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Conditions for $A times B$ to be countable
Clash Royale CLAN TAG#URR8PPP
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Suppose $A$ and $B$ are both sets, and $B$ is for sure countably infinite. What are the conditions that $A$ must have so that $Atimes B$ is countable? The possible answers (I suspect more than one is true) are
- necessary that $A$ is countably infinite
- necessary that $A$ is just countable
- sufficient that $A$ is countably infinite
- sufficient that $A$ is just countable
I don't quite understand the difference between "necessary" and "sufficient", and I'm also not quite clear on how being countable vs countably infinite changes the problem.
For example, I know that if $A, B$ are both countably infinite, $Atimes B$ is countably infinite. But does that mean that the fact that $A$ is countably infinite is "necessary" or just "sufficient"? I really don't understand. Thank you so much for your help!
elementary-set-theory
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
asked Aug 2 at 19:06
D.R.
1,190519
add a comment |Â
up vote
1
down vote
favorite
Suppose $A$ and $B$ are both sets, and $B$ is for sure countably infinite. What are the conditions that $A$ must have so that $Atimes B$ is countable? The possible answers (I suspect more than one is true) are
- necessary that $A$ is countably infinite
- necessary that $A$ is just countable
- sufficient that $A$ is countably infinite
- sufficient that $A$ is just countable
I don't quite understand the difference between "necessary" and "sufficient", and I'm also not quite clear on how being countable vs countably infinite changes the problem.
For example, I know that if $A, B$ are both countably infinite, $Atimes B$ is countably infinite. But does that mean that the fact that $A$ is countably infinite is "necessary" or just "sufficient"? I really don't understand. Thank you so much for your help!
elementary-set-theory
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
asked Aug 2 at 19:06
D.R.
1,190519
Countable set is a set with the same cardinality as some subset of the set of natural numbers. Now if a set has the same cardinality as the set of natural numbers, it is called countably infinite, but because of the fact that any set is a subset of itself it means that any countably infinite set is countable.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:11
1
$P$ being "sufficient" for $Q$ means that $Prightarrow Q$. On the other hand, $P$ being "necessary" for $Q$ means that $Qrightarrow P$. In other words, "sufficient" is short for "$P$ is sufficient to imply $Q$", and "necessary" is short for "$P$ necessarily occurs whenever $Q$ does".
– anakhro
Aug 2 at 19:12
About "necessary" and "sufficient". Suppose you know that if some proposition P is true than proposition Q is true. Than, you say that P is sufficient for Q, and Q is necessary for P.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:15
2
To win the lottery it is necessary to buy a ticket. But it is not sufficient; many people buy tickets and do not win. To become ill, it is sufficient to drink bleach, but it is not necessary; many people become ill without drinking bleach.
– MJD
Aug 2 at 19:22
add a comment |Â
up vote
1
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up vote
1
down vote
favorite
Suppose $A$ and $B$ are both sets, and $B$ is for sure countably infinite. What are the conditions that $A$ must have so that $Atimes B$ is countable? The possible answers (I suspect more than one is true) are
- necessary that $A$ is countably infinite
- necessary that $A$ is just countable
- sufficient that $A$ is countably infinite
- sufficient that $A$ is just countable
I don't quite understand the difference between "necessary" and "sufficient", and I'm also not quite clear on how being countable vs countably infinite changes the problem.
For example, I know that if $A, B$ are both countably infinite, $Atimes B$ is countably infinite. But does that mean that the fact that $A$ is countably infinite is "necessary" or just "sufficient"? I really don't understand. Thank you so much for your help!
elementary-set-theory
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
asked Aug 2 at 19:06
D.R.
1,190519
Suppose $A$ and $B$ are both sets, and $B$ is for sure countably infinite. What are the conditions that $A$ must have so that $Atimes B$ is countable? The possible answers (I suspect more than one is true) are
- necessary that $A$ is countably infinite
- necessary that $A$ is just countable
- sufficient that $A$ is countably infinite
- sufficient that $A$ is just countable
I don't quite understand the difference between "necessary" and "sufficient", and I'm also not quite clear on how being countable vs countably infinite changes the problem.
