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Elementary set theory involving partition

Elementary Set theory proofsError in book's definition of open sets in terms of neighborhoods?Number of size 1 partitions of the empty setVery basic question about set theory: unions and intersectionFinding counterexamples in elementary set theory.Limit of closed set with lower boundary $a_k$ converging to $a$ but $a_k$ > a is open.Definition of Partition in Analysis vs TopologyProof of Elementary Resulting Concerning Finite Set CardinalityPartition an infinite set into countable sets$tau_3 = mathbfR,varnothingcup[n,infty):ninmathbfZ^+$ is a topology on $mathbfR$.

0

$begingroup$

Hello could someone verify and correct if necessary my answer to the following question :

Is the given collection P a partition of the set A?Justify your answers.

With: $A=BbbR$ and $P=$$[n,n+1)$ : $ nin BbbZ$ where [n,n+1) is the half-open interval.

Then my attempt:

For the collection of nonempty subsets P to be a partition of A the following conditions must hold true.

(i)For all elements $ain A$ , there exists a set $S in P$ such that $a in S$ .

[case : $a in BbbZ$]

• 
  

  For every $a in BbbZ$ one can find a $n in BbbZ$ such that
  $nle alt n+1$, simply by setting $n=a$.
  In interval and set notations we have the following $S:=$ $[a,a+1)$= r $in BbbR:ale r lt a+1$.

  
  
  

  By the defintion of P,this half-open interval S is an element of the
  collection P that contains $a$.

  
  

[case : $a notin BbbZ$]

• 
  

  Therefore $a in BbbR$ $BbbZ$ or equivalently $a$ has numbers
  following the decimal point that are not all zero.Suppose $a_rounded$ is the next lowest integer to which we can round $a$,then $a_roundedle a lt a_rounded+1$ is assured.Letting $S:=[a_rounded,a_rouned+1)=$ $r in BbbR :a_rounded le r lt a_rounded+1 $.

  
  
  

  By the defintion of P,this half-open interval S is an element of
  the collection P that contains $a_rounded$.

  
  

Therefore condition (i) has been met for all possible elements of A.

(ii) For all $S,Tin P$, if $S neq T$ then $S cap T = varnothing$.

Let $m,n in BbbZ$ where $m neq n$.Further define $S:=[n,n+1)$ and
$T:=[m,m+1)$.

[case : $mgt n$]

• 
  

  If $mgt n $ then $mge n+1$ and hence $n+1 le m lt n+2 le m+1$ .Also $T=[m,m+1)$=$rin BbbR : n+1le
  r lt m+1$$=[n+1,m+1)$.

  
  
  

  Consequently with $S:=[n,n+1)$ and $T=[n+1,m+1)$ the following result
  is assured $S cap T=varnothing$

  
  

[case : $mlt n$]

• 
  

  If $m lt n$ then $mle n-1$ and hence $m le n-1 lt m+1 le n$. Also $T=[m,m+1)$=$rin BbbR : mle
  r lt n$$=[m,n)$.

  
  
  

  Consequently with $S:=[n,n+1)$ and $T=[m,n)$ the following result
  is assured $S cap T=varnothing$

  
  

Therefore the second condtion (ii) holds true for any pair $S,T in P$.

Since both conditions have been verified,the collection P is a valid partition of set A.

Does this make sense?

Thank you for your time.

proof-verification elementary-set-theory proof-explanation

share|cite|improve this question

edited Mar 12 at 21:49

HalfAFoot

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect proof!!
  $endgroup$
  – asdf
  Mar 12 at 11:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  honest or jokeing?
  $endgroup$
  – HalfAFoot
  Mar 12 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your proof should be much worse for jokes about it to be funny! Why would I post an unfunny joke on a serious site?
  $endgroup$
  – asdf
  Mar 13 at 7:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh ok thank you..i honestly do not know.!!..gotta hate textbooks witout answers!!
  $endgroup$
  – HalfAFoot
  Mar 13 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you start from the premises and reached the conclusion without cheating, that's it!
  $endgroup$
  – asdf
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Hello could someone verify and correct if necessary my answer to the following question :

Is the given collection P a partition of the set A?Justify your answers.

With: $A=BbbR$ and $P=$$[n,n+1)$ : $ nin BbbZ$ where [n,n+1) is the half-open interval.

Then my attempt:

For the collection of nonempty subsets P to be a partition of A the following conditions must hold true.

(i)For all elements $ain A$ , there exists a set $S in P$ such that $a in S$ .

[case : $a in BbbZ$]

• 
  

  For every $a in BbbZ$ one can find a $n in BbbZ$ such that
  $nle alt n+1$, simply by setting $n=a$.
  In interval and set notations we have the following $S:=$ $[a,a+1)$= r $in BbbR:ale r lt a+1$.

  
  
  

  By the defintion of P,this half-open interval S is an element of the
  collection P that contains $a$.

  
  

[case : $a notin BbbZ$]

• 
  

  Therefore $a in BbbR$ $BbbZ$ or equivalently $a$ has numbers
  following the decimal point that are not all zero.Suppose $a_rounded$ is the next lowest integer to which we can round $a$,then $a_roundedle a lt a_rounded+1$ is assured.Letting $S:=[a_rounded,a_rouned+1)=$ $r in BbbR :a_rounded le r lt a_rounded+1 $.

