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A countable open base implies that any open covering has a countable subcovering.
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Proposition: If metric space $M$ has a countable open base, then any open covering of $M$ admits a countable subcovering.
Definition: A collection of open subsets $U_i$ of $M$ is called an open base of $M$, if every of open set $Osubseteq M$ is expressible as $O=bigcup_i U_i$.
Proof Attempt:
Let $Omega_i_iin I$, where $I$ is an arbitrary indexing set, be an open cover of $M$; that is $M subseteq bigcup_i=1^infty Omega_i$ and for all $i in I$, $Omega_i$ is an open set. First, we show that $forall iin I$, $Omega_i cap M$ is open relative to $M$. Let $xin Omega_i cap M$, then there exists $epsilon gt 0$ such that $N_epsilon (x)subseteq Omega_i$. But, if $N_epsilon (x)nsubseteq Omega_i cap M$, then it must be that there exists $nin N_epsilon (x)$ such that $n notin M$. Relative to $M$, it does not affect whether $Omega_i cap M$ is open in $M$. What cannot be is $n in M$ and $n notin Omega_i$. So, at any rate, $Omega_i cap M$ is open in $M$. Now, suppose $M$ has a countable open base $mathscrO_k_k=1^infty$. Thus, $Omega_i cap M=bigcup_k=1^infty mathscrO_k$. Consider that $Omega_icap McapmathscrO_k_k=1^infty$ is countable $forall i in I$, it covers $M$, and $forall iin I$, $Omega_icap McapmathscrO_ksubseteqOmega_k$. Therefore, $bigcup_i=1^infty(bigcup_k=1^inftyOmega_icap McapmathscrO_k)$ is a countable subcovering of an arbitrary open covering of $M$, because the union of countable set is countable.
proof-verification metric-spaces separable-spaces
asked 2 days ago
TheLast Cipher
529314
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Proposition: If metric space $M$ has a countable open base, then any open covering of $M$ admits a countable subcovering.
Definition: A collection of open subsets $U_i$ of $M$ is called an open base of $M$, if every of open set $Osubseteq M$ is expressible as $O=bigcup_i U_i$.
Proof Attempt:
Let $Omega_i_iin I$, where $I$ is an arbitrary indexing set, be an open cover of $M$; that is $M subseteq bigcup_i=1^infty Omega_i$ and for all $i in I$, $Omega_i$ is an open set. First, we show that $forall iin I$, $Omega_i cap M$ is open relative to $M$. Let $xin Omega_i cap M$, then there exists $epsilon gt 0$ such that $N_epsilon (x)subseteq Omega_i$. But, if $N_epsilon (x)nsubseteq Omega_i cap M$, then it must be that there exists $nin N_epsilon (x)$ such that $n notin M$. Relative to $M$, it does not affect whether $Omega_i cap M$ is open in $M$. What cannot be is $n in M$ and $n notin Omega_i$. So, at any rate, $Omega_i cap M$ is open in $M$. Now, suppose $M$ has a countable open base $mathscrO_k_k=1^infty$. Thus, $Omega_i cap M=bigcup_k=1^infty mathscrO_k$. Consider that $Omega_icap McapmathscrO_k_k=1^infty$ is countable $forall i in I$, it covers $M$, and $forall iin I$, $Omega_icap McapmathscrO_ksubseteqOmega_k$. Therefore, $bigcup_i=1^infty(bigcup_k=1^inftyOmega_icap McapmathscrO_k)$ is a countable subcovering of an arbitrary open covering of $M$, because the union of countable set is countable.
proof-verification metric-spaces separable-spaces
asked 2 days ago
TheLast Cipher
529314
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Proposition: If metric space $M$ has a countable open base, then any open covering of $M$ admits a countable subcovering.
Definition: A collection of open subsets $U_i$ of $M$ is called an open base of $M$, if every of open set $Osubseteq M$ is expressible as $O=bigcup_i U_i$.