For example, I know that if $A, B$ are both countably infinite, $Atimes B$ is countably infinite. But does that mean that the fact that $A$ is countably infinite is "necessary" or just "sufficient"? I really don't understand. Thank you so much for your help!
elementary-set-theory
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
asked Aug 2 at 19:06
D.R.
1,190519
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
edited Aug 2 at 19:08
Eric Wofsey
161k12188297
161k12188297
asked Aug 2 at 19:06
D.R.
1,190519
asked Aug 2 at 19:06
D.R.
1,190519
asked Aug 2 at 19:06
D.R.
1,190519
1,190519
Countable set is a set with the same cardinality as some subset of the set of natural numbers. Now if a set has the same cardinality as the set of natural numbers, it is called countably infinite, but because of the fact that any set is a subset of itself it means that any countably infinite set is countable.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:11
1
$P$ being "sufficient" for $Q$ means that $Prightarrow Q$. On the other hand, $P$ being "necessary" for $Q$ means that $Qrightarrow P$. In other words, "sufficient" is short for "$P$ is sufficient to imply $Q$", and "necessary" is short for "$P$ necessarily occurs whenever $Q$ does".
– anakhro
Aug 2 at 19:12
About "necessary" and "sufficient". Suppose you know that if some proposition P is true than proposition Q is true. Than, you say that P is sufficient for Q, and Q is necessary for P.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:15
2
To win the lottery it is necessary to buy a ticket. But it is not sufficient; many people buy tickets and do not win. To become ill, it is sufficient to drink bleach, but it is not necessary; many people become ill without drinking bleach.
– MJD
Aug 2 at 19:22
add a comment |Â
Countable set is a set with the same cardinality as some subset of the set of natural numbers. Now if a set has the same cardinality as the set of natural numbers, it is called countably infinite, but because of the fact that any set is a subset of itself it means that any countably infinite set is countable.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:11
1
$P$ being "sufficient" for $Q$ means that $Prightarrow Q$. On the other hand, $P$ being "necessary" for $Q$ means that $Qrightarrow P$. In other words, "sufficient" is short for "$P$ is sufficient to imply $Q$", and "necessary" is short for "$P$ necessarily occurs whenever $Q$ does".
– anakhro
Aug 2 at 19:12
About "necessary" and "sufficient". Suppose you know that if some proposition P is true than proposition Q is true. Than, you say that P is sufficient for Q, and Q is necessary for P.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:15
2
To win the lottery it is necessary to buy a ticket. But it is not sufficient; many people buy tickets and do not win. To become ill, it is sufficient to drink bleach, but it is not necessary; many people become ill without drinking bleach.
– MJD
Aug 2 at 19:22
Countable set is a set with the same cardinality as some subset of the set of natural numbers. Now if a set has the same cardinality as the set of natural numbers, it is called countably infinite, but because of the fact that any set is a subset of itself it means that any countably infinite set is countable.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:11
Countable set is a set with the same cardinality as some subset of the set of natural numbers. Now if a set has the same cardinality as the set of natural numbers, it is called countably infinite, but because of the fact that any set is a subset of itself it means that any countably infinite set is countable.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:11
1
1
$P$ being "sufficient" for $Q$ means that $Prightarrow Q$. On the other hand, $P$ being "necessary" for $Q$ means that $Qrightarrow P$. In other words, "sufficient" is short for "$P$ is sufficient to imply $Q$", and "necessary" is short for "$P$ necessarily occurs whenever $Q$ does".
