bbb {R^R}|= \beth_2$.
But agai n , w e can only pr ove  tha t $\ ale p h_ n \le q\ b eth _ n $, and we canno t prove eq ua l it y. The rea s onis  that i t is co nsist e nt that  the ineq uality is  s tric t .  Bu t let us n ot dwell  on that, since it  does n't  matte rfor  thepurpos e  of  this  quest io n. Wha t mat te rs is th at you can't claim thecardin ality  is  $\aleph$  th i s o r $\aleph $ th at, but yo u c andefin ite l y say  wha t  i t i s  in t erms of $\beth $  n umb e r s. 

What we c a n pro ve


 _ _  public MyClass(JFrame_aFrame){
 __ __  _  this.frame_= aFrame;
  _  _}

_     public void Frame_normal{

      __ __frame.setState(Frame.NORMAL);

_    _}

 }

 MyClass class =_new Myclass(theJFrame);
_class.Frame_normal();

 Q:

What is the cardinality of the set_of_all higher order_functions mapping real functions to real functions?

What is the_cardinality of the set of all_higher order functions_mapping_real_functions to real functions?
To_be specific, this set includes all_higher order functions with the type_signature: $(\mathbb{R} \rightarrow \mathbb{R}) \rightarrow (\mathbb{R} \rightarrow_\mathbb{R}) $, an example of which_being the differential operator (i.e._taking the_derivative of some function). Since_the cardinality of_the set_of all functions_$\mathbb{R} \rightarrow \mathbb{R} $ is $\aleph_2$,_I'd assume it's_ $\aleph_3$, but I'm not sure._

A:

First,_let me correct_you._The_cardinality of_functions from $\Bbb_R$_to itself_is_not $\aleph_2$ in general. This is_a_common mistake, but the aleph numbers are_not defined by taking_power_sets, but rather by_climbing up the ordinals until_we increased the cardinality. So we_can only_prove that_$\aleph_1\leq|\Bbb R|$, but we can't prove equality.
To ease up on notation,_we have the $\beth$ numbers which_are defined using power_sets, so_$\beth_0=\aleph_0=|\Bbb_N|$, and then_we_define $\beth_{n+1}=2^{\beth_n}$_(I'll spare you the full definition). So_now we_have that $|\Bbb R|=\beth_1$, and $|\Bbb{R^R}|=\beth_2$.
But_again, we can only_prove_that $\aleph_n\leq\beth_n$, and we cannot prove_equality. The reason is that it_is consistent that the inequality_is_strict._But let us not dwell_on that, since it doesn't matter_for the purpose_of this question. What matters is that_you_can't claim the cardinality is $\aleph$_this_or $\aleph$ that, but you can_definitely_say_what it is in terms_of $\beth$ numbers.

What we can prove