Does partition of unity implies second countable?
$begingroup$
Reading the definition of partition of unity:
Let $Asubset Bbb R^n$ and let $mathcal{O}$ be an open cover of $A$. Then there is a collection $Phi$ of $C^infty$ functions $varphi$ defined in an open set containing $A$, with the following properties:
For each $x in A$ we have $0 leq varphi(x) leq 1$.
For each $x in A$ there is an open set $V$ containing $x$ such that all but finitely many $varphi in Phi$ are $0$ on $V$.
For each $x in A$ we have $sum_{varphi in Phi}varphi(x)=1$ (by 2 for each $x$ their sum is finite in some open set containing $x$).
For each $varphi in Phi$ there is an open set $U$ in $mathcal{O}$ such that $varphi = 0$ outside of some closed set contained in $U$.
Make me feel that implies second countable because of condition (2), but I am no quite sure if this hold.
real-analysis manifolds second-countable
$endgroup$
add a comment |
$begingroup$
Reading the definition of partition of unity:
Let $Asubset Bbb R^n$ and let $mathcal{O}$ be an open cover of $A$. Then there is a collection $Phi$ of $C^infty$ functions $varphi$ defined in an open set containing $A$, with the following properties:
For each $x in A$ we have $0 leq varphi(x) leq 1$.
For each $x in A$ there is an open set $V$ containing $x$ such that all but finitely many $varphi in Phi$ are $0$ on $V$.
For each $x in A$ we have $sum_{varphi in Phi}varphi(x)=1$ (by 2 for each $x$ their sum is finite in some open set containing $x$).
For each $varphi in Phi$ there is an open set $U$ in $mathcal{O}$ such that $varphi = 0$ outside of some closed set contained in $U$.
Make me feel that implies second countable because of condition (2), but I am no quite sure if this hold.
real-analysis manifolds second-countable
$endgroup$
add a comment |
$begingroup$
Reading the definition of partition of unity:
Let $Asubset Bbb R^n$ and let $mathcal{O}$ be an open cover of $A$. Then there is a collection $Phi$ of $C^infty$ functions $varphi$ defined in an open set containing $A$, with the following properties:
For each $x in A$ we have $0 leq varphi(x) leq 1$.
For each $x in A$ there is an open set $V$ containing $x$ such that all but finitely many $varphi in Phi$ are $0$ on $V$.
For each $x in A$ we have $sum_{varphi in Phi}varphi(x)=1$ (by 2 for each $x$ their sum is finite in some open set containing $x$).
For each $varphi in Phi$ there is an open set $U$ in $mathcal{O}$ such that $varphi = 0$ outside of some closed set contained in $U$.
Make me feel that implies second countable because of condition (2), but I am no quite sure if this hold.
real-analysis manifolds second-countable
$endgroup$
Reading the definition of partition of unity:
Let $Asubset Bbb R^n$ and let $mathcal{O}$ be an open cover of $A$. Then there is a collection $Phi$ of $C^infty$ functions $varphi$ defined in an open set containing $A$, with the following properties:
For each $x in A$ we have $0 leq varphi(x) leq 1$.
For each $x in A$ there is an open set $V$ containing $x$ such that all but finitely many $varphi in Phi$ are $0$ on $V$.
For each $x in A$ we have $sum_{varphi in Phi}varphi(x)=1$ (by 2 for each $x$ their sum is finite in some open set containing $x$).
For each $varphi in Phi$ there is an open set $U$ in $mathcal{O}$ such that $varphi = 0$ outside of some closed set contained in $U$.
Make me feel that implies second countable because of condition (2), but I am no quite sure if this hold.
real-analysis manifolds second-countable
real-analysis manifolds second-countable
asked Apr 13 '18 at 8:46
user97512
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
There is a property called paracompactness stating that every open cover has an open locally finite refinement. So to a given open cover $mathcal{U}$ there is a refining open cover $mathcal{V}$ such that every point $x$ has a neighborhood $W$ that intersects only finitely many members of $mathcal{V}$.
The existence of partition of unity is equivalent to paracompactness. And paracompactness does not imply second countability. Also, every metrizable space is paracompact. Note that second countability is a global property, but paracompactness and metrizability is not. Any topological sum of paracompact or metrizable spaces is paracompact or metrizable, but if you have uncountably many nontrivial summands, then the sum won't be second countable.
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
$endgroup$
3
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
add a comment |
$begingroup$
Remark: If a topological space $X$ is connected and has a $C^infty$ atlas, then:
$$X text{ admits partitions of unity}Leftrightarrow X text{ is Hausdorff and second-countable}$$
The proof of $(Leftarrow)$ can be found in any book about smooth manifolds (for example, in Lee's Introduction to Smooth Manifolds, theorem 2.23).
