If A is dense in Q, then it must be dense in R.Defining dense subsets of $mathbbR$Why aren't all dense subsets of $mathbbR$ uncountable?Show that $mathbbQ$ is dense in the real numbers. (Using Supremum)Characterization of dense open subsets of the real numbersProof of the infinitude of rational and irrational numbersMust a comeager set be dense?countable dense subset of R^kGiven that the rationals are countable and the denseness of $mathbbQ$ in $mathbbR$ how can $mathbbR$ be uncountable?Proof technique for between any two real numbers is an irrational numberprove that $mathbbQ^n$is dense subset of $mathbbR^n$

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If A is dense in Q, then it must be dense in R.


Defining dense subsets of $mathbbR$Why aren't all dense subsets of $mathbbR$ uncountable?Show that $mathbbQ$ is dense in the real numbers. (Using Supremum)Characterization of dense open subsets of the real numbersProof of the infinitude of rational and irrational numbersMust a comeager set be dense?countable dense subset of R^kGiven that the rationals are countable and the denseness of $mathbbQ$ in $mathbbR$ how can $mathbbR$ be uncountable?Proof technique for between any two real numbers is an irrational numberprove that $mathbbQ^n$is dense subset of $mathbbR^n$













5












$begingroup$


I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.










share|cite|improve this question









New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
    $endgroup$
    – user334732
    5 hours ago










  • $begingroup$
    And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
    $endgroup$
    – user334732
    4 hours ago
















5












$begingroup$


I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.










share|cite|improve this question









New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
    $endgroup$
    – user334732
    5 hours ago










  • $begingroup$
    And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
    $endgroup$
    – user334732
    4 hours ago














5












5








5


1



$begingroup$


I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.










share|cite|improve this question









New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.







real-analysis real-numbers






share|cite|improve this question









New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Floris Claassens

1,04016




1,04016






New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









Priti DPriti D

262




262




New contributor




Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
    $endgroup$
    – user334732
    5 hours ago










  • $begingroup$
    And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
    $endgroup$
    – user334732
    4 hours ago

















  • $begingroup$
    Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
    $endgroup$
    – user334732
    5 hours ago










  • $begingroup$
    And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
    $endgroup$
    – user334732
    4 hours ago
















$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
5 hours ago




$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
5 hours ago












$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago





$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago











2 Answers
2






active

oldest

votes


















4












$begingroup$

Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
    $endgroup$
    – Mason
    5 hours ago











  • $begingroup$
    This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
    $endgroup$
    – Dbchatto67
    4 hours ago



















4












$begingroup$

$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.



We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    @CiaPan. I think you are right. This was written hastily.
    $endgroup$
    – Mason
    4 hours ago










  • $begingroup$
    :) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
    $endgroup$
    – Mason
    4 hours ago











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
    $endgroup$
    – Mason
    5 hours ago











  • $begingroup$
    This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
    $endgroup$
    – Dbchatto67
    4 hours ago
















4












$begingroup$

Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
    $endgroup$
    – Mason
    5 hours ago











  • $begingroup$
    This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
    $endgroup$
    – Dbchatto67
    4 hours ago














4












4








4





$begingroup$

Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$






share|cite|improve this answer









$endgroup$



Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









Dbchatto67Dbchatto67

1,975319




1,975319







  • 1




    $begingroup$
    Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
    $endgroup$
    – Mason
    5 hours ago











  • $begingroup$
    This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
    $endgroup$
    – Dbchatto67
    4 hours ago













  • 1




    $begingroup$
    Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
    $endgroup$
    – Mason
    5 hours ago











  • $begingroup$
    This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
    $endgroup$
    – Dbchatto67
    4 hours ago








1




1




$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
5 hours ago





$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
5 hours ago













$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago





$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago












4












$begingroup$

$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.



We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    @CiaPan. I think you are right. This was written hastily.
    $endgroup$
    – Mason
    4 hours ago










  • $begingroup$
    :) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
    $endgroup$
    – Mason
    4 hours ago
















4












$begingroup$

$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.



We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    @CiaPan. I think you are right. This was written hastily.
    $endgroup$
    – Mason
    4 hours ago










  • $begingroup$
    :) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
    $endgroup$
    – Mason
    4 hours ago














4












4








4





$begingroup$

$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.



We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$






share|cite|improve this answer











$endgroup$



$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.



We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 4 hours ago

























answered 5 hours ago









MasonMason

1,7651630




1,7651630







  • 1




    $begingroup$
    Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    @CiaPan. I think you are right. This was written hastily.
    $endgroup$
    – Mason
    4 hours ago










  • $begingroup$
    :) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
    $endgroup$
    – Mason
    4 hours ago













  • 1




    $begingroup$
    Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    @CiaPan. I think you are right. This was written hastily.
    $endgroup$
    – Mason
    4 hours ago










  • $begingroup$
    :) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
    $endgroup$
    – CiaPan
    4 hours ago










  • $begingroup$
    I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
    $endgroup$
    – Mason
    4 hours ago








1




1




$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago




$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago












$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago




$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago












$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago




$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago












$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago





$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago











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ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result