Visualizing $E=Ycap G$ geometrically for some open set $Gsubset X$
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I came across this theorem(2.30) in Rudin's text on analysis.
Suppose $Ysubset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E=Ycap G$ for some open subset $G$ of $X$.
(A set $Esubset Y$ is open relative to $Y$ when for every $pin E$, there is an $r_p>0$ such that $qin Y$ and $d(p, q)<r_p$ implies $qin E$.)
I can understand the proof he has provided. He has tried to construct such a $G$ by defining open sets $V_p$ as follows: $V_p={qin X|d(p,q)}<r_p$ and has taken $G= cup_{pin E}V_p$ which is good enough to prove that $Esubset Gcap Y$.
But I am still unable to understand the geometric intuition behind this; in particular, what is the geometric motivation behind such a construction? I am interested in a better way to visualize this which will allow me to construct a formal proof by looking at the picture.
real-analysis general-topology
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I came across this theorem(2.30) in Rudin's text on analysis.
Suppose $Ysubset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E=Ycap G$ for some open subset $G$ of $X$.
(A set $Esubset Y$ is open relative to $Y$ when for every $pin E$, there is an $r_p>0$ such that $qin Y$ and $d(p, q)<r_p$ implies $qin E$.)
I can understand the proof he has provided. He has tried to construct such a $G$ by defining open sets $V_p$ as follows: $V_p={qin X|d(p,q)}<r_p$ and has taken $G= cup_{pin E}V_p$ which is good enough to prove that $Esubset Gcap Y$.
But I am still unable to understand the geometric intuition behind this; in particular, what is the geometric motivation behind such a construction? I am interested in a better way to visualize this which will allow me to construct a formal proof by looking at the picture.
real-analysis general-topology
$endgroup$
add a comment |
$begingroup$
I came across this theorem(2.30) in Rudin's text on analysis.
Suppose $Ysubset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E=Ycap G$ for some open subset $G$ of $X$.
(A set $Esubset Y$ is open relative to $Y$ when for every $pin E$, there is an $r_p>0$ such that $qin Y$ and $d(p, q)<r_p$ implies $qin E$.)
I can understand the proof he has provided. He has tried to construct such a $G$ by defining open sets $V_p$ as follows: $V_p={qin X|d(p,q)}<r_p$ and has taken $G= cup_{pin E}V_p$ which is good enough to prove that $Esubset Gcap Y$.
But I am still unable to understand the geometric intuition behind this; in particular, what is the geometric motivation behind such a construction? I am interested in a better way to visualize this which will allow me to construct a formal proof by looking at the picture.
real-analysis general-topology
$endgroup$
I came across this theorem(2.30) in Rudin's text on analysis.
Suppose $Ysubset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E=Ycap G$ for some open subset $G$ of $X$.
(A set $Esubset Y$ is open relative to $Y$ when for every $pin E$, there is an $r_p>0$ such that $qin Y$ and $d(p, q)<r_p$ implies $qin E$.)
I can understand the proof he has provided. He has tried to construct such a $G$ by defining open sets $V_p$ as follows: $V_p={qin X|d(p,q)}<r_p$ and has taken $G= cup_{pin E}V_p$ which is good enough to prove that $Esubset Gcap Y$.
But I am still unable to understand the geometric intuition behind this; in particular, what is the geometric motivation behind such a construction? I am interested in a better way to visualize this which will allow me to construct a formal proof by looking at the picture.
real-analysis general-topology
real-analysis general-topology
asked Mar 28 '13 at 10:51
user67449
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1 Answer
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To understand this construction you should see it in the general context of constructing new topological spaces from old ones. In the following picture (taken from Topology and Groupoids)
we have the big space $X=mathbb R^2$, the usual plane, and the subset $A$, which is the circle. You also see a point $P$ on the circle and the intersection of a neighbourhood of $P$ with the circle. This "square""neighbourhood would be given by a distance if the distance used on the plane is given by the norm $||(x,y)||=max(|x|,|y|)$. The picture also shows a point $Q$ on $A$ and an open set in $X$ whose intersection with $A$ is not a neighbourhood of $Q$.
Drawing a picture of this type should give you a key idea of the proof which you have written out.
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
To understand this construction you should see it in the general context of constructing new topological spaces from old ones. In the following picture (taken from Topology and Groupoids)
we have the big space $X=mathbb R^2$, the usual plane, and the subset $A$, which is the circle. You also see a point $P$ on the circle and the intersection of a neighbourhood of $P$ with the circle. This "square""neighbourhood would be given by a distance if the distance used on the plane is given by the norm $||(x,y)||=max(|x|,|y|)$. The picture also shows a point $Q$ on $A$ and an open set in $X$ whose intersection with $A$ is not a neighbourhood of $Q$.
Drawing a picture of this type should give you a key idea of the proof which you have written out.
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
To understand this construction you should see it in the general context of constructing new topological spaces from old ones. In the following picture (taken from Topology and Groupoids)
we have the big space $X=mathbb R^2$, the usual plane, and the subset $A$, which is the circle. You also see a point $P$ on the circle and the intersection of a neighbourhood of $P$ with the circle. This "square""neighbourhood would be given by a distance if the distance used on the plane is given by the norm $||(x,y)||=max(|x|,|y|)$. The picture also shows a point $Q$ on $A$ and an open set in $X$ whose intersection with $A$ is not a neighbourhood of $Q$.
Drawing a picture of this type should give you a key idea of the proof which you have written out.
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
To understand this construction you should see it in the general context of constructing new topological spaces from old ones. In the following picture (taken from Topology and Groupoids)
we have the big space $X=mathbb R^2$, the usual plane, and the subset $A$, which is the circle. You also see a point $P$ on the circle and the intersection of a neighbourhood of $P$ with the circle. This "square""neighbourhood would be given by a distance if the distance used on the plane is given by the norm $||(x,y)||=max(|x|,|y|)$. The picture also shows a point $Q$ on $A$ and an open set in $X$ whose intersection with $A$ is not a neighbourhood of $Q$.
Drawing a picture of this type should give you a key idea of the proof which you have written out.
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
To understand this construction you should see it in the general context of constructing new topological spaces from old ones. In the following picture (taken from Topology and Groupoids)
we have the big space $X=mathbb R^2$, the usual plane, and the subset $A$, which is the circle. You also see a point $P$ on the circle and the intersection of a neighbourhood of $P$ with the circle. This "square""neighbourhood would be given by a distance if the distance used on the plane is given by the norm $||(x,y)||=max(|x|,|y|)$. The picture also shows a point $Q$ on $A$ and an open set in $X$ whose intersection with $A$ is not a neighbourhood of $Q$.
Drawing a picture of this type should give you a key idea of the proof which you have written out.
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
edited Dec 25 '18 at 12:19
Glorfindel
3,42981830
3,42981830
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
answered Mar 28 '13 at 12:28
Ronnie BrownRonnie Brown
12.1k12939
12.1k12939
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