Show that a disjoint family $(I_lambda)_{lambda in Lambda}$ of not degenerate intervals of $Bbb{R}$ is at...
$begingroup$
I found this problem in a book:
Let $(I_lambda)_{lambda in Lambda}$ be a not degenerate family of disjoint sets of $Bbb{R}$. Prove that $card((I_lambda)) leq aleph_0$ (That is, the family is at most denumerable
I have proved it by taking a rational number $q_lambda$ such that $forall lambda in Lambda: q_lambda in I_lambda cap Bbb{Q}$. Such $q_lambda$ exists because $Bbb{Q}$ is dense in every non degenerate interval of $Bbb{R}$. Then I've built the function $f:(I_lambda)_{lambda in Lambda} rightarrow Bbb{Q}$ such that $I_lambda longmapsto q_lambda$. The function is obviously injective because $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} = emptyset$ by hypothesis, hence showing that $card((I_lambda)_{lambda in Lambda}) leq card(Bbb{Q}) = aleph_0$
But now I'm wondering whether the hypothesis are too restrictive: I can replace the disjoint part of the theorem with the following $$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} text{is non degenerate interval of $Bbb{R}$}$$
EDIT2: neither of the weaker hypothesis below works. See accepted answer and comments
EDIT: As pointed out by JDMan4444 and Hagen von Eitzen this weaker condition does not hold but then I produced another condition weaker than the one given in the theorem but stronger than the one I've given:
$$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2} ; text{and} ; I_{lambda_1} cap I_{lambda_2} text{is not degenerate interval of $Bbb{R}$}$$
And the proof would be similar, but the $q_lambda$ is taken from the interval that is unique to each set.
Actually, while I'm writing this question, I'm thinking that if we have a family of non degenerate intervals of $Bbb{R}$ defined with the weaker hypothesis, we could define another family $$(J_lambda)_{lambda in Lambda} = bigcap_{lambda_1 neq lambda_2 in Lambda} (Bbb{R} smallsetminus (I_{lambda_1} cap I_{lambda_2}))$$
of disjoint intervals of $Bbb{R}$ which is denumerable as proven above... I don't know if this helps nor if it is correct
real-analysis general-topology
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
$endgroup$
add a comment |
$begingroup$
I found this problem in a book:
Let $(I_lambda)_{lambda in Lambda}$ be a not degenerate family of disjoint sets of $Bbb{R}$. Prove that $card((I_lambda)) leq aleph_0$ (That is, the family is at most denumerable
I have proved it by taking a rational number $q_lambda$ such that $forall lambda in Lambda: q_lambda in I_lambda cap Bbb{Q}$. Such $q_lambda$ exists because $Bbb{Q}$ is dense in every non degenerate interval of $Bbb{R}$. Then I've built the function $f:(I_lambda)_{lambda in Lambda} rightarrow Bbb{Q}$ such that $I_lambda longmapsto q_lambda$. The function is obviously injective because $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} = emptyset$ by hypothesis, hence showing that $card((I_lambda)_{lambda in Lambda}) leq card(Bbb{Q}) = aleph_0$
But now I'm wondering whether the hypothesis are too restrictive: I can replace the disjoint part of the theorem with the following $$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} text{is non degenerate interval of $Bbb{R}$}$$
EDIT2: neither of the weaker hypothesis below works. See accepted answer and comments
EDIT: As pointed out by JDMan4444 and Hagen von Eitzen this weaker condition does not hold but then I produced another condition weaker than the one given in the theorem but stronger than the one I've given:
$$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2} ; text{and} ; I_{lambda_1} cap I_{lambda_2} text{is not degenerate interval of $Bbb{R}$}$$
And the proof would be similar, but the $q_lambda$ is taken from the interval that is unique to each set.
