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Any finite intersection of countable sets is countable. (By induction for any amount of countable sets) [on hold]

-2

$begingroup$

I know how to do the base case I'm just struggling with the inductive hypothesis for this question.

elementary-set-theory logic

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edited 2 days ago

Andrés E. Caicedo

65.8k8160252

asked Mar 28 at 8:51

user652437user652437

22

$endgroup$

put on hold as off-topic by Shailesh, Claude Leibovici, José Carlos Santos, Especially Lime, Javi Mar 28 at 12:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Claude Leibovici, José Carlos Santos, Especially Lime, Javi

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Any subset of a countable set is countable (trusting that you consider finite, or empty, sets to be countable...not everyone does).
  $endgroup$
  – lulu
  Mar 28 at 8:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you perhaps mean union?
  $endgroup$
  – Arthur
  Mar 28 at 8:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no i meant interception and i need to do it with induction
  $endgroup$
  – user652437
  Mar 28 at 8:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you need to do it by induction? I think it would actually be difficult to write a proof for the induction step which is not basically a full proof on its own, without any induction mechanism (i.e. a proof that has to actually use the induction hypothesis).
  $endgroup$
  – Arthur
  Mar 28 at 9:00
  
  

  
  
  
  

add a comment | 

-2

$begingroup$

I know how to do the base case I'm just struggling with the inductive hypothesis for this question.

elementary-set-theory logic

share|cite|improve this question

edited 2 days ago

Andrés E. Caicedo

65.8k8160252

asked Mar 28 at 8:51

user652437user652437

22

$endgroup$

put on hold as off-topic by Shailesh, Claude Leibovici, José Carlos Santos, Especially Lime, Javi Mar 28 at 12:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Claude Leibovici, José Carlos Santos, Especially Lime, Javi

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Any subset of a countable set is countable (trusting that you consider finite, or empty, sets to be countable...not everyone does).
  $endgroup$
  – lulu
  Mar 28 at 8:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you perhaps mean union?
  $endgroup$
  – Arthur
  Mar 28 at 8:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no i meant interception and i need to do it with induction
  $endgroup$
  – user652437
  Mar 28 at 8:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you need to do it by induction? I think it would actually be difficult to write a proof for the induction step which is not basically a full proof on its own, without any induction mechanism (i.e. a proof that has to actually use the induction hypothesis).
  $endgroup$
  – Arthur
  Mar 28 at 9:00
  
  

  
  
  
  

add a comment | 

$begingroup$

I know how to do the base case I'm just struggling with the inductive hypothesis for this question.

elementary-set-theory logic

share|cite|improve this question

edited 2 days ago

Andrés E. Caicedo

65.8k8160252

asked Mar 28 at 8:51

user652437user652437

22

$endgroup$

share|cite|improve this question

edited 2 days ago

Andrés E. Caicedo

65.8k8160252

asked Mar 28 at 8:51

user652437user652437

22

share|cite|improve this question

edited 2 days ago

Andrés E. Caicedo

65.8k8160252

asked Mar 28 at 8:51

user652437user652437

22

edited 2 days ago

Andrés E. Caicedo

65.8k8160252

edited 2 days ago

Andrés E. Caicedo

65.8k8160252

asked Mar 28 at 8:51

user652437user652437

22

asked Mar 28 at 8:51

user652437user652437

22

1 Answer
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0

$begingroup$

The argument for the base case ($n = 2$) already proves the whole claim, so this is pointless, but if you really want, here we go:

Induction hypothesis: For any $n$ countable sets $A_1, ldots, A_n$ the intersection $bigcaplimits_k = 1^n A_k$ is countable.

Induction step: Let $A_1, ldots, A_n+1$ be $n+1$ countable sets. Then
$bigcaplimits_k = 1^n+1 A_k = left(bigcaplimits_k = 1^n A_kright) cap A_n+1$. By the induction hypothesis the first set is countable. Hence $bigcaplimits_k = 1^n+1 A_k$ is the intesection of two countable sets and therefore by the base case countable as well.

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answered Mar 28 at 9:39

KlausKlaus

2,832213

$endgroup$

add a comment | 

0

$begingroup$

The argument for the base case ($n = 2$) already proves the whole claim, so this is pointless, but if you really want, here we go:

Induction hypothesis: For any $n$ countable sets $A_1, ldots, A_n$ the intersection $bigcaplimits_k = 1^n A_k$ is countable.

Induction step: Let $A_1, ldots, A_n+1$ be $n+1$ countable sets. Then
$bigcaplimits_k = 1^n+1 A_k = left(bigcaplimits_k = 1^n A_kright) cap A_n+1$. By the induction hypothesis the first set is countable. Hence $bigcaplimits_k = 1^n+1 A_k$ is the intesection of two countable sets and therefore by the base case countable as well.

share|cite|improve this answer

answered Mar 28 at 9:39

KlausKlaus

2,832213

$endgroup$

add a comment | 

0

$begingroup$

The argument for the base case ($n = 2$) already proves the whole claim, so this is pointless, but if you really want, here we go:

Induction hypothesis: For any $n$ countable sets $A_1, ldots, A_n$ the intersection $bigcaplimits_k = 1^n A_k$ is countable.

Induction step: Let $A_1, ldots, A_n+1$ be $n+1$ countable sets. Then
$bigcaplimits_k = 1^n+1 A_k = left(bigcaplimits_k = 1^n A_kright) cap A_n+1$. By the induction hypothesis the first set is countable. Hence $bigcaplimits_k = 1^n+1 A_k$ is the intesection of two countable sets and therefore by the base case countable as well.

share|cite|improve this answer

answered Mar 28 at 9:39

KlausKlaus

2,832213

$endgroup$

add a comment | 

$begingroup$

The argument for the base case ($n = 2$) already proves the whole claim, so this is pointless, but if you really want, here we go:

Induction hypothesis: For any $n$ countable sets $A_1, ldots, A_n$ the intersection $bigcaplimits_k = 1^n A_k$ is countable.

Induction step: Let $A_1, ldots, A_n+1$ be $n+1$ countable sets. Then
$bigcaplimits_k = 1^n+1 A_k = left(bigcaplimits_k = 1^n A_kright) cap A_n+1$. By the induction hypothesis the first set is countable. Hence $bigcaplimits_k = 1^n+1 A_k$ is the intesection of two countable sets and therefore by the base case countable as well.

share|cite|improve this answer

answered Mar 28 at 9:39

KlausKlaus

2,832213

$endgroup$

share|cite|improve this answer

answered Mar 28 at 9:39

KlausKlaus

2,832213

answered Mar 28 at 9:39

KlausKlaus

2,832213

answered Mar 28 at 9:39

KlausKlaus

2,832213