Mixing
An element of a set with a finite cover must be an element of at most two open intervals in a subcover?
up vote
2
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Prove:
If a set $Asubseteqmathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $xin A$, x is an element of at most two of the open intervals in the subcover.
My attempt:
To be honest, I have grappled with this problem for too long; I have no idea how to approach this proof. I only have the definitions of cover, subcover, and compact sets and the Heine-Borel Theorem at my disposal. I am having difficulty connecting this ideas to prove what needs to be proven. Could someone give me an idea on how to begin this proof?
real-analysis general-topology proof-verification continuity uniform-continuity
edited yesterday
freakish
10.3k1526
asked yesterday
SebastianLinde
978
add a comment |
up vote
2
down vote
favorite
Prove:
If a set $Asubseteqmathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $xin A$, x is an element of at most two of the open intervals in the subcover.
My attempt:
To be honest, I have grappled with this problem for too long; I have no idea how to approach this proof. I only have the definitions of cover, subcover, and compact sets and the Heine-Borel Theorem at my disposal. I am having difficulty connecting this ideas to prove what needs to be proven. Could someone give me an idea on how to begin this proof?
real-analysis general-topology proof-verification continuity uniform-continuity
edited yesterday
freakish
10.3k1526
asked yesterday
SebastianLinde
978
@ViktorGlombik Yes. I am sorry I forgot to mention it.
– SebastianLinde
yesterday
Can you prove it in the special case where your finite cover has only $3$ members? If some point $x$ is in all $3$ of the intervals, can you prove that one of the $3$ intervals is covered by the other $2$, so that interval can be discarded?
– bof
yesterday
@bof I cannot. I can picture it, and I know it must be true, but I have trouble proving it with the given information.
– SebastianLinde
yesterday
You can have a refinement with that property not always a subcover. The refinement is part of the definition of one-dimensional, in the covering sense.
– Henno Brandsma
yesterday
@HennoBrandsma He is starting with a finite cover, so in this case he can get a subcover with that property.
– bof
23 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove:
If a set $Asubseteqmathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $xin A$, x is an element of at most two of the open intervals in the subcover.
My attempt:
To be honest, I have grappled with this problem for too long; I have no idea how to approach this proof. I only have the definitions of cover, subcover, and compact sets and the Heine-Borel Theorem at my disposal. I am having difficulty connecting this ideas to prove what needs to be proven. Could someone give me an idea on how to begin this proof?
real-analysis general-topology proof-verification continuity uniform-continuity
edited yesterday
freakish
10.3k1526
asked yesterday
SebastianLinde
978
Prove:
If a set $Asubseteqmathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $xin A$, x is an element of at most two of the open intervals in the subcover.
My attempt:
To be honest, I have grappled with this problem for too long; I have no idea how to approach this proof. I only have the definitions of cover, subcover, and compact sets and the Heine-Borel Theorem at my disposal. I am having difficulty connecting this ideas to prove what needs to be proven. Could someone give me an idea on how to begin this proof?
real-analysis general-topology proof-verification continuity uniform-continuity
real-analysis general-topology proof-verification continuity uniform-continuity
edited yesterday
freakish
10.3k1526
asked yesterday
SebastianLinde
978
edited yesterday
freakish
10.3k1526
asked yesterday
SebastianLinde
978
edited yesterday
freakish
10.3k1526
edited yesterday
freakish
10.3k1526
edited yesterday
freakish
10.3k1526
10.3k1526
asked yesterday
SebastianLinde
978
asked yesterday
SebastianLinde
978
asked yesterday
SebastianLinde
978
978
@ViktorGlombik Yes. I am sorry I forgot to mention it.
– SebastianLinde
yesterday
Can you prove it in the special case where your finite cover has only $3$ members? If some point $x$ is in all $3$ of the intervals, can you prove that one of the $3$ intervals is covered by the other $2$, so that interval can be discarded?
– bof
yesterday
@bof I cannot. I can picture it, and I know it must be true, but I have trouble proving it with the given information.
– SebastianLinde
yesterday
You can have a refinement with that property not always a subcover. The refinement is part of the definition of one-dimensional, in the covering sense.
