Any finite intersection of countable sets is countable. (By induction for any amount of countable sets) [on hold] The Next CEO of Stack OverflowSets induction problem (complement of intersection equals union of complements)intersection of infinite collection of finite sets?Prove by induction that $Gamma vdash varphi Rightarrow Gamma[x/c] vdash varphi[x/c]$.Finite Union of Countable sets is countableSlowly being introduced to Propositional Logic…did I do this right?Slowly being introduced to Propositional Logic…did I get this right?Using mathematical induction to prove a generalized form of DeMorgan's Law for setsInductive first-order formula constructionFor any two finite sets $X$ and $Y$, prove that $#(Y^X)=#(Y)^#(X)$ by inductionDiscrete Math - Proving Distributive Laws for Sets by induction
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Any finite intersection of countable sets is countable. (By induction for any amount of countable sets) [on hold]
The Next CEO of Stack OverflowSets induction problem (complement of intersection equals union of complements)intersection of infinite collection of finite sets?Prove by induction that $Gamma vdash varphi Rightarrow Gamma[x/c] vdash varphi[x/c]$.Finite Union of Countable sets is countableSlowly being introduced to Propositional Logic…did I do this right?Slowly being introduced to Propositional Logic…did I get this right?Using mathematical induction to prove a generalized form of DeMorgan's Law for setsInductive first-order formula constructionFor any two finite sets $X$ and $Y$, prove that $#(Y^X)=#(Y)^#(X)$ by inductionDiscrete Math - Proving Distributive Laws for Sets by induction
$begingroup$
I know how to do the base case I'm just struggling with the inductive hypothesis for this question.
elementary-set-theory logic
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
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
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
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
edited 2 days ago
Andrés E. Caicedo
65.8k8160252
asked Mar 28 at 8:51
user652437user652437
22
$endgroup$
I know how to do the base case I'm just struggling with the inductive hypothesis for this question.
elementary-set-theory logic
elementary-set-theory logic
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
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
edited 2 days ago
Andrés E. Caicedo
65.8k8160252
65.8k8160252
asked Mar 28 at 8:51
user652437user652437
22
asked Mar 28 at 8:51
user652437user652437
22
asked Mar 28 at 8:51
user652437user652437
22
22
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
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
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 |
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
1
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
$begingroup$
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
$begingroup$
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
$begingroup$
no i meant interception and i need to do it with induction
$endgroup$
– user652437
Mar 28 at 8:55
1
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
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 28 at 9:39
KlausKlaus
2,832213
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered Mar 28 at 9:39
KlausKlaus
2,832213
$endgroup$
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.
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
answered Mar 28 at 9:39
KlausKlaus
2,832213
2,832213
add a comment |
add a comment |
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