Distinct Uncountable Dense Subsets of R
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I'm working on Question #5.v in chapter 3.1 of Sidney Morris's Toplogy without Tears.
I'm supposed to prove that $mathbb{R}$ has $2^{mathfrak{c}}$ distinct uncountable dense subsets.
I think my method gets at least halfway there.
Here's what I've done:
Note that $|mathbb{I}| = |mathbb{I}^{+}| = mathfrak{c}$
$forallalphain mathbb{I}^{+}, $ let the set $B_alpha = {mathbb{R}^{-}cupmathbb{Q}^{+}cup{alpha}}$
Then, each $B_alpha$ is distinct, uncountable, and dense in $mathbb{R}$, and there are $mathfrak{c}$ of them indexed to the positive irrationals.
Now, consider that $|P(mathbb{I}^{+})| = 2^{mathfrak{c}}$, where each element of the power set is a distinct set of positive irrationals.
Since each element of the power set can be mapped 1-1 to some union of $B_alpha$'s, we have $2^{mathfrak{c}}$ distinct uncountable dense subsets of $mathbb{R}$.
If this is correct, I have two further questions:
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
general-topology elementary-set-theory cardinals
asked Nov 12 at 16:54
Cassius12
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up vote
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I'm working on Question #5.v in chapter 3.1 of Sidney Morris's Toplogy without Tears.
I'm supposed to prove that $mathbb{R}$ has $2^{mathfrak{c}}$ distinct uncountable dense subsets.
I think my method gets at least halfway there.
Here's what I've done:
Note that $|mathbb{I}| = |mathbb{I}^{+}| = mathfrak{c}$
$forallalphain mathbb{I}^{+}, $ let the set $B_alpha = {mathbb{R}^{-}cupmathbb{Q}^{+}cup{alpha}}$
Then, each $B_alpha$ is distinct, uncountable, and dense in $mathbb{R}$, and there are $mathfrak{c}$ of them indexed to the positive irrationals.
Now, consider that $|P(mathbb{I}^{+})| = 2^{mathfrak{c}}$, where each element of the power set is a distinct set of positive irrationals.
Since each element of the power set can be mapped 1-1 to some union of $B_alpha$'s, we have $2^{mathfrak{c}}$ distinct uncountable dense subsets of $mathbb{R}$.
If this is correct, I have two further questions:
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
general-topology elementary-set-theory cardinals
asked Nov 12 at 16:54
Cassius12
255
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$mathbb{R}$ has only $2^mathfrak{c}$ subsets in total, because the cardinality of $mathbb{R}$ is $mathfrak{c}$ by definition.
– Connor Harris
Nov 12 at 17:14
This resembles math.stackexchange.com/q/2991576.
– Paul Frost
Nov 12 at 17:26
Oh, right. Thanks, Connor. I was overthinking it.
– Cassius12
Nov 12 at 18:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm working on Question #5.v in chapter 3.1 of Sidney Morris's Toplogy without Tears.
I'm supposed to prove that $mathbb{R}$ has $2^{mathfrak{c}}$ distinct uncountable dense subsets.
I think my method gets at least halfway there.
Here's what I've done:
Note that $|mathbb{I}| = |mathbb{I}^{+}| = mathfrak{c}$
$forallalphain mathbb{I}^{+}, $ let the set $B_alpha = {mathbb{R}^{-}cupmathbb{Q}^{+}cup{alpha}}$
Then, each $B_alpha$ is distinct, uncountable, and dense in $mathbb{R}$, and there are $mathfrak{c}$ of them indexed to the positive irrationals.
Now, consider that $|P(mathbb{I}^{+})| = 2^{mathfrak{c}}$, where each element of the power set is a distinct set of positive irrationals.
Since each element of the power set can be mapped 1-1 to some union of $B_alpha$'s, we have $2^{mathfrak{c}}$ distinct uncountable dense subsets of $mathbb{R}$.
If this is correct, I have two further questions:
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
general-topology elementary-set-theory cardinals
asked Nov 12 at 16:54
Cassius12
255
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm working on Question #5.v in chapter 3.1 of Sidney Morris's Toplogy without Tears.