– Henno Brandsma
yesterday
@HennoBrandsma He is starting with a finite cover, so in this case he can get a subcover with that property.
– bof
23 hours ago
add a comment |
@ViktorGlombik Yes. I am sorry I forgot to mention it.
– SebastianLinde
yesterday
Can you prove it in the special case where your finite cover has only $3$ members? If some point $x$ is in all $3$ of the intervals, can you prove that one of the $3$ intervals is covered by the other $2$, so that interval can be discarded?
– bof
yesterday
@bof I cannot. I can picture it, and I know it must be true, but I have trouble proving it with the given information.
– SebastianLinde
yesterday
You can have a refinement with that property not always a subcover. The refinement is part of the definition of one-dimensional, in the covering sense.
– Henno Brandsma
yesterday
@HennoBrandsma He is starting with a finite cover, so in this case he can get a subcover with that property.
– bof
23 hours ago
@ViktorGlombik Yes. I am sorry I forgot to mention it.
– SebastianLinde
yesterday
@ViktorGlombik Yes. I am sorry I forgot to mention it.
– SebastianLinde
yesterday
Can you prove it in the special case where your finite cover has only $3$ members? If some point $x$ is in all $3$ of the intervals, can you prove that one of the $3$ intervals is covered by the other $2$, so that interval can be discarded?
– bof
yesterday
Can you prove it in the special case where your finite cover has only $3$ members? If some point $x$ is in all $3$ of the intervals, can you prove that one of the $3$ intervals is covered by the other $2$, so that interval can be discarded?
– bof
yesterday
@bof I cannot. I can picture it, and I know it must be true, but I have trouble proving it with the given information.
– SebastianLinde
yesterday
@bof I cannot. I can picture it, and I know it must be true, but I have trouble proving it with the given information.
– SebastianLinde
yesterday
You can have a refinement with that property not always a subcover. The refinement is part of the definition of one-dimensional, in the covering sense.
– Henno Brandsma
yesterday
You can have a refinement with that property not always a subcover. The refinement is part of the definition of one-dimensional, in the covering sense.
– Henno Brandsma
yesterday
@HennoBrandsma He is starting with a finite cover, so in this case he can get a subcover with that property.
– bof
23 hours ago
@HennoBrandsma He is starting with a finite cover, so in this case he can get a subcover with that property.
– bof
23 hours ago
add a comment |
3 Answers
3
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up vote
1
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Denote by $mathcal{C}$ the open cover. Let $I_1inmathcal{C}$ denote an interval containing $inf A$ (or with boundary $inf A$, if necessary). Next, let $I_2inmathcal{C}$ be the interval containing $inf (A- I_1)$ with maximal upper bound. Then let $I_3inmathcal{C}$ be the interval containing $inf(A- I_2-I_1)$ with maximal upper bound. Since $sup I_3>sup I_2$ and $I_2$ has maximal upper bound for intervals containing $inf(A-I_1)$, it follows that $I_3>inf(A- I_1)$. Continue inductively until finished.
answered yesterday
Ben W
1,079510
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
add a comment |
up vote
1
down vote
The following is based on ideas from bof's comment and Black8Mamba23's answer. Since your given cover is finite, consider a minimal subcover $C$. Suppose, toward a contradiction, that some point $x$ is in more than two of the intervals in $C$. Among those (finitely many but more than $2$) intervals from $C$ that contain $x$, let $I$ be one with the smallest left endpoint, and let $J$ be one with the largest right endpoint. There is at least one other interval $K$ from $C$ that contains $x$, because of our "more than $2$" assumption. Any such $K$ is included in $Icup J$, because (1) our choice of $I$ ensures that the part of $K$ to the left of $x$ is included in $I$ and (2) our choice of $J$ ensures that the part of $K$ to the right of $x$ is included in $J$. So $C-{K}$ is still a cover, and that contradicts our choice of $C$ as a minimal subcover.
answered 8 hours ago
Andreas Blass
48.8k350106
add a comment |
up vote
0
down vote
What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:
Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $xin A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$.
Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $xin A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover.
answered 22 hours ago
Black8Mamba23
11
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Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Denote by $mathcal{C}$ the open cover. Let $I_1inmathcal{C}$ denote an interval containing $inf A$ (or with boundary $inf A$, if necessary). Next, let $I_2inmathcal{C}$ be the interval containing $inf (A- I_1)$ with maximal upper bound. Then let $I_3inmathcal{C}$ be the interval containing $inf(A- I_2-I_1)$ with maximal upper bound. Since $sup I_3>sup I_2$ and $I_2$ has maximal upper bound for intervals containing $inf(A-I_1)$, it follows that $I_3>inf(A- I_1)$. Continue inductively until finished.
answered yesterday
Ben W
1,079510
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
add a comment |
up vote
1
down vote
Denote by $mathcal{C}$ the open cover. Let $I_1inmathcal{C}$ denote an interval containing $inf A$ (or with boundary $inf A$, if necessary). Next, let $I_2inmathcal{C}$ be the interval containing $inf (A- I_1)$ with maximal upper bound. Then let $I_3inmathcal{C}$ be the interval containing $inf(A- I_2-I_1)$ with maximal upper bound. Since $sup I_3>sup I_2$ and $I_2$ has maximal upper bound for intervals containing $inf(A-I_1)$, it follows that $I_3>inf(A- I_1)$. Continue inductively until finished.
answered yesterday
Ben W
1,079510
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Denote by $mathcal{C}$ the open cover. Let $I_1inmathcal{C}$ denote an interval containing $inf A$ (or with boundary $inf A$, if necessary). Next, let $I_2inmathcal{C}$ be the interval containing $inf (A- I_1)$ with maximal upper bound. Then let $I_3inmathcal{C}$ be the interval containing $inf(A- I_2-I_1)$ with maximal upper bound. Since $sup I_3>sup I_2$ and $I_2$ has maximal upper bound for intervals containing $inf(A-I_1)$, it follows that $I_3>inf(A- I_1)$. Continue inductively until finished.
answered yesterday
Ben W
1,079510
Denote by $mathcal{C}$ the open cover. Let $I_1inmathcal{C}$ denote an interval containing $inf A$ (or with boundary $inf A$, if necessary). Next, let $I_2inmathcal{C}$ be the interval containing $inf (A- I_1)$ with maximal upper bound. Then let $I_3inmathcal{C}$ be the interval containing $inf(A- I_2-I_1)$ with maximal upper bound. Since $sup I_3>sup I_2$ and $I_2$ has maximal upper bound for intervals containing $inf(A-I_1)$, it follows that $I_3>inf(A- I_1)$. Continue inductively until finished.
answered yesterday
Ben W
1,079510
edited 9 hours ago
answered yesterday
Ben W
1,079510
answered yesterday
Ben W
1,079510
answered yesterday
Ben W
1,079510
1,079510
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
add a comment |
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
There may be no interval containing $inf A$. (You might say $I_1$ extending to $inf A$, regardless whether $inf A=-infty$ or a finite number.)
– Mirko
21 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
Oops! Fixed it.
– Ben W
9 hours ago
add a comment |
up vote
1
down vote
The following is based on ideas from bof's comment and Black8Mamba23's answer. Since your given cover is finite, consider a minimal subcover $C$. Suppose, toward a contradiction, that some point $x$ is in more than two of the intervals in $C$. Among those (finitely many but more than $2$) intervals from $C$ that contain $x$, let $I$ be one with the smallest left endpoint, and let $J$ be one with the largest right endpoint. There is at least one other interval $K$ from $C$ that contains $x$, because of our "more than $2$" assumption. Any such $K$ is included in $Icup J$, because (1) our choice of $I$ ensures that the part of $K$ to the left of $x$ is included in $I$ and (2) our choice of $J$ ensures that the part of $K$ to the right of $x$ is included in $J$. So $C-{K}$ is still a cover, and that contradicts our choice of $C$ as a minimal subcover.