Proof Attempt:
Let $Omega_i_iin I$, where $I$ is an arbitrary indexing set, be an open cover of $M$; that is $M subseteq bigcup_i=1^infty Omega_i$ and for all $i in I$, $Omega_i$ is an open set. First, we show that $forall iin I$, $Omega_i cap M$ is open relative to $M$. Let $xin Omega_i cap M$, then there exists $epsilon gt 0$ such that $N_epsilon (x)subseteq Omega_i$. But, if $N_epsilon (x)nsubseteq Omega_i cap M$, then it must be that there exists $nin N_epsilon (x)$ such that $n notin M$. Relative to $M$, it does not affect whether $Omega_i cap M$ is open in $M$. What cannot be is $n in M$ and $n notin Omega_i$. So, at any rate, $Omega_i cap M$ is open in $M$. Now, suppose $M$ has a countable open base $mathscrO_k_k=1^infty$. Thus, $Omega_i cap M=bigcup_k=1^infty mathscrO_k$. Consider that $Omega_icap McapmathscrO_k_k=1^infty$ is countable $forall i in I$, it covers $M$, and $forall iin I$, $Omega_icap McapmathscrO_ksubseteqOmega_k$. Therefore, $bigcup_i=1^infty(bigcup_k=1^inftyOmega_icap McapmathscrO_k)$ is a countable subcovering of an arbitrary open covering of $M$, because the union of countable set is countable.
proof-verification metric-spaces separable-spaces
asked 2 days ago
TheLast Cipher
529314
Proposition: If metric space $M$ has a countable open base, then any open covering of $M$ admits a countable subcovering.
Definition: A collection of open subsets $U_i$ of $M$ is called an open base of $M$, if every of open set $Osubseteq M$ is expressible as $O=bigcup_i U_i$.
Proof Attempt:
Let $Omega_i_iin I$, where $I$ is an arbitrary indexing set, be an open cover of $M$; that is $M subseteq bigcup_i=1^infty Omega_i$ and for all $i in I$, $Omega_i$ is an open set. First, we show that $forall iin I$, $Omega_i cap M$ is open relative to $M$. Let $xin Omega_i cap M$, then there exists $epsilon gt 0$ such that $N_epsilon (x)subseteq Omega_i$. But, if $N_epsilon (x)nsubseteq Omega_i cap M$, then it must be that there exists $nin N_epsilon (x)$ such that $n notin M$. Relative to $M$, it does not affect whether $Omega_i cap M$ is open in $M$. What cannot be is $n in M$ and $n notin Omega_i$. So, at any rate, $Omega_i cap M$ is open in $M$. Now, suppose $M$ has a countable open base $mathscrO_k_k=1^infty$. Thus, $Omega_i cap M=bigcup_k=1^infty mathscrO_k$. Consider that $Omega_icap McapmathscrO_k_k=1^infty$ is countable $forall i in I$, it covers $M$, and $forall iin I$, $Omega_icap McapmathscrO_ksubseteqOmega_k$. Therefore, $bigcup_i=1^infty(bigcup_k=1^inftyOmega_icap McapmathscrO_k)$ is a countable subcovering of an arbitrary open covering of $M$, because the union of countable set is countable.
proof-verification metric-spaces separable-spaces
asked 2 days ago
TheLast Cipher
529314
asked 2 days ago
TheLast Cipher
529314
asked 2 days ago
TheLast Cipher
529314
asked 2 days ago
TheLast Cipher
529314
529314
add a comment |Â
add a comment |Â
1 Answer
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I noticed some inaccuracies in your attempt that you should think about:
You 'waste' a lot of time proving things related to relative open vs. open. This is not necessary here, since $M$ is the only metric space you are considering and everything happens inside $M$. In particular $Omega_i cap M = Omega_i$ etc.
Also
$$ Omega_i cap M=bigcup_k=1^infty mathscrO_k $$ will not be correct in general, because the right side is equal to $M$.Moreover, $Omega_i cap M cap mathscrO_k_k$ does not cover $M$ but $Omega_i$.