– anakhro
Aug 2 at 19:12
$P$ being "sufficient" for $Q$ means that $Prightarrow Q$. On the other hand, $P$ being "necessary" for $Q$ means that $Qrightarrow P$. In other words, "sufficient" is short for "$P$ is sufficient to imply $Q$", and "necessary" is short for "$P$ necessarily occurs whenever $Q$ does".
– anakhro
Aug 2 at 19:12
About "necessary" and "sufficient". Suppose you know that if some proposition P is true than proposition Q is true. Than, you say that P is sufficient for Q, and Q is necessary for P.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:15
About "necessary" and "sufficient". Suppose you know that if some proposition P is true than proposition Q is true. Than, you say that P is sufficient for Q, and Q is necessary for P.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:15
2
2
To win the lottery it is necessary to buy a ticket. But it is not sufficient; many people buy tickets and do not win. To become ill, it is sufficient to drink bleach, but it is not necessary; many people become ill without drinking bleach.
– MJD
Aug 2 at 19:22
To win the lottery it is necessary to buy a ticket. But it is not sufficient; many people buy tickets and do not win. To become ill, it is sufficient to drink bleach, but it is not necessary; many people become ill without drinking bleach.
– MJD
Aug 2 at 19:22
add a comment |Â
4 Answers
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Suppose we have propositions $P$ and $Q$. We say that $P$ is necessary for $Q$ if you can never have $Q$ without $P$ (so whenever you have $Q$ you have $P$, i.e. $Q Rightarrow P$). Conversely, $P$ is sufficient for $Q$ if whenever you have $P$ you have $Q$, i.e. $Q Leftarrow P$.
So let's suppose that $B$ is countably infinite, and take $Q$ to be the statement "$Atimes B$ is countable.
- Is it true that $A times B$ being countable implies that $A$ is countably infinite? No, $A$ could be a finite set.
- Is it possible for $A times B$ to be countable but $A$ not countable? No (try to prove this!). So it is necessary that $A$ is countable.
3) and 4) proceed similarly.
Re: Countable vs countably infinite. A set is countably infinite if it is in bijection with $mathbbN$, and countable if it's countably infinite or finite.
answered Aug 2 at 19:14
Daniel Mroz
851314
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1
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...I'm also not quite clear on how being countable vs countably infinite changes the problem.
For a set to be countably infinite it has to be infinite, but a countable set can also be finite.
answered Aug 2 at 19:25
Bram28
54.5k33880
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
add a comment |Â
up vote
1
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If $P$ and $Q$ are two propositions, then you have the following equivalences:
$$beginmatrix
textIf P text then Q & & & P implies Q & & & P text is a sufficient condition for Q, \[6pt]
textIf Q text then P & & & P impliedby Q & & & P text is a necessary condition for Q. \[6pt]
endmatrix$$
For example, I know that if A,B are both countably infinite, A×B is countably infinite. But does that mean that the fact that A is countably infinite is "necessary" or just "sufficient"?
That means that $A$ and $B$ being countably infinite is sufficient for $A times B$ to be countably infinite. In this case, necessity would mean that if $A times B$ is countably infinite then $A$ and $B$ are countably infinite, which is false.
answered Aug 3 at 6:52
Ben
78911
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1
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I will call infinite countable sets simply countable. Sets with finite number of elements are finite.
It is necessary and sufficient that the set $A$ is non-empty and at most countable.
If the set $A$ is empty, then $Atimes B=emptyset$ obviously cannot be countable, therefore it is necessary that $Aneqemptyset$.
It suffices for $Aneqemptyset$ to be finite. It's also known that the direct product of countable sets is countable. Option 3 is definitely correct. With option 4, it must be specified that $Aneqemptyset$.
It is not necessary for $A$ to be countable (=countably infinite). If it were necessary then the following implication would be true:
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies |A| = |mathbb N|.quad text(why is this false?) $$
Neither is it necessary for $A$ to be finite (because it can be countable).