answered 8 hours ago
Andreas Blass
48.8k350106
add a comment |
up vote
1
down vote
The following is based on ideas from bof's comment and Black8Mamba23's answer. Since your given cover is finite, consider a minimal subcover $C$. Suppose, toward a contradiction, that some point $x$ is in more than two of the intervals in $C$. Among those (finitely many but more than $2$) intervals from $C$ that contain $x$, let $I$ be one with the smallest left endpoint, and let $J$ be one with the largest right endpoint. There is at least one other interval $K$ from $C$ that contains $x$, because of our "more than $2$" assumption. Any such $K$ is included in $Icup J$, because (1) our choice of $I$ ensures that the part of $K$ to the left of $x$ is included in $I$ and (2) our choice of $J$ ensures that the part of $K$ to the right of $x$ is included in $J$. So $C-{K}$ is still a cover, and that contradicts our choice of $C$ as a minimal subcover.
answered 8 hours ago
Andreas Blass
48.8k350106
add a comment |
up vote
1
down vote
up vote
1
down vote
The following is based on ideas from bof's comment and Black8Mamba23's answer. Since your given cover is finite, consider a minimal subcover $C$. Suppose, toward a contradiction, that some point $x$ is in more than two of the intervals in $C$. Among those (finitely many but more than $2$) intervals from $C$ that contain $x$, let $I$ be one with the smallest left endpoint, and let $J$ be one with the largest right endpoint. There is at least one other interval $K$ from $C$ that contains $x$, because of our "more than $2$" assumption. Any such $K$ is included in $Icup J$, because (1) our choice of $I$ ensures that the part of $K$ to the left of $x$ is included in $I$ and (2) our choice of $J$ ensures that the part of $K$ to the right of $x$ is included in $J$. So $C-{K}$ is still a cover, and that contradicts our choice of $C$ as a minimal subcover.
answered 8 hours ago
Andreas Blass
48.8k350106
The following is based on ideas from bof's comment and Black8Mamba23's answer. Since your given cover is finite, consider a minimal subcover $C$. Suppose, toward a contradiction, that some point $x$ is in more than two of the intervals in $C$. Among those (finitely many but more than $2$) intervals from $C$ that contain $x$, let $I$ be one with the smallest left endpoint, and let $J$ be one with the largest right endpoint. There is at least one other interval $K$ from $C$ that contains $x$, because of our "more than $2$" assumption. Any such $K$ is included in $Icup J$, because (1) our choice of $I$ ensures that the part of $K$ to the left of $x$ is included in $I$ and (2) our choice of $J$ ensures that the part of $K$ to the right of $x$ is included in $J$. So $C-{K}$ is still a cover, and that contradicts our choice of $C$ as a minimal subcover.
answered 8 hours ago
Andreas Blass
48.8k350106
answered 8 hours ago
Andreas Blass
48.8k350106
answered 8 hours ago
Andreas Blass
48.8k350106
answered 8 hours ago
Andreas Blass
48.8k350106
48.8k350106
add a comment |
add a comment |
up vote
0
down vote
What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:
Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $xin A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$.
Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $xin A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover.
answered 22 hours ago
Black8Mamba23
11
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Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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up vote
0
down vote
What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:
Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $xin A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$.
Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $xin A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover.
answered 22 hours ago
Black8Mamba23
11
New contributor
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:
Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $xin A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$.
Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $xin A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover.
answered 22 hours ago
Black8Mamba23
11
New contributor
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:
Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $xin A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$.
Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $xin A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover.
answered 22 hours ago
Black8Mamba23
11
New contributor
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
answered 22 hours ago
Black8Mamba23
11
New contributor
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 22 hours ago
Black8Mamba23
11
answered 22 hours ago
Black8Mamba23
11
11
New contributor
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Black8Mamba23 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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@ViktorGlombik Yes. I am sorry I forgot to mention it.
– SebastianLinde
yesterday
Can you prove it in the special case where your finite cover has only $3$ members? If some point $x$ is in all $3$ of the intervals, can you prove that one of the $3$ intervals is covered by the other $2$, so that interval can be discarded?
– bof
yesterday
@bof I cannot. I can picture it, and I know it must be true, but I have trouble proving it with the given information.
– SebastianLinde
yesterday
You can have a refinement with that property not always a subcover. The refinement is part of the definition of one-dimensional, in the covering sense.
– Henno Brandsma
yesterday
@HennoBrandsma He is starting with a finite cover, so in this case he can get a subcover with that property.
– bof
23 hours ago