Actually, while I'm writing this question, I'm thinking that if we have a family of non degenerate intervals of $Bbb{R}$ defined with the weaker hypothesis, we could define another family $$(J_lambda)_{lambda in Lambda} = bigcap_{lambda_1 neq lambda_2 in Lambda} (Bbb{R} smallsetminus (I_{lambda_1} cap I_{lambda_2}))$$
of disjoint intervals of $Bbb{R}$ which is denumerable as proven above... I don't know if this helps nor if it is correct
real-analysis general-topology
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
$endgroup$
$begingroup$
But: For $lambda>0$ let $I_lambda=(0,lambda)$, Then for $lambda_1nelambda_2$, $I_{lambda_1}cap I_{lambda_2}$ is a non-degenerate interal.
$endgroup$
– Hagen von Eitzen
Nov 28 '18 at 19:36
$begingroup$
"every non degenerate interval of $mathbb R$ is dense in $mathbb Q$". Rather, $mathbb Q$ is dense in every non degenerate interval.
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 19:50
add a comment |
$begingroup$
I found this problem in a book:
Let $(I_lambda)_{lambda in Lambda}$ be a not degenerate family of disjoint sets of $Bbb{R}$. Prove that $card((I_lambda)) leq aleph_0$ (That is, the family is at most denumerable
I have proved it by taking a rational number $q_lambda$ such that $forall lambda in Lambda: q_lambda in I_lambda cap Bbb{Q}$. Such $q_lambda$ exists because $Bbb{Q}$ is dense in every non degenerate interval of $Bbb{R}$. Then I've built the function $f:(I_lambda)_{lambda in Lambda} rightarrow Bbb{Q}$ such that $I_lambda longmapsto q_lambda$. The function is obviously injective because $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} = emptyset$ by hypothesis, hence showing that $card((I_lambda)_{lambda in Lambda}) leq card(Bbb{Q}) = aleph_0$
But now I'm wondering whether the hypothesis are too restrictive: I can replace the disjoint part of the theorem with the following $$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} text{is non degenerate interval of $Bbb{R}$}$$
EDIT2: neither of the weaker hypothesis below works. See accepted answer and comments
EDIT: As pointed out by JDMan4444 and Hagen von Eitzen this weaker condition does not hold but then I produced another condition weaker than the one given in the theorem but stronger than the one I've given:
$$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2} ; text{and} ; I_{lambda_1} cap I_{lambda_2} text{is not degenerate interval of $Bbb{R}$}$$
And the proof would be similar, but the $q_lambda$ is taken from the interval that is unique to each set.
Actually, while I'm writing this question, I'm thinking that if we have a family of non degenerate intervals of $Bbb{R}$ defined with the weaker hypothesis, we could define another family $$(J_lambda)_{lambda in Lambda} = bigcap_{lambda_1 neq lambda_2 in Lambda} (Bbb{R} smallsetminus (I_{lambda_1} cap I_{lambda_2}))$$
of disjoint intervals of $Bbb{R}$ which is denumerable as proven above... I don't know if this helps nor if it is correct
real-analysis general-topology
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
$endgroup$
I found this problem in a book:
Let $(I_lambda)_{lambda in Lambda}$ be a not degenerate family of disjoint sets of $Bbb{R}$. Prove that $card((I_lambda)) leq aleph_0$ (That is, the family is at most denumerable
I have proved it by taking a rational number $q_lambda$ such that $forall lambda in Lambda: q_lambda in I_lambda cap Bbb{Q}$. Such $q_lambda$ exists because $Bbb{Q}$ is dense in every non degenerate interval of $Bbb{R}$. Then I've built the function $f:(I_lambda)_{lambda in Lambda} rightarrow Bbb{Q}$ such that $I_lambda longmapsto q_lambda$. The function is obviously injective because $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} = emptyset$ by hypothesis, hence showing that $card((I_lambda)_{lambda in Lambda}) leq card(Bbb{Q}) = aleph_0$
But now I'm wondering whether the hypothesis are too restrictive: I can replace the disjoint part of the theorem with the following $$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} cap I_{lambda_2} text{is non degenerate interval of $Bbb{R}$}$$
EDIT2: neither of the weaker hypothesis below works. See accepted answer and comments
EDIT: As pointed out by JDMan4444 and Hagen von Eitzen this weaker condition does not hold but then I produced another condition weaker than the one given in the theorem but stronger than the one I've given:
$$forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2} ; text{and} ; I_{lambda_1} cap I_{lambda_2} text{is not degenerate interval of $Bbb{R}$}$$
And the proof would be similar, but the $q_lambda$ is taken from the interval that is unique to each set.