In case of 2, if it is specified that $Aneqemptyset$ then it is necessary that $A$ is at most countable i.e
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies exists ninmathbb Nsetminus0 :|A|= nquadmbox orquad |A| = |mathbb N|. $$
answered Aug 2 at 19:17
Alvin Lepik
2,035819
Could you please elaborate more?
– peterh
Aug 2 at 22:07
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Suppose we have propositions $P$ and $Q$. We say that $P$ is necessary for $Q$ if you can never have $Q$ without $P$ (so whenever you have $Q$ you have $P$, i.e. $Q Rightarrow P$). Conversely, $P$ is sufficient for $Q$ if whenever you have $P$ you have $Q$, i.e. $Q Leftarrow P$.
So let's suppose that $B$ is countably infinite, and take $Q$ to be the statement "$Atimes B$ is countable.
- Is it true that $A times B$ being countable implies that $A$ is countably infinite? No, $A$ could be a finite set.
- Is it possible for $A times B$ to be countable but $A$ not countable? No (try to prove this!). So it is necessary that $A$ is countable.
3) and 4) proceed similarly.
Re: Countable vs countably infinite. A set is countably infinite if it is in bijection with $mathbbN$, and countable if it's countably infinite or finite.
answered Aug 2 at 19:14
Daniel Mroz
851314
add a comment |Â
up vote
1
down vote
Suppose we have propositions $P$ and $Q$. We say that $P$ is necessary for $Q$ if you can never have $Q$ without $P$ (so whenever you have $Q$ you have $P$, i.e. $Q Rightarrow P$). Conversely, $P$ is sufficient for $Q$ if whenever you have $P$ you have $Q$, i.e. $Q Leftarrow P$.
So let's suppose that $B$ is countably infinite, and take $Q$ to be the statement "$Atimes B$ is countable.
- Is it true that $A times B$ being countable implies that $A$ is countably infinite? No, $A$ could be a finite set.
- Is it possible for $A times B$ to be countable but $A$ not countable? No (try to prove this!). So it is necessary that $A$ is countable.
3) and 4) proceed similarly.
Re: Countable vs countably infinite. A set is countably infinite if it is in bijection with $mathbbN$, and countable if it's countably infinite or finite.
answered Aug 2 at 19:14
Daniel Mroz
851314
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose we have propositions $P$ and $Q$. We say that $P$ is necessary for $Q$ if you can never have $Q$ without $P$ (so whenever you have $Q$ you have $P$, i.e. $Q Rightarrow P$). Conversely, $P$ is sufficient for $Q$ if whenever you have $P$ you have $Q$, i.e. $Q Leftarrow P$.
So let's suppose that $B$ is countably infinite, and take $Q$ to be the statement "$Atimes B$ is countable.
- Is it true that $A times B$ being countable implies that $A$ is countably infinite? No, $A$ could be a finite set.
- Is it possible for $A times B$ to be countable but $A$ not countable? No (try to prove this!). So it is necessary that $A$ is countable.
3) and 4) proceed similarly.
Re: Countable vs countably infinite. A set is countably infinite if it is in bijection with $mathbbN$, and countable if it's countably infinite or finite.
answered Aug 2 at 19:14
Daniel Mroz
851314
Suppose we have propositions $P$ and $Q$. We say that $P$ is necessary for $Q$ if you can never have $Q$ without $P$ (so whenever you have $Q$ you have $P$, i.e. $Q Rightarrow P$). Conversely, $P$ is sufficient for $Q$ if whenever you have $P$ you have $Q$, i.e. $Q Leftarrow P$.
So let's suppose that $B$ is countably infinite, and take $Q$ to be the statement "$Atimes B$ is countable.
- Is it true that $A times B$ being countable implies that $A$ is countably infinite? No, $A$ could be a finite set.
- Is it possible for $A times B$ to be countable but $A$ not countable? No (try to prove this!). So it is necessary that $A$ is countable.
3) and 4) proceed similarly.
