Example of quotient map from Munkres book
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I was reading the notion of quotient map, topology and space but ran into the following example.
In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.
Would be thankful if anyone will show the rigorous proof.
general-topology
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up vote
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favorite
I was reading the notion of quotient map, topology and space but ran into the following example.
In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.
Would be thankful if anyone will show the rigorous proof.
general-topology
It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
â Rushabh Mehta
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was reading the notion of quotient map, topology and space but ran into the following example.
In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.
Would be thankful if anyone will show the rigorous proof.
general-topology
I was reading the notion of quotient map, topology and space but ran into the following example.
In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.
Would be thankful if anyone will show the rigorous proof.
general-topology
general-topology
asked 3 hours ago
RFZ
4,77031337
4,77031337
It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
â Rushabh Mehta
3 hours ago
add a comment |Â
It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
â Rushabh Mehta
3 hours ago
It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
â Rushabh Mehta
3 hours ago
It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
â Rushabh Mehta
3 hours ago
add a comment |Â
2 Answers
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Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.
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If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).
Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that
$Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,
$p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$
$[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.
Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.
add a comment |Â
up vote
3
down vote
Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.
Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.
answered 3 hours ago
Kavi Rama Murthy
32.2k31644
32.2k31644
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add a comment |Â
up vote
1
down vote
If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).
Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that
$Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,
$p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$
$[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.
Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.
add a comment |Â
up vote
1
down vote
If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).
Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that
$Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,
$p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$
$[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.
Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).
Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that
$Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,
$p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$
$[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.
Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.
If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).
Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that
$Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,
$p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$
$[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.
Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.
answered 2 hours ago
Matematleta
8,6442918
8,6442918
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It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
â Rushabh Mehta
3 hours ago