Here are an outline for $(Rightarrow)$:
(a) Hausdorff: Let $p,qin X$ be distinct. Take a partition of unity ${varphi_U,varphi_V}$ subordinate to the open cover ${U:=M-{p},V:=M-{q}}$. In that case, $varphi_U(p)=varphi_V(q)=0$ and $varphi_U(q)=varphi_V(p)=1$. The idea is to take small neighbourhood $W_p$ such that $varphi_U(x)$ is close to $0$ and $varphi_V(x)$ is close to $1$ for $xin W_p$. Similarly, take a small $W_q$ at which $varphi_U(y)$ is close to $1$ and $varphi_V(y)$ is close to $0$ for $yin W_q$. Then $W_p,W_q$ separate $p,q$.
(b) Second-countable: Discussed here.
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
$endgroup$
1
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
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Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
add a comment |
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2 Answers
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2 Answers
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$begingroup$
There is a property called paracompactness stating that every open cover has an open locally finite refinement. So to a given open cover $mathcal{U}$ there is a refining open cover $mathcal{V}$ such that every point $x$ has a neighborhood $W$ that intersects only finitely many members of $mathcal{V}$.
The existence of partition of unity is equivalent to paracompactness. And paracompactness does not imply second countability. Also, every metrizable space is paracompact. Note that second countability is a global property, but paracompactness and metrizability is not. Any topological sum of paracompact or metrizable spaces is paracompact or metrizable, but if you have uncountably many nontrivial summands, then the sum won't be second countable.
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
$endgroup$
3
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
add a comment |
$begingroup$
There is a property called paracompactness stating that every open cover has an open locally finite refinement. So to a given open cover $mathcal{U}$ there is a refining open cover $mathcal{V}$ such that every point $x$ has a neighborhood $W$ that intersects only finitely many members of $mathcal{V}$.
The existence of partition of unity is equivalent to paracompactness. And paracompactness does not imply second countability. Also, every metrizable space is paracompact. Note that second countability is a global property, but paracompactness and metrizability is not. Any topological sum of paracompact or metrizable spaces is paracompact or metrizable, but if you have uncountably many nontrivial summands, then the sum won't be second countable.
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
$endgroup$
3
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
add a comment |
$begingroup$
There is a property called paracompactness stating that every open cover has an open locally finite refinement. So to a given open cover $mathcal{U}$ there is a refining open cover $mathcal{V}$ such that every point $x$ has a neighborhood $W$ that intersects only finitely many members of $mathcal{V}$.
The existence of partition of unity is equivalent to paracompactness. And paracompactness does not imply second countability. Also, every metrizable space is paracompact. Note that second countability is a global property, but paracompactness and metrizability is not. Any topological sum of paracompact or metrizable spaces is paracompact or metrizable, but if you have uncountably many nontrivial summands, then the sum won't be second countable.
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
$endgroup$
There is a property called paracompactness stating that every open cover has an open locally finite refinement. So to a given open cover $mathcal{U}$ there is a refining open cover $mathcal{V}$ such that every point $x$ has a neighborhood $W$ that intersects only finitely many members of $mathcal{V}$.
The existence of partition of unity is equivalent to paracompactness. And paracompactness does not imply second countability. Also, every metrizable space is paracompact. Note that second countability is a global property, but paracompactness and metrizability is not. Any topological sum of paracompact or metrizable spaces is paracompact or metrizable, but if you have uncountably many nontrivial summands, then the sum won't be second countable.
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
answered Apr 13 '18 at 8:59
user87690user87690
6,6741825
6,6741825
3
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
add a comment |
3
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
3
3
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
$begingroup$
As a specific example, the Sorgenfrey line (the real line under the lower limit topology, where open sets are $[a,b)$, $[a, infty)$, or $(-infty, b)$) is paracompact Hausdorff, but not second countable.
$endgroup$
– Fargle
Apr 13 '18 at 9:03
add a comment |
$begingroup$
Remark: If a topological space $X$ is connected and has a $C^infty$ atlas, then:
$$X text{ admits partitions of unity}Leftrightarrow X text{ is Hausdorff and second-countable}$$
The proof of $(Leftarrow)$ can be found in any book about smooth manifolds (for example, in Lee's Introduction to Smooth Manifolds, theorem 2.23).
Here are an outline for $(Rightarrow)$:
(a) Hausdorff: Let $p,qin X$ be distinct. Take a partition of unity ${varphi_U,varphi_V}$ subordinate to the open cover ${U:=M-{p},V:=M-{q}}$. In that case, $varphi_U(p)=varphi_V(q)=0$ and $varphi_U(q)=varphi_V(p)=1$. The idea is to take small neighbourhood $W_p$ such that $varphi_U(x)$ is close to $0$ and $varphi_V(x)$ is close to $1$ for $xin W_p$. Similarly, take a small $W_q$ at which $varphi_U(y)$ is close to $1$ and $varphi_V(y)$ is close to $0$ for $yin W_q$. Then $W_p,W_q$ separate $p,q$.