You correctly noted in the beginning that $I$ is arbitrary, but in the end you index the $Omega_i$ by $i in BbbN$, in which case there was nothing to prove from the start.
$Omega_i cap M cap mathscrO_k_k,i$ is not a subcover of $Omega_i_i$ in general. Recall that a subcover of $Omega_i_i in I$ is of the form $Omega_l_l in L$ for some $L subseteq I$.
A better way to proceed is to prove the following statements:
- Every $Omega_i$ is the union of some of the $mathscrO_k_k$, say $Omega_i = bigcup_k in J_imathscrO_k$ where $J_i subseteq BbbN$.
- The set $J = bigcup_i in I J_i$ is countable.
- The set $Omega_k :$ is a countable cover of $M$.
- For each $k in J$ there exists $i_k in I$ such that $mathscrO_k subseteq Omega_i_k$. Choose one $i_k$ for each $k in J$.
- The set $Omega_i_k :$ is a countable subcover of $Omega_i_i$.
answered 2 days ago
Matthias Klupsch
6,0341127
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I noticed some inaccuracies in your attempt that you should think about:
You 'waste' a lot of time proving things related to relative open vs. open. This is not necessary here, since $M$ is the only metric space you are considering and everything happens inside $M$. In particular $Omega_i cap M = Omega_i$ etc.
Also
$$ Omega_i cap M=bigcup_k=1^infty mathscrO_k $$ will not be correct in general, because the right side is equal to $M$.Moreover, $Omega_i cap M cap mathscrO_k_k$ does not cover $M$ but $Omega_i$.
You correctly noted in the beginning that $I$ is arbitrary, but in the end you index the $Omega_i$ by $i in BbbN$, in which case there was nothing to prove from the start.
$Omega_i cap M cap mathscrO_k_k,i$ is not a subcover of $Omega_i_i$ in general. Recall that a subcover of $Omega_i_i in I$ is of the form $Omega_l_l in L$ for some $L subseteq I$.
A better way to proceed is to prove the following statements:
- Every $Omega_i$ is the union of some of the $mathscrO_k_k$, say $Omega_i = bigcup_k in J_imathscrO_k$ where $J_i subseteq BbbN$.
- The set $J = bigcup_i in I J_i$ is countable.
- The set $Omega_k :$ is a countable cover of $M$.
- For each $k in J$ there exists $i_k in I$ such that $mathscrO_k subseteq Omega_i_k$. Choose one $i_k$ for each $k in J$.
- The set $Omega_i_k :$ is a countable subcover of $Omega_i_i$.
answered 2 days ago
Matthias Klupsch
6,0341127
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
add a comment |Â
up vote
0
down vote
accepted
I noticed some inaccuracies in your attempt that you should think about:
You 'waste' a lot of time proving things related to relative open vs. open. This is not necessary here, since $M$ is the only metric space you are considering and everything happens inside $M$. In particular $Omega_i cap M = Omega_i$ etc.
Also
$$ Omega_i cap M=bigcup_k=1^infty mathscrO_k $$ will not be correct in general, because the right side is equal to $M$.Moreover, $Omega_i cap M cap mathscrO_k_k$ does not cover $M$ but $Omega_i$.
You correctly noted in the beginning that $I$ is arbitrary, but in the end you index the $Omega_i$ by $i in BbbN$, in which case there was nothing to prove from the start.
$Omega_i cap M cap mathscrO_k_k,i$ is not a subcover of $Omega_i_i$ in general. Recall that a subcover of $Omega_i_i in I$ is of the form $Omega_l_l in L$ for some $L subseteq I$.
A better way to proceed is to prove the following statements:
- Every $Omega_i$ is the union of some of the $mathscrO_k_k$, say $Omega_i = bigcup_k in J_imathscrO_k$ where $J_i subseteq BbbN$.
- The set $J = bigcup_i in I J_i$ is countable.