Re: Countable vs countably infinite. A set is countably infinite if it is in bijection with $mathbbN$, and countable if it's countably infinite or finite.
answered Aug 2 at 19:14
Daniel Mroz
851314
answered Aug 2 at 19:14
Daniel Mroz
851314
answered Aug 2 at 19:14
Daniel Mroz
851314
answered Aug 2 at 19:14
Daniel Mroz
851314
851314
add a comment |Â
add a comment |Â
up vote
1
down vote
...I'm also not quite clear on how being countable vs countably infinite changes the problem.
For a set to be countably infinite it has to be infinite, but a countable set can also be finite.
answered Aug 2 at 19:25
Bram28
54.5k33880
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
add a comment |Â
up vote
1
down vote
...I'm also not quite clear on how being countable vs countably infinite changes the problem.
For a set to be countably infinite it has to be infinite, but a countable set can also be finite.
answered Aug 2 at 19:25
Bram28
54.5k33880
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
...I'm also not quite clear on how being countable vs countably infinite changes the problem.
For a set to be countably infinite it has to be infinite, but a countable set can also be finite.
answered Aug 2 at 19:25
Bram28
54.5k33880
...I'm also not quite clear on how being countable vs countably infinite changes the problem.
For a set to be countably infinite it has to be infinite, but a countable set can also be finite.
answered Aug 2 at 19:25
Bram28
54.5k33880
answered Aug 2 at 19:25
Bram28
54.5k33880
answered Aug 2 at 19:25
Bram28
54.5k33880
answered Aug 2 at 19:25
Bram28
54.5k33880
54.5k33880
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
add a comment |Â
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
Is that universal? I thought that some used countable only for the infinite case. Rather like rings and unity, it is advisable to check what the author means.
– badjohn
Aug 3 at 9:51
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
@badjohn I don't doubt that some authors use countable only for infinite sets, but given that the question in the OP makes a difference between countable and countably infinite, I am pretty sure that this is the difference the quesrion is getting at.
– Bram28
Aug 3 at 12:58
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
Yes it seems so in his case but I thought that a warning may be appropriate (unless I am wrong and no one uses countable only for infinite sets).
– badjohn
Aug 3 at 13:02
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
@badjohn OK, fair enough! Thanks.
– Bram28
Aug 3 at 13:51
add a comment |Â
up vote
1
down vote
If $P$ and $Q$ are two propositions, then you have the following equivalences:
$$beginmatrix
textIf P text then Q & & & P implies Q & & & P text is a sufficient condition for Q, \[6pt]
textIf Q text then P & & & P impliedby Q & & & P text is a necessary condition for Q. \[6pt]
endmatrix$$
For example, I know that if A,B are both countably infinite, A×B is countably infinite. But does that mean that the fact that A is countably infinite is "necessary" or just "sufficient"?
That means that $A$ and $B$ being countably infinite is sufficient for $A times B$ to be countably infinite. In this case, necessity would mean that if $A times B$ is countably infinite then $A$ and $B$ are countably infinite, which is false.
answered Aug 3 at 6:52
Ben
78911
add a comment |Â
up vote
1
down vote
If $P$ and $Q$ are two propositions, then you have the following equivalences:
$$beginmatrix
textIf P text then Q & & & P implies Q & & & P text is a sufficient condition for Q, \[6pt]
textIf Q text then P & & & P impliedby Q & & & P text is a necessary condition for Q. \[6pt]
endmatrix$$
For example, I know that if A,B are both countably infinite, A×B is countably infinite. But does that mean that the fact that A is countably infinite is "necessary" or just "sufficient"?