(b) Second-countable: Discussed here.
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
$endgroup$
1
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
$begingroup$
Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
add a comment |
$begingroup$
Remark: If a topological space $X$ is connected and has a $C^infty$ atlas, then:
$$X text{ admits partitions of unity}Leftrightarrow X text{ is Hausdorff and second-countable}$$
The proof of $(Leftarrow)$ can be found in any book about smooth manifolds (for example, in Lee's Introduction to Smooth Manifolds, theorem 2.23).
Here are an outline for $(Rightarrow)$:
(a) Hausdorff: Let $p,qin X$ be distinct. Take a partition of unity ${varphi_U,varphi_V}$ subordinate to the open cover ${U:=M-{p},V:=M-{q}}$. In that case, $varphi_U(p)=varphi_V(q)=0$ and $varphi_U(q)=varphi_V(p)=1$. The idea is to take small neighbourhood $W_p$ such that $varphi_U(x)$ is close to $0$ and $varphi_V(x)$ is close to $1$ for $xin W_p$. Similarly, take a small $W_q$ at which $varphi_U(y)$ is close to $1$ and $varphi_V(y)$ is close to $0$ for $yin W_q$. Then $W_p,W_q$ separate $p,q$.
(b) Second-countable: Discussed here.
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
$endgroup$
1
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
$begingroup$
Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
add a comment |
$begingroup$
Remark: If a topological space $X$ is connected and has a $C^infty$ atlas, then:
$$X text{ admits partitions of unity}Leftrightarrow X text{ is Hausdorff and second-countable}$$
The proof of $(Leftarrow)$ can be found in any book about smooth manifolds (for example, in Lee's Introduction to Smooth Manifolds, theorem 2.23).
Here are an outline for $(Rightarrow)$:
(a) Hausdorff: Let $p,qin X$ be distinct. Take a partition of unity ${varphi_U,varphi_V}$ subordinate to the open cover ${U:=M-{p},V:=M-{q}}$. In that case, $varphi_U(p)=varphi_V(q)=0$ and $varphi_U(q)=varphi_V(p)=1$. The idea is to take small neighbourhood $W_p$ such that $varphi_U(x)$ is close to $0$ and $varphi_V(x)$ is close to $1$ for $xin W_p$. Similarly, take a small $W_q$ at which $varphi_U(y)$ is close to $1$ and $varphi_V(y)$ is close to $0$ for $yin W_q$. Then $W_p,W_q$ separate $p,q$.
(b) Second-countable: Discussed here.
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
$endgroup$
Remark: If a topological space $X$ is connected and has a $C^infty$ atlas, then:
$$X text{ admits partitions of unity}Leftrightarrow X text{ is Hausdorff and second-countable}$$
The proof of $(Leftarrow)$ can be found in any book about smooth manifolds (for example, in Lee's Introduction to Smooth Manifolds, theorem 2.23).
Here are an outline for $(Rightarrow)$:
(a) Hausdorff: Let $p,qin X$ be distinct. Take a partition of unity ${varphi_U,varphi_V}$ subordinate to the open cover ${U:=M-{p},V:=M-{q}}$. In that case, $varphi_U(p)=varphi_V(q)=0$ and $varphi_U(q)=varphi_V(p)=1$. The idea is to take small neighbourhood $W_p$ such that $varphi_U(x)$ is close to $0$ and $varphi_V(x)$ is close to $1$ for $xin W_p$. Similarly, take a small $W_q$ at which $varphi_U(y)$ is close to $1$ and $varphi_V(y)$ is close to $0$ for $yin W_q$. Then $W_p,W_q$ separate $p,q$.
(b) Second-countable: Discussed here.
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
edited Jan 20 at 23:33
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
answered Jan 17 at 14:52
rmdmc89rmdmc89
2,2421923
2,2421923
1
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
$begingroup$
Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
add a comment |
1
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
$begingroup$
Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
1
1
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
$begingroup$
Your claim is not true, unless you also assume $X$ has at most countably many connected components. For example, a disjoint union of uncountably many copies of $mathbb R$ admits partitions of unity, but is not second-countable.
$endgroup$
– Jack Lee
Jan 20 at 16:30
$begingroup$
Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
$begingroup$
Thanks for that, @JackLee. I was implicitly assuming $X$ to be connected, but forgot to mention it. I'll correct it.
$endgroup$
– rmdmc89
Jan 20 at 23:32
add a comment |
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