Actually, while I'm writing this question, I'm thinking that if we have a family of non degenerate intervals of $Bbb{R}$ defined with the weaker hypothesis, we could define another family $$(J_lambda)_{lambda in Lambda} = bigcap_{lambda_1 neq lambda_2 in Lambda} (Bbb{R} smallsetminus (I_{lambda_1} cap I_{lambda_2}))$$
of disjoint intervals of $Bbb{R}$ which is denumerable as proven above... I don't know if this helps nor if it is correct
real-analysis general-topology
real-analysis general-topology
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
edited Nov 28 '18 at 19:55
LuxGiammi
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
asked Nov 28 '18 at 19:29
LuxGiammiLuxGiammi
17410
17410
$begingroup$
But: For $lambda>0$ let $I_lambda=(0,lambda)$, Then for $lambda_1nelambda_2$, $I_{lambda_1}cap I_{lambda_2}$ is a non-degenerate interal.
$endgroup$
– Hagen von Eitzen
Nov 28 '18 at 19:36
$begingroup$
"every non degenerate interval of $mathbb R$ is dense in $mathbb Q$". Rather, $mathbb Q$ is dense in every non degenerate interval.
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 19:50
add a comment |
$begingroup$
But: For $lambda>0$ let $I_lambda=(0,lambda)$, Then for $lambda_1nelambda_2$, $I_{lambda_1}cap I_{lambda_2}$ is a non-degenerate interal.
$endgroup$
– Hagen von Eitzen
Nov 28 '18 at 19:36
$begingroup$
"every non degenerate interval of $mathbb R$ is dense in $mathbb Q$". Rather, $mathbb Q$ is dense in every non degenerate interval.
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 19:50
$begingroup$
But: For $lambda>0$ let $I_lambda=(0,lambda)$, Then for $lambda_1nelambda_2$, $I_{lambda_1}cap I_{lambda_2}$ is a non-degenerate interal.
$endgroup$
– Hagen von Eitzen
Nov 28 '18 at 19:36
$begingroup$
But: For $lambda>0$ let $I_lambda=(0,lambda)$, Then for $lambda_1nelambda_2$, $I_{lambda_1}cap I_{lambda_2}$ is a non-degenerate interal.
$endgroup$
– Hagen von Eitzen
Nov 28 '18 at 19:36
$begingroup$
"every non degenerate interval of $mathbb R$ is dense in $mathbb Q$". Rather, $mathbb Q$ is dense in every non degenerate interval.
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 19:50
$begingroup$
"every non degenerate interval of $mathbb R$ is dense in $mathbb Q$". Rather, $mathbb Q$ is dense in every non degenerate interval.
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 19:50
add a comment |
1 Answer
1
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$begingroup$
I don't think you can weaken the hypothesis as you have suggested.
Take $Lambda=mathbb{R}^{+}$, the non-negative reals, and define $I_{lambda}=(-infty,lambda)$ for $lambdainLambda$. Pairwise intersections are non-degenerate, and there are uncountably many of them.
As I was typing this a comment was made with the same basic idea.
As per the comment below and edit above, we can define $I_lambda=(-1/lambda,1+lambda^2)$ for $lambdainmathbb{R}_{>0}$, the positive reals, showing that the issue is not that the intervals are nested.
