Mixing
Connectedness of complements of vector subspaces of $mathbb{R}^n$
$begingroup$
I want to prove the following:
Let $E subset mathbb{R}^n$ be a vector subspace of $mathbb{R}^n$ which is not equal to $mathbb{R}^n$. Then $mathbb{R}^n setminus E$ is connected if, and only if, the dimension of $E$ is at most $n - 2$ (or equivalently, codimension at least $2$).
For the $implies$ direction, here's what I've got: Let $E$ be a vector subspace with dimension $n - 1$. Then: $$mathbb{R}^n setminus E = A cup B$$
where $A = {v in mathbb{R}^n vert langle v, n rangle < 0 }$ and $B = {v in mathbb{R}^n vert langle v, n rangle > 0 }$ and $n$ is a normal vector to $E$ (which certainly exists). Now, intuitively, I know that $A$ and $B$ are disjoint open sets, but I haven't been able to prove it.
I couldn't think of anything yet regarding the $impliedby$ direction. My strategy was proving that any two points in $mathbb{R}^n setminus E$ can be joined by a line or a union of broken lines, but I have been unsuccessful in getting anywhere with that approach.
real-analysis linear-algebra general-topology
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
$endgroup$
add a comment |
$begingroup$
I want to prove the following:
Let $E subset mathbb{R}^n$ be a vector subspace of $mathbb{R}^n$ which is not equal to $mathbb{R}^n$. Then $mathbb{R}^n setminus E$ is connected if, and only if, the dimension of $E$ is at most $n - 2$ (or equivalently, codimension at least $2$).
For the $implies$ direction, here's what I've got: Let $E$ be a vector subspace with dimension $n - 1$. Then: $$mathbb{R}^n setminus E = A cup B$$
where $A = {v in mathbb{R}^n vert langle v, n rangle < 0 }$ and $B = {v in mathbb{R}^n vert langle v, n rangle > 0 }$ and $n$ is a normal vector to $E$ (which certainly exists). Now, intuitively, I know that $A$ and $B$ are disjoint open sets, but I haven't been able to prove it.
I couldn't think of anything yet regarding the $impliedby$ direction. My strategy was proving that any two points in $mathbb{R}^n setminus E$ can be joined by a line or a union of broken lines, but I have been unsuccessful in getting anywhere with that approach.
real-analysis linear-algebra general-topology
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
$endgroup$
$begingroup$
This cannot be strictly true because $mathbb{R}^n setminus E$ is connected when $E=mathbb{R}^n$.
$endgroup$
– lhf
Jan 30 at 23:03
$begingroup$
Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $varnothing$ is not connected?
$endgroup$
– GEdgar
Jan 30 at 23:04
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@GEdgar Notice I stated codimension $geq 2$.
$endgroup$
– Matheus Andrade
Jan 30 at 23:07
$begingroup$
So: the same question I asked becomes: Why state codimension ${} ge 2$ instead of codimension ${}ne 1$?
$endgroup$
– GEdgar
Jan 30 at 23:12
$begingroup$
No particular reason. They mean the same thing.
$endgroup$
– Matheus Andrade
Jan 30 at 23:14
add a comment |
$begingroup$
I want to prove the following:
Let $E subset mathbb{R}^n$ be a vector subspace of $mathbb{R}^n$ which is not equal to $mathbb{R}^n$. Then $mathbb{R}^n setminus E$ is connected if, and only if, the dimension of $E$ is at most $n - 2$ (or equivalently, codimension at least $2$).
For the $implies$ direction, here's what I've got: Let $E$ be a vector subspace with dimension $n - 1$. Then: $$mathbb{R}^n setminus E = A cup B$$
where $A = {v in mathbb{R}^n vert langle v, n rangle < 0 }$ and $B = {v in mathbb{R}^n vert langle v, n rangle > 0 }$ and $n$ is a normal vector to $E$ (which certainly exists). Now, intuitively, I know that $A$ and $B$ are disjoint open sets, but I haven't been able to prove it.