That means that $A$ and $B$ being countably infinite is sufficient for $A times B$ to be countably infinite. In this case, necessity would mean that if $A times B$ is countably infinite then $A$ and $B$ are countably infinite, which is false.
answered Aug 3 at 6:52
Ben
78911
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $P$ and $Q$ are two propositions, then you have the following equivalences:
$$beginmatrix
textIf P text then Q & & & P implies Q & & & P text is a sufficient condition for Q, \[6pt]
textIf Q text then P & & & P impliedby Q & & & P text is a necessary condition for Q. \[6pt]
endmatrix$$
For example, I know that if A,B are both countably infinite, A×B is countably infinite. But does that mean that the fact that A is countably infinite is "necessary" or just "sufficient"?
That means that $A$ and $B$ being countably infinite is sufficient for $A times B$ to be countably infinite. In this case, necessity would mean that if $A times B$ is countably infinite then $A$ and $B$ are countably infinite, which is false.
answered Aug 3 at 6:52
Ben
78911
If $P$ and $Q$ are two propositions, then you have the following equivalences:
$$beginmatrix
textIf P text then Q & & & P implies Q & & & P text is a sufficient condition for Q, \[6pt]
textIf Q text then P & & & P impliedby Q & & & P text is a necessary condition for Q. \[6pt]
endmatrix$$
For example, I know that if A,B are both countably infinite, A×B is countably infinite. But does that mean that the fact that A is countably infinite is "necessary" or just "sufficient"?
That means that $A$ and $B$ being countably infinite is sufficient for $A times B$ to be countably infinite. In this case, necessity would mean that if $A times B$ is countably infinite then $A$ and $B$ are countably infinite, which is false.
answered Aug 3 at 6:52
Ben
78911
answered Aug 3 at 6:52
Ben
78911
answered Aug 3 at 6:52
Ben
78911
answered Aug 3 at 6:52
Ben
78911
78911
add a comment |Â
add a comment |Â
up vote
1
down vote
I will call infinite countable sets simply countable. Sets with finite number of elements are finite.
It is necessary and sufficient that the set $A$ is non-empty and at most countable.
If the set $A$ is empty, then $Atimes B=emptyset$ obviously cannot be countable, therefore it is necessary that $Aneqemptyset$.
It suffices for $Aneqemptyset$ to be finite. It's also known that the direct product of countable sets is countable. Option 3 is definitely correct. With option 4, it must be specified that $Aneqemptyset$.
It is not necessary for $A$ to be countable (=countably infinite). If it were necessary then the following implication would be true:
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies |A| = |mathbb N|.quad text(why is this false?) $$
Neither is it necessary for $A$ to be finite (because it can be countable).
In case of 2, if it is specified that $Aneqemptyset$ then it is necessary that $A$ is at most countable i.e
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies exists ninmathbb Nsetminus0 :|A|= nquadmbox orquad |A| = |mathbb N|. $$
answered Aug 2 at 19:17
Alvin Lepik
2,035819
Could you please elaborate more?
– peterh
Aug 2 at 22:07
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
add a comment |Â
up vote
1
down vote
I will call infinite countable sets simply countable. Sets with finite number of elements are finite.
It is necessary and sufficient that the set $A$ is non-empty and at most countable.
If the set $A$ is empty, then $Atimes B=emptyset$ obviously cannot be countable, therefore it is necessary that $Aneqemptyset$.
It suffices for $Aneqemptyset$ to be finite. It's also known that the direct product of countable sets is countable. Option 3 is definitely correct. With option 4, it must be specified that $Aneqemptyset$.
It is not necessary for $A$ to be countable (=countably infinite). If it were necessary then the following implication would be true:
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies |A| = |mathbb N|.quad text(why is this false?) $$
Neither is it necessary for $A$ to be finite (because it can be countable).
In case of 2, if it is specified that $Aneqemptyset$ then it is necessary that $A$ is at most countable i.e
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies exists ninmathbb Nsetminus0 :|A|= nquadmbox orquad |A| = |mathbb N|. $$
answered Aug 2 at 19:17
Alvin Lepik
2,035819
Could you please elaborate more?
– peterh
Aug 2 at 22:07
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I will call infinite countable sets simply countable. Sets with finite number of elements are finite.