I'm supposed to prove that $mathbb{R}$ has $2^{mathfrak{c}}$ distinct uncountable dense subsets.
I think my method gets at least halfway there.
Here's what I've done:
Note that $|mathbb{I}| = |mathbb{I}^{+}| = mathfrak{c}$
$forallalphain mathbb{I}^{+}, $ let the set $B_alpha = {mathbb{R}^{-}cupmathbb{Q}^{+}cup{alpha}}$
Then, each $B_alpha$ is distinct, uncountable, and dense in $mathbb{R}$, and there are $mathfrak{c}$ of them indexed to the positive irrationals.
Now, consider that $|P(mathbb{I}^{+})| = 2^{mathfrak{c}}$, where each element of the power set is a distinct set of positive irrationals.
Since each element of the power set can be mapped 1-1 to some union of $B_alpha$'s, we have $2^{mathfrak{c}}$ distinct uncountable dense subsets of $mathbb{R}$.
If this is correct, I have two further questions:
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
general-topology elementary-set-theory cardinals
general-topology elementary-set-theory cardinals
asked Nov 12 at 16:54
Cassius12
255
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:54
Cassius12
255
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:54
Cassius12
255
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:54
Cassius12
255
asked Nov 12 at 16:54
Cassius12
255
255
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Cassius12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$mathbb{R}$ has only $2^mathfrak{c}$ subsets in total, because the cardinality of $mathbb{R}$ is $mathfrak{c}$ by definition.
– Connor Harris
Nov 12 at 17:14
This resembles math.stackexchange.com/q/2991576.
– Paul Frost
Nov 12 at 17:26
Oh, right. Thanks, Connor. I was overthinking it.
– Cassius12
Nov 12 at 18:01
add a comment |
1
$mathbb{R}$ has only $2^mathfrak{c}$ subsets in total, because the cardinality of $mathbb{R}$ is $mathfrak{c}$ by definition.
– Connor Harris
Nov 12 at 17:14
This resembles math.stackexchange.com/q/2991576.
– Paul Frost
Nov 12 at 17:26
Oh, right. Thanks, Connor. I was overthinking it.
– Cassius12
Nov 12 at 18:01
1
1
$mathbb{R}$ has only $2^mathfrak{c}$ subsets in total, because the cardinality of $mathbb{R}$ is $mathfrak{c}$ by definition.
– Connor Harris
Nov 12 at 17:14
$mathbb{R}$ has only $2^mathfrak{c}$ subsets in total, because the cardinality of $mathbb{R}$ is $mathfrak{c}$ by definition.
– Connor Harris
Nov 12 at 17:14
This resembles math.stackexchange.com/q/2991576.
– Paul Frost
Nov 12 at 17:26
This resembles math.stackexchange.com/q/2991576.
– Paul Frost
Nov 12 at 17:26
Oh, right. Thanks, Connor. I was overthinking it.
– Cassius12
Nov 12 at 18:01
Oh, right. Thanks, Connor. I was overthinking it.
– Cassius12
Nov 12 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
$mathbb{R}$ has $2^{mathfrak{c}}$ subsets. So yeah, there can't be more in the sense of cardinality.
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
By "required type" you mean "dense"? If not then obviously that totally depends on what "required type" means and it is a case-by-case study.
If yes, then your construction can be simplified: for any subset $Asubseteqmathbb{R}backslashmathbb{Q}$ the subset $mathbb{Q}cup A$ is dense. And $mathbb{Q}cup A=mathbb{Q}cup B$ if and only if $A=B$ (under the assumption that both $A,Bsubseteqmathbb{R}backslashmathbb{Q}$). Since $mathbb{R}backslashmathbb{Q}$ is of cardinality $mathfrak{c}$ then you're done.
answered Nov 12 at 17:42
freakish
10.1k1526
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
$mathbb{R}$ has $2^{mathfrak{c}}$ subsets. So yeah, there can't be more in the sense of cardinality.
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
By "required type" you mean "dense"? If not then obviously that totally depends on what "required type" means and it is a case-by-case study.
