Finite subcover with special property
Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.
Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have
$$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$
How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?
general-topology metric-spaces compactness
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Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.
Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have
$$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$
How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?
general-topology metric-spaces compactness
add a comment |
Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.
Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have
$$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$
How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?
general-topology metric-spaces compactness
Let $M$ compact and $mathcal{U}$ an open covering of $M$ that satisfies: each $p in M$ is contained in at least two members of $mathcal{U}$. Show that $mathcal{U}$ has a finite subcovering with the same property.
Writing $mathcal{U} = {U_{i}}$, since $M$ is compact, we have
$$M subset U_{alpha_{1}} cup cdots cup U_{alpha_{k}}.$$
How to ensure that taken $p in M$, $p in U_{alpha_{n}}cap U_{alpha_{m}}$? Intuitively, could it not be $p in U_{alpha_{1}}cap U_{alpha_{k+1}}$? I really have no idea how to prove it. Can someone give me a hint?
general-topology metric-spaces compactness
general-topology metric-spaces compactness
edited 6 hours ago
Kenny Wong
17k21135
17k21135
asked 6 hours ago
Lucas Corrêa
1,3451321
1,3451321
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add a comment |
1 Answer
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Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.
Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?
Then use the compactness of $M$ to find a finite refinement of $mathcal V$...
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
1
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.
Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?
Then use the compactness of $M$ to find a finite refinement of $mathcal V$...
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
1
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
add a comment |
Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.
Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?
Then use the compactness of $M$ to find a finite refinement of $mathcal V$...
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
1
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
add a comment |
Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.
Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?
Then use the compactness of $M$ to find a finite refinement of $mathcal V$...
Hint: Let $mathcal U = { U_i : iin I}$ be the original open covering of $M$.
Can you show that $mathcal V = { U_i cap U_j : i, j in I, i neq j }$ is also an open covering of $M$?
Then use the compactness of $M$ to find a finite refinement of $mathcal V$...
answered 6 hours ago
Kenny Wong
17k21135
17k21135
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
1
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
add a comment |
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
1
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Since every $p in M$ is contained in at least two members of $mathcal{U}$, each $p$ is contained in $U_{i} cap U_{j}$ for $i neq j$, so $displaystyle bigcup_{i,jmid i neq j} U_{i} cap U_{j}$ cover $M$. Since $M$ is compact, we can extract a finite subcovering that satisfies the property.
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
Is this a good idea?
– Lucas Corrêa
6 hours ago
1
1
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
@LucasCorrêa Yes, that's the right idea.
– Kenny Wong
6 hours ago
add a comment |
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