  
  
  

  By the defintion of P,this half-open interval S is an element of
  the collection P that contains $a_rounded$.

  
  

Therefore condition (i) has been met for all possible elements of A.

(ii) For all $S,Tin P$, if $S neq T$ then $S cap T = varnothing$.

Let $m,n in BbbZ$ where $m neq n$.Further define $S:=[n,n+1)$ and
$T:=[m,m+1)$.

[case : $mgt n$]

• 
  

  If $mgt n $ then $mge n+1$ and hence $n+1 le m lt n+2 le m+1$ .Also $T=[m,m+1)$=$rin BbbR : n+1le
  r lt m+1$$=[n+1,m+1)$.

  
  
  

  Consequently with $S:=[n,n+1)$ and $T=[n+1,m+1)$ the following result
  is assured $S cap T=varnothing$

  
  

[case : $mlt n$]

• 
  

  If $m lt n$ then $mle n-1$ and hence $m le n-1 lt m+1 le n$. Also $T=[m,m+1)$=$rin BbbR : mle
  r lt n$$=[m,n)$.

  
  
  

  Consequently with $S:=[n,n+1)$ and $T=[m,n)$ the following result
  is assured $S cap T=varnothing$

  
  

Therefore the second condtion (ii) holds true for any pair $S,T in P$.

Since both conditions have been verified,the collection P is a valid partition of set A.

Does this make sense?

Thank you for your time.

proof-verification elementary-set-theory proof-explanation

share|cite|improve this question

edited Mar 12 at 21:49

HalfAFoot

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect proof!!
  $endgroup$
  – asdf
  Mar 12 at 11:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  honest or jokeing?
  $endgroup$
  – HalfAFoot
  Mar 12 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your proof should be much worse for jokes about it to be funny! Why would I post an unfunny joke on a serious site?
  $endgroup$
  – asdf
  Mar 13 at 7:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh ok thank you..i honestly do not know.!!..gotta hate textbooks witout answers!!
  $endgroup$
  – HalfAFoot
  Mar 13 at 16:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you start from the premises and reached the conclusion without cheating, that's it!
  $endgroup$
  – asdf
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Hello could someone verify and correct if necessary my answer to the following question :

Is the given collection P a partition of the set A?Justify your answers.

With: $A=BbbR$ and $P=$$[n,n+1)$ : $ nin BbbZ$ where [n,n+1) is the half-open interval.

Then my attempt:

For the collection of nonempty subsets P to be a partition of A the following conditions must hold true.

(i)For all elements $ain A$ , there exists a set $S in P$ such that $a in S$ .

[case : $a in BbbZ$]

• 
  

  For every $a in BbbZ$ one can find a $n in BbbZ$ such that
  $nle alt n+1$, simply by setting $n=a$.
  In interval and set notations we have the following $S:=$ $[a,a+1)$= r $in BbbR:ale r lt a+1$.

  
  
  

  By the defintion of P,this half-open interval S is an element of the
  collection P that contains $a$.

  
  

[case : $a notin BbbZ$]

• 
  

  Therefore $a in BbbR$ $BbbZ$ or equivalently $a$ has numbers
  following the decimal point that are not all zero.Suppose $a_rounded$ is the next lowest integer to which we can round $a$,then $a_roundedle a lt a_rounded+1$ is assured.Letting $S:=[a_rounded,a_rouned+1)=$ $r in BbbR :a_rounded le r lt a_rounded+1 $.

  
  
  

  By the defintion of P,this half-open interval S is an element of
  the collection P that contains $a_rounded$.

  
  

Therefore condition (i) has been met for all possible elements of A.

(ii) For all $S,Tin P$, if $S neq T$ then $S cap T = varnothing$.

Let $m,n in BbbZ$ where $m neq n$.Further define $S:=[n,n+1)$ and
$T:=[m,m+1)$.

[case : $mgt n$]

• 
  

  If $mgt n $ then $mge n+1$ and hence $n+1 le m lt n+2 le m+1$ .Also $T=[m,m+1)$=$rin BbbR : n+1le
  r lt m+1$$=[n+1,m+1)$.

  
  
  

  Consequently with $S:=[n,n+1)$ and $T=[n+1,m+1)$ the following result
  is assured $S cap T=varnothing$

  
  

[case : $mlt n$]

• 
  

  If $m lt n$ then $mle n-1$ and hence $m le n-1 lt m+1 le n$. Also $T=[m,m+1)$=$rin BbbR : mle
  r lt n$$=[m,n)$.

  
  
  

  Consequently with $S:=[n,n+1)$ and $T=[m,n)$ the following result
  is assured $S cap T=varnothing$

  
  

Therefore the second condtion (ii) holds true for any pair $S,T in P$.

Since both conditions have been verified,the collection P is a valid partition of set A.

Does this make sense?

Thank you for your time.

proof-verification elementary-set-theory proof-explanation

share|cite|improve this question

edited Mar 12 at 21:49

HalfAFoot

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226

$endgroup$

share|cite|improve this question

edited Mar 12 at 21:49

HalfAFoot

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226

share|cite|improve this question

edited Mar 12 at 21:49

HalfAFoot

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226

asked Mar 12 at 11:11

HalfAFootHalfAFoot

226