If yes, then your construction can be simplified: for any subset $Asubseteqmathbb{R}backslashmathbb{Q}$ the subset $mathbb{Q}cup A$ is dense. And $mathbb{Q}cup A=mathbb{Q}cup B$ if and only if $A=B$ (under the assumption that both $A,Bsubseteqmathbb{R}backslashmathbb{Q}$). Since $mathbb{R}backslashmathbb{Q}$ is of cardinality $mathfrak{c}$ then you're done.
answered Nov 12 at 17:42
freakish
10.1k1526
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
add a comment |
up vote
3
down vote
accepted
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
$mathbb{R}$ has $2^{mathfrak{c}}$ subsets. So yeah, there can't be more in the sense of cardinality.
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
By "required type" you mean "dense"? If not then obviously that totally depends on what "required type" means and it is a case-by-case study.
If yes, then your construction can be simplified: for any subset $Asubseteqmathbb{R}backslashmathbb{Q}$ the subset $mathbb{Q}cup A$ is dense. And $mathbb{Q}cup A=mathbb{Q}cup B$ if and only if $A=B$ (under the assumption that both $A,Bsubseteqmathbb{R}backslashmathbb{Q}$). Since $mathbb{R}backslashmathbb{Q}$ is of cardinality $mathfrak{c}$ then you're done.
answered Nov 12 at 17:42
freakish
10.1k1526
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
$mathbb{R}$ has $2^{mathfrak{c}}$ subsets. So yeah, there can't be more in the sense of cardinality.
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
By "required type" you mean "dense"? If not then obviously that totally depends on what "required type" means and it is a case-by-case study.
If yes, then your construction can be simplified: for any subset $Asubseteqmathbb{R}backslashmathbb{Q}$ the subset $mathbb{Q}cup A$ is dense. And $mathbb{Q}cup A=mathbb{Q}cup B$ if and only if $A=B$ (under the assumption that both $A,Bsubseteqmathbb{R}backslashmathbb{Q}$). Since $mathbb{R}backslashmathbb{Q}$ is of cardinality $mathfrak{c}$ then you're done.
answered Nov 12 at 17:42
freakish
10.1k1526
1) My construction identifies $2^{mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{mathfrak{c}}$ or more?
$mathbb{R}$ has $2^{mathfrak{c}}$ subsets. So yeah, there can't be more in the sense of cardinality.
2) Is there a cleaner/more elegant way to construct $2^{mathfrak{c}}$ subsets of the required type?
By "required type" you mean "dense"? If not then obviously that totally depends on what "required type" means and it is a case-by-case study.
If yes, then your construction can be simplified: for any subset $Asubseteqmathbb{R}backslashmathbb{Q}$ the subset $mathbb{Q}cup A$ is dense. And $mathbb{Q}cup A=mathbb{Q}cup B$ if and only if $A=B$ (under the assumption that both $A,Bsubseteqmathbb{R}backslashmathbb{Q}$). Since $mathbb{R}backslashmathbb{Q}$ is of cardinality $mathfrak{c}$ then you're done.
answered Nov 12 at 17:42
freakish
10.1k1526
answered Nov 12 at 17:42
freakish
10.1k1526
answered Nov 12 at 17:42
freakish
10.1k1526
answered Nov 12 at 17:42
freakish
10.1k1526
10.1k1526
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
add a comment |
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
Sorry for being unclear. "Required" for this problem meant "distinct, countable, dense". Your answer makes sense to me--Thanks! I think I was trying too hard to explicitly construct an example, when I should have been thinking more about the characteristics of general subsets.
– Cassius12
Nov 12 at 18:06
add a comment |
Cassius12 is a new contributor. Be nice, and check out our Code of Conduct.
Cassius12 is a new contributor. Be nice, and check out our Code of Conduct.
Cassius12 is a new contributor. Be nice, and check out our Code of Conduct.
Cassius12 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$mathbb{R}$ has only $2^mathfrak{c}$ subsets in total, because the cardinality of $mathbb{R}$ is $mathfrak{c}$ by definition.
– Connor Harris
Nov 12 at 17:14
This resembles math.stackexchange.com/q/2991576.
– Paul Frost
Nov 12 at 17:26
Oh, right. Thanks, Connor. I was overthinking it.
– Cassius12
Nov 12 at 18:01