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
$endgroup$
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
add a comment |
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$begingroup$
I don't think you can weaken the hypothesis as you have suggested.
Take $Lambda=mathbb{R}^{+}$, the non-negative reals, and define $I_{lambda}=(-infty,lambda)$ for $lambdainLambda$. Pairwise intersections are non-degenerate, and there are uncountably many of them.
As I was typing this a comment was made with the same basic idea.
As per the comment below and edit above, we can define $I_lambda=(-1/lambda,1+lambda^2)$ for $lambdainmathbb{R}_{>0}$, the positive reals, showing that the issue is not that the intervals are nested.
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
$endgroup$
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
add a comment |
$begingroup$
I don't think you can weaken the hypothesis as you have suggested.
Take $Lambda=mathbb{R}^{+}$, the non-negative reals, and define $I_{lambda}=(-infty,lambda)$ for $lambdainLambda$. Pairwise intersections are non-degenerate, and there are uncountably many of them.
As I was typing this a comment was made with the same basic idea.
As per the comment below and edit above, we can define $I_lambda=(-1/lambda,1+lambda^2)$ for $lambdainmathbb{R}_{>0}$, the positive reals, showing that the issue is not that the intervals are nested.
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
$endgroup$
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
add a comment |
$begingroup$
I don't think you can weaken the hypothesis as you have suggested.
Take $Lambda=mathbb{R}^{+}$, the non-negative reals, and define $I_{lambda}=(-infty,lambda)$ for $lambdainLambda$. Pairwise intersections are non-degenerate, and there are uncountably many of them.
As I was typing this a comment was made with the same basic idea.
As per the comment below and edit above, we can define $I_lambda=(-1/lambda,1+lambda^2)$ for $lambdainmathbb{R}_{>0}$, the positive reals, showing that the issue is not that the intervals are nested.
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
$endgroup$
I don't think you can weaken the hypothesis as you have suggested.
Take $Lambda=mathbb{R}^{+}$, the non-negative reals, and define $I_{lambda}=(-infty,lambda)$ for $lambdainLambda$. Pairwise intersections are non-degenerate, and there are uncountably many of them.
As I was typing this a comment was made with the same basic idea.
As per the comment below and edit above, we can define $I_lambda=(-1/lambda,1+lambda^2)$ for $lambdainmathbb{R}_{>0}$, the positive reals, showing that the issue is not that the intervals are nested.
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
edited Nov 28 '18 at 19:48
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
answered Nov 28 '18 at 19:38
JDMan4444JDMan4444
22714
22714
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
add a comment |
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
I didn't think of it... but what if we weaken it by saying that $forall lambda_1, lambda_2 in Lambda, lambda_1 neq lambda_2, I_{lambda_1} nsubseteq I_{lambda_2}$? The problem should be solved, but it is a more restrictive hypothesis that the one I've given.
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:45
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
@LuxGiammi I have added work that I think shows that this new hypothesis is still not strong enough.
$endgroup$
– JDMan4444
Nov 28 '18 at 19:49
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
Ok. That was just an idea that came up to my mind and I just would have liked to check whether it was correct. Thanks a lot for your help. :-)
$endgroup$
– LuxGiammi
Nov 28 '18 at 19:55
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
$begingroup$
@LuxGiammi You are welcome!
$endgroup$
– JDMan4444
Nov 28 '18 at 19:56
add a comment |
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$begingroup$
But: For $lambda>0$ let $I_lambda=(0,lambda)$, Then for $lambda_1nelambda_2$, $I_{lambda_1}cap I_{lambda_2}$ is a non-degenerate interal.
$endgroup$
– Hagen von Eitzen
Nov 28 '18 at 19:36
$begingroup$
"every non degenerate interval of $mathbb R$ is dense in $mathbb Q$". Rather, $mathbb Q$ is dense in every non degenerate interval.
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 19:50