Visualizing $E=Ycap G$ geometrically for some open set $Gsubset X$
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
add a comment |
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
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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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 yesterday
Glorfindel
3,41981830
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 yesterday
Glorfindel
3,41981830
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
add a comment |
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 yesterday
Glorfindel
3,41981830
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
add a comment |
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 yesterday
Glorfindel
3,41981830
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
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 yesterday
Glorfindel
3,41981830
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
edited yesterday
Glorfindel
3,41981830
edited yesterday
Glorfindel
3,41981830
edited yesterday
Glorfindel
3,41981830
3,41981830
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
answered Mar 28 '13 at 12:28
Ronnie Brown
11.9k12938
11.9k12938
add a comment |
add a comment |
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