- The set $Omega_k :$ is a countable cover of $M$.
- For each $k in J$ there exists $i_k in I$ such that $mathscrO_k subseteq Omega_i_k$. Choose one $i_k$ for each $k in J$.
- The set $Omega_i_k :$ is a countable subcover of $Omega_i_i$.
answered 2 days ago
Matthias Klupsch
6,0341127
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I noticed some inaccuracies in your attempt that you should think about:
You 'waste' a lot of time proving things related to relative open vs. open. This is not necessary here, since $M$ is the only metric space you are considering and everything happens inside $M$. In particular $Omega_i cap M = Omega_i$ etc.
Also
$$ Omega_i cap M=bigcup_k=1^infty mathscrO_k $$ will not be correct in general, because the right side is equal to $M$.Moreover, $Omega_i cap M cap mathscrO_k_k$ does not cover $M$ but $Omega_i$.
You correctly noted in the beginning that $I$ is arbitrary, but in the end you index the $Omega_i$ by $i in BbbN$, in which case there was nothing to prove from the start.
$Omega_i cap M cap mathscrO_k_k,i$ is not a subcover of $Omega_i_i$ in general. Recall that a subcover of $Omega_i_i in I$ is of the form $Omega_l_l in L$ for some $L subseteq I$.
A better way to proceed is to prove the following statements:
- Every $Omega_i$ is the union of some of the $mathscrO_k_k$, say $Omega_i = bigcup_k in J_imathscrO_k$ where $J_i subseteq BbbN$.
- The set $J = bigcup_i in I J_i$ is countable.
- The set $Omega_k :$ is a countable cover of $M$.
- For each $k in J$ there exists $i_k in I$ such that $mathscrO_k subseteq Omega_i_k$. Choose one $i_k$ for each $k in J$.
- The set $Omega_i_k :$ is a countable subcover of $Omega_i_i$.
answered 2 days ago
Matthias Klupsch
6,0341127
I noticed some inaccuracies in your attempt that you should think about:
You 'waste' a lot of time proving things related to relative open vs. open. This is not necessary here, since $M$ is the only metric space you are considering and everything happens inside $M$. In particular $Omega_i cap M = Omega_i$ etc.
Also
$$ Omega_i cap M=bigcup_k=1^infty mathscrO_k $$ will not be correct in general, because the right side is equal to $M$.Moreover, $Omega_i cap M cap mathscrO_k_k$ does not cover $M$ but $Omega_i$.
You correctly noted in the beginning that $I$ is arbitrary, but in the end you index the $Omega_i$ by $i in BbbN$, in which case there was nothing to prove from the start.
$Omega_i cap M cap mathscrO_k_k,i$ is not a subcover of $Omega_i_i$ in general. Recall that a subcover of $Omega_i_i in I$ is of the form $Omega_l_l in L$ for some $L subseteq I$.
A better way to proceed is to prove the following statements:
- Every $Omega_i$ is the union of some of the $mathscrO_k_k$, say $Omega_i = bigcup_k in J_imathscrO_k$ where $J_i subseteq BbbN$.
- The set $J = bigcup_i in I J_i$ is countable.
- The set $Omega_k :$ is a countable cover of $M$.
- For each $k in J$ there exists $i_k in I$ such that $mathscrO_k subseteq Omega_i_k$. Choose one $i_k$ for each $k in J$.
- The set $Omega_i_k :$ is a countable subcover of $Omega_i_i$.
answered 2 days ago
Matthias Klupsch
6,0341127
answered 2 days ago
Matthias Klupsch
6,0341127
answered 2 days ago
Matthias Klupsch
6,0341127
answered 2 days ago
Matthias Klupsch
6,0341127
6,0341127
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
add a comment |Â
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
at (1), I thought about assuming everything happens in $M$, but decided not to. at (2),(3),(4),(5).. I don't know what I was thinking. Thanks for pointing it out. I'll try the outline you provided.
– TheLast Cipher
2 days ago
add a comment |Â
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