I couldn't think of anything yet regarding the $impliedby$ direction. My strategy was proving that any two points in $mathbb{R}^n setminus E$ can be joined by a line or a union of broken lines, but I have been unsuccessful in getting anywhere with that approach.
real-analysis linear-algebra general-topology
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
$endgroup$
I want to prove the following:
Let $E subset mathbb{R}^n$ be a vector subspace of $mathbb{R}^n$ which is not equal to $mathbb{R}^n$. Then $mathbb{R}^n setminus E$ is connected if, and only if, the dimension of $E$ is at most $n - 2$ (or equivalently, codimension at least $2$).
For the $implies$ direction, here's what I've got: Let $E$ be a vector subspace with dimension $n - 1$. Then: $$mathbb{R}^n setminus E = A cup B$$
where $A = {v in mathbb{R}^n vert langle v, n rangle < 0 }$ and $B = {v in mathbb{R}^n vert langle v, n rangle > 0 }$ and $n$ is a normal vector to $E$ (which certainly exists). Now, intuitively, I know that $A$ and $B$ are disjoint open sets, but I haven't been able to prove it.
I couldn't think of anything yet regarding the $impliedby$ direction. My strategy was proving that any two points in $mathbb{R}^n setminus E$ can be joined by a line or a union of broken lines, but I have been unsuccessful in getting anywhere with that approach.
real-analysis linear-algebra general-topology
real-analysis linear-algebra general-topology
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
edited Jan 30 at 23:07
Matheus Andrade
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
asked Jan 30 at 22:59
Matheus AndradeMatheus Andrade
1,380418
1,380418
$begingroup$
This cannot be strictly true because $mathbb{R}^n setminus E$ is connected when $E=mathbb{R}^n$.
$endgroup$
– lhf
Jan 30 at 23:03
$begingroup$
Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $varnothing$ is not connected?
$endgroup$
– GEdgar
Jan 30 at 23:04
$begingroup$
@GEdgar Notice I stated codimension $geq 2$.
$endgroup$
– Matheus Andrade
Jan 30 at 23:07
$begingroup$
So: the same question I asked becomes: Why state codimension ${} ge 2$ instead of codimension ${}ne 1$?
$endgroup$
– GEdgar
Jan 30 at 23:12
$begingroup$
No particular reason. They mean the same thing.
$endgroup$
– Matheus Andrade
Jan 30 at 23:14
add a comment |
$begingroup$
This cannot be strictly true because $mathbb{R}^n setminus E$ is connected when $E=mathbb{R}^n$.
$endgroup$
– lhf
Jan 30 at 23:03
$begingroup$
Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $varnothing$ is not connected?
$endgroup$
– GEdgar
Jan 30 at 23:04
$begingroup$
@GEdgar Notice I stated codimension $geq 2$.
$endgroup$
– Matheus Andrade
Jan 30 at 23:07
$begingroup$
So: the same question I asked becomes: Why state codimension ${} ge 2$ instead of codimension ${}ne 1$?
$endgroup$
– GEdgar
Jan 30 at 23:12
$begingroup$
No particular reason. They mean the same thing.
$endgroup$
– Matheus Andrade
Jan 30 at 23:14
$begingroup$
This cannot be strictly true because $mathbb{R}^n setminus E$ is connected when $E=mathbb{R}^n$.
$endgroup$
– lhf
Jan 30 at 23:03
$begingroup$
This cannot be strictly true because $mathbb{R}^n setminus E$ is connected when $E=mathbb{R}^n$.
$endgroup$
– lhf
Jan 30 at 23:03
$begingroup$
Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $varnothing$ is not connected?
$endgroup$
– GEdgar
Jan 30 at 23:04
$begingroup$
Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $varnothing$ is not connected?
$endgroup$
– GEdgar
Jan 30 at 23:04
$begingroup$
@GEdgar Notice I stated codimension $geq 2$.
$endgroup$
– Matheus Andrade
Jan 30 at 23:07
$begingroup$
@GEdgar Notice I stated codimension $geq 2$.
$endgroup$
– Matheus Andrade
Jan 30 at 23:07
$begingroup$
So: the same question I asked becomes: Why state codimension ${} ge 2$ instead of codimension ${}ne 1$?
$endgroup$
– GEdgar
Jan 30 at 23:12
$begingroup$
So: the same question I asked becomes: Why state codimension ${} ge 2$ instead of codimension ${}ne 1$?