It is necessary and sufficient that the set $A$ is non-empty and at most countable.
If the set $A$ is empty, then $Atimes B=emptyset$ obviously cannot be countable, therefore it is necessary that $Aneqemptyset$.
It suffices for $Aneqemptyset$ to be finite. It's also known that the direct product of countable sets is countable. Option 3 is definitely correct. With option 4, it must be specified that $Aneqemptyset$.
It is not necessary for $A$ to be countable (=countably infinite). If it were necessary then the following implication would be true:
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies |A| = |mathbb N|.quad text(why is this false?) $$
Neither is it necessary for $A$ to be finite (because it can be countable).
In case of 2, if it is specified that $Aneqemptyset$ then it is necessary that $A$ is at most countable i.e
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies exists ninmathbb Nsetminus0 :|A|= nquadmbox orquad |A| = |mathbb N|. $$
answered Aug 2 at 19:17
Alvin Lepik
2,035819
I will call infinite countable sets simply countable. Sets with finite number of elements are finite.
It is necessary and sufficient that the set $A$ is non-empty and at most countable.
If the set $A$ is empty, then $Atimes B=emptyset$ obviously cannot be countable, therefore it is necessary that $Aneqemptyset$.
It suffices for $Aneqemptyset$ to be finite. It's also known that the direct product of countable sets is countable. Option 3 is definitely correct. With option 4, it must be specified that $Aneqemptyset$.
It is not necessary for $A$ to be countable (=countably infinite). If it were necessary then the following implication would be true:
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies |A| = |mathbb N|.quad text(why is this false?) $$
Neither is it necessary for $A$ to be finite (because it can be countable).
In case of 2, if it is specified that $Aneqemptyset$ then it is necessary that $A$ is at most countable i.e
$$ |Atimes B| = |mathbb N| quadmboxandquad |B| = |mathbb N|implies exists ninmathbb Nsetminus0 :|A|= nquadmbox orquad |A| = |mathbb N|. $$
answered Aug 2 at 19:17
Alvin Lepik
2,035819
edited Aug 3 at 7:19
answered Aug 2 at 19:17
Alvin Lepik
2,035819
answered Aug 2 at 19:17
Alvin Lepik
2,035819
answered Aug 2 at 19:17
Alvin Lepik
2,035819
2,035819
Could you please elaborate more?
– peterh
Aug 2 at 22:07
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
add a comment |Â
Could you please elaborate more?
– peterh
Aug 2 at 22:07
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
Could you please elaborate more?
– peterh
Aug 2 at 22:07
Could you please elaborate more?
– peterh
Aug 2 at 22:07
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
@peterh I added some details, but I am used to different terminology, please pay attention to what I say at the very start
– Alvin Lepik
Aug 3 at 6:38
add a comment |Â
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Countable set is a set with the same cardinality as some subset of the set of natural numbers. Now if a set has the same cardinality as the set of natural numbers, it is called countably infinite, but because of the fact that any set is a subset of itself it means that any countably infinite set is countable.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:11
1
$P$ being "sufficient" for $Q$ means that $Prightarrow Q$. On the other hand, $P$ being "necessary" for $Q$ means that $Qrightarrow P$. In other words, "sufficient" is short for "$P$ is sufficient to imply $Q$", and "necessary" is short for "$P$ necessarily occurs whenever $Q$ does".
– anakhro
Aug 2 at 19:12
About "necessary" and "sufficient". Suppose you know that if some proposition P is true than proposition Q is true. Than, you say that P is sufficient for Q, and Q is necessary for P.
– Ã®Ñ€Ñ–ù ïрþш
Aug 2 at 19:15
2
To win the lottery it is necessary to buy a ticket. But it is not sufficient; many people buy tickets and do not win. To become ill, it is sufficient to drink bleach, but it is not necessary; many people become ill without drinking bleach.
– MJD
Aug 2 at 19:22