$operatorname{card}(bigcuplimits_{n in mathbb{N}} underbrace{Atimes…times A}_{n})=k$ if...
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I was reading a proof of a theorem that goes like this:
Let $A$ be an infinite set of cardinality $k$ and $A^{<omega}$ the set of finite sequences of elements of A. Then $operatorname{card}(A^{<omega})=k$.
In the proof, we note that $A^{<omega}=bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}$. So at some point, we have to show $operatorname{card}(A times A)=k$, and for this the author uses the fact that $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$.
I am not super sure why this is the case.
My thoughts: We have by definition a bijection $A leftrightarrow operatorname{card}(A)$. So we have a bijection $A times A leftrightarrow operatorname{card}(A) times operatorname{card}(A)$. So in particular since these sets are bijective they have the same cardinal $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$. Is that it?
So from this we can prove $operatorname{card}(underbrace{Atimes...times A}_{n})=k$ $forall n$.
Then the author says $kle operatorname{card}(bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}) le operatorname{card}(A)cdotomega$ where the $cdot$ is the cardinal product. Why is that?
cardinals ordinals
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
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add a comment |
$begingroup$
I was reading a proof of a theorem that goes like this:
Let $A$ be an infinite set of cardinality $k$ and $A^{<omega}$ the set of finite sequences of elements of A. Then $operatorname{card}(A^{<omega})=k$.
In the proof, we note that $A^{<omega}=bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}$. So at some point, we have to show $operatorname{card}(A times A)=k$, and for this the author uses the fact that $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$.
I am not super sure why this is the case.
My thoughts: We have by definition a bijection $A leftrightarrow operatorname{card}(A)$. So we have a bijection $A times A leftrightarrow operatorname{card}(A) times operatorname{card}(A)$. So in particular since these sets are bijective they have the same cardinal $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$. Is that it?
So from this we can prove $operatorname{card}(underbrace{Atimes...times A}_{n})=k$ $forall n$.
Then the author says $kle operatorname{card}(bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}) le operatorname{card}(A)cdotomega$ where the $cdot$ is the cardinal product. Why is that?
cardinals ordinals
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
$endgroup$
add a comment |
$begingroup$
I was reading a proof of a theorem that goes like this:
Let $A$ be an infinite set of cardinality $k$ and $A^{<omega}$ the set of finite sequences of elements of A. Then $operatorname{card}(A^{<omega})=k$.
In the proof, we note that $A^{<omega}=bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}$. So at some point, we have to show $operatorname{card}(A times A)=k$, and for this the author uses the fact that $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$.
I am not super sure why this is the case.
My thoughts: We have by definition a bijection $A leftrightarrow operatorname{card}(A)$. So we have a bijection $A times A leftrightarrow operatorname{card}(A) times operatorname{card}(A)$. So in particular since these sets are bijective they have the same cardinal $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$. Is that it?
So from this we can prove $operatorname{card}(underbrace{Atimes...times A}_{n})=k$ $forall n$.
Then the author says $kle operatorname{card}(bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}) le operatorname{card}(A)cdotomega$ where the $cdot$ is the cardinal product. Why is that?
cardinals ordinals
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
$endgroup$
I was reading a proof of a theorem that goes like this:
Let $A$ be an infinite set of cardinality $k$ and $A^{<omega}$ the set of finite sequences of elements of A. Then $operatorname{card}(A^{<omega})=k$.
In the proof, we note that $A^{<omega}=bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}$. So at some point, we have to show $operatorname{card}(A times A)=k$, and for this the author uses the fact that $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$.
I am not super sure why this is the case.
My thoughts: We have by definition a bijection $A leftrightarrow operatorname{card}(A)$. So we have a bijection $A times A leftrightarrow operatorname{card}(A) times operatorname{card}(A)$. So in particular since these sets are bijective they have the same cardinal $operatorname{card}(A times A)=operatorname{card}(operatorname{card}(A) times operatorname{card}(A))$. Is that it?
So from this we can prove $operatorname{card}(underbrace{Atimes...times A}_{n})=k$ $forall n$.
Then the author says $kle operatorname{card}(bigcuplimits_{n in mathbb{N}} underbrace{Atimes...times A}_{n}) le operatorname{card}(A)cdotomega$ where the $cdot$ is the cardinal product. Why is that?
cardinals ordinals
cardinals ordinals
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
edited Jan 23 at 21:36
J. W. Tanner
2,8111217
2,8111217
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
asked Jan 23 at 20:28
roi_saumonroi_saumon
57738
57738
add a comment |
add a comment |
1 Answer
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So far, so good. Next you need $k^2=k$ to prove by induction that each of the products has cardinality $k$. Then their union has at most size $kaleph_0le k^2=k$. But the proof there are at least $k$ sequences is trivial.
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
$endgroup$
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
1
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
add a comment |
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1 Answer
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So far, so good. Next you need $k^2=k$ to prove by induction that each of the products has cardinality $k$. Then their union has at most size $kaleph_0le k^2=k$. But the proof there are at least $k$ sequences is trivial.
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
$endgroup$
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
1
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
add a comment |
$begingroup$
So far, so good. Next you need $k^2=k$ to prove by induction that each of the products has cardinality $k$. Then their union has at most size $kaleph_0le k^2=k$. But the proof there are at least $k$ sequences is trivial.
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
$endgroup$
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
1
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
add a comment |
$begingroup$
So far, so good. Next you need $k^2=k$ to prove by induction that each of the products has cardinality $k$. Then their union has at most size $kaleph_0le k^2=k$. But the proof there are at least $k$ sequences is trivial.
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
$endgroup$
So far, so good. Next you need $k^2=k$ to prove by induction that each of the products has cardinality $k$. Then their union has at most size $kaleph_0le k^2=k$. But the proof there are at least $k$ sequences is trivial.
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
answered Jan 23 at 22:17
J.G.J.G.
28.7k22845
28.7k22845
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
1
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
add a comment |
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
1
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
$begingroup$
In general, what bound do we have : $operatorname{card}(bigcuplimits_{i in I} B_i) le ?$
$endgroup$
– roi_saumon
Jan 23 at 23:56
1
1
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
$begingroup$
@rol_saumon We have lower bound $max_ioperatorname{card}(B_i)$, upper bound $sum_ioperatorname{card}(B_i,)$.
$endgroup$
– J.G.
Jan 24 at 6:02
add a comment |
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