$endgroup$
– GEdgar
Jan 30 at 23:12
$begingroup$
No particular reason. They mean the same thing.
$endgroup$
– Matheus Andrade
Jan 30 at 23:14
$begingroup$
No particular reason. They mean the same thing.
$endgroup$
– Matheus Andrade
Jan 30 at 23:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Let $F$ be a linear complement for $E$, that is, $mathbb{R}^n = E oplus F$. Take $x,y in mathbb{R}^n setminus E$.
One path joining $x$ and $y$ is $ x to bar x to bar y to y$, where the bar means projection onto $F$. In the path $bar x to bar y$, you need to avoid the origin.
answered Jan 30 at 23:09
lhflhf
167k11172404
$endgroup$
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
add a comment |
$begingroup$
If $dim (E)=n-1$ let ${e_j:1le jle n-1}$ be an orthonormal basis for $E$ and let $Ecup {e_n}$ be an orthonormal basis for $Bbb R^n.$ In other words $E^{perp}={re_n:rin Bbb R}.$ The projection $P(v)=langle v|e_nrangle$ is continuous from $Bbb R^n$ to $Bbb R$ so $A=P^{-1}(-infty,0)$ and $B=P^{-1}(0,infty)$ are open.
$P$ is Lipschitz-continuous: $|P(v)-P(v')|=|langle (v-v')|e_nrangle|,le |v-v'|cdot |e_n|=|v-v'|.$
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
$endgroup$
– DanielWainfleet
Jan 31 at 0:01
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $F$ be a linear complement for $E$, that is, $mathbb{R}^n = E oplus F$. Take $x,y in mathbb{R}^n setminus E$.
One path joining $x$ and $y$ is $ x to bar x to bar y to y$, where the bar means projection onto $F$. In the path $bar x to bar y$, you need to avoid the origin.
answered Jan 30 at 23:09
lhflhf
167k11172404
$endgroup$
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
add a comment |
$begingroup$
Hint: Let $F$ be a linear complement for $E$, that is, $mathbb{R}^n = E oplus F$. Take $x,y in mathbb{R}^n setminus E$.
One path joining $x$ and $y$ is $ x to bar x to bar y to y$, where the bar means projection onto $F$. In the path $bar x to bar y$, you need to avoid the origin.
answered Jan 30 at 23:09
lhflhf
167k11172404
$endgroup$
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
add a comment |
$begingroup$
Hint: Let $F$ be a linear complement for $E$, that is, $mathbb{R}^n = E oplus F$. Take $x,y in mathbb{R}^n setminus E$.
One path joining $x$ and $y$ is $ x to bar x to bar y to y$, where the bar means projection onto $F$. In the path $bar x to bar y$, you need to avoid the origin.
answered Jan 30 at 23:09
lhflhf
167k11172404
$endgroup$
Hint: Let $F$ be a linear complement for $E$, that is, $mathbb{R}^n = E oplus F$. Take $x,y in mathbb{R}^n setminus E$.
One path joining $x$ and $y$ is $ x to bar x to bar y to y$, where the bar means projection onto $F$. In the path $bar x to bar y$, you need to avoid the origin.
answered Jan 30 at 23:09
lhflhf
167k11172404
answered Jan 30 at 23:09
lhflhf
167k11172404
answered Jan 30 at 23:09
lhflhf
167k11172404
answered Jan 30 at 23:09
lhflhf
167k11172404
167k11172404
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
add a comment |
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
Thanks! Is my $implies$ direction correct then? Also, it's not really clear what you mean by (could you make that more explicit?) $x to bar{x} to bar{y} to y$. Does each $to$ mean the line joining the two points?
$endgroup$
– Matheus Andrade
Jan 30 at 23:21
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
@MatheusAndrade, yes, they are line segments, unless $bar x to bar y$ passes through the origin, but that's easily fixed.
$endgroup$
– lhf
Jan 31 at 0:01
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
$begingroup$
Given any two points $x, y$ we can make a path joining them that does not pass through the origin: choose a line that passes through $x$ that does not go through the origin with inclination $alpha$ and take another line passing through $y$ with inclination $beta neq alpha$ that does not pass through the origin. It intersects the other line and so we can make a continuous path joining $x$ and $y$ that does not pass through the origin. Is that it? Also, this may be obvious but why is each of the $to$ contained in $mathbb{R}^n setminus E$?
$endgroup$
– Matheus Andrade
Jan 31 at 0:05
add a comment |
$begingroup$
If $dim (E)=n-1$ let ${e_j:1le jle n-1}$ be an orthonormal basis for $E$ and let $Ecup {e_n}$ be an orthonormal basis for $Bbb R^n.$ In other words $E^{perp}={re_n:rin Bbb R}.$ The projection $P(v)=langle v|e_nrangle$ is continuous from $Bbb R^n$ to $Bbb R$ so $A=P^{-1}(-infty,0)$ and $B=P^{-1}(0,infty)$ are open.
$P$ is Lipschitz-continuous: $|P(v)-P(v')|=|langle (v-v')|e_nrangle|,le |v-v'|cdot |e_n|=|v-v'|.$
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
$endgroup$
– DanielWainfleet
Jan 31 at 0:01
add a comment |
$begingroup$
If $dim (E)=n-1$ let ${e_j:1le jle n-1}$ be an orthonormal basis for $E$ and let $Ecup {e_n}$ be an orthonormal basis for $Bbb R^n.$ In other words $E^{perp}={re_n:rin Bbb R}.$ The projection $P(v)=langle v|e_nrangle$ is continuous from $Bbb R^n$ to $Bbb R$ so $A=P^{-1}(-infty,0)$ and $B=P^{-1}(0,infty)$ are open.
$P$ is Lipschitz-continuous: $|P(v)-P(v')|=|langle (v-v')|e_nrangle|,le |v-v'|cdot |e_n|=|v-v'|.$
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
$endgroup$
– DanielWainfleet
Jan 31 at 0:01
add a comment |
$begingroup$
If $dim (E)=n-1$ let ${e_j:1le jle n-1}$ be an orthonormal basis for $E$ and let $Ecup {e_n}$ be an orthonormal basis for $Bbb R^n.$ In other words $E^{perp}={re_n:rin Bbb R}.$ The projection $P(v)=langle v|e_nrangle$ is continuous from $Bbb R^n$ to $Bbb R$ so $A=P^{-1}(-infty,0)$ and $B=P^{-1}(0,infty)$ are open.
$P$ is Lipschitz-continuous: $|P(v)-P(v')|=|langle (v-v')|e_nrangle|,le |v-v'|cdot |e_n|=|v-v'|.$
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
If $dim (E)=n-1$ let ${e_j:1le jle n-1}$ be an orthonormal basis for $E$ and let $Ecup {e_n}$ be an orthonormal basis for $Bbb R^n.$ In other words $E^{perp}={re_n:rin Bbb R}.$ The projection $P(v)=langle v|e_nrangle$ is continuous from $Bbb R^n$ to $Bbb R$ so $A=P^{-1}(-infty,0)$ and $B=P^{-1}(0,infty)$ are open.
$P$ is Lipschitz-continuous: $|P(v)-P(v')|=|langle (v-v')|e_nrangle|,le |v-v'|cdot |e_n|=|v-v'|.$
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
answered Jan 30 at 23:49
DanielWainfleetDanielWainfleet
35.7k31648
35.7k31648
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
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– DanielWainfleet
Jan 31 at 0:01
add a comment |
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
$endgroup$
– DanielWainfleet
Jan 31 at 0:01
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
$endgroup$
– DanielWainfleet
Jan 31 at 0:01
$begingroup$
An intuitive explanation of why this does not apply when $dim Ele n-2$ is that $E^{perp}$ minus the origin is still connected so the projection of $Bbb R^n$ onto $E^{perp}$ maps $Bbb R^nsetminus E$ onto a connected set.
$endgroup$
– DanielWainfleet
Jan 31 at 0:01
add a comment |
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This cannot be strictly true because $mathbb{R}^n setminus E$ is connected when $E=mathbb{R}^n$.
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– lhf
Jan 30 at 23:03
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Why say codimenison ${} < 2$ instead of codimension $1$? To allow for codimension zero? Then the claim is that $varnothing$ is not connected?
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– GEdgar
Jan 30 at 23:04
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@GEdgar Notice I stated codimension $geq 2$.
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– Matheus Andrade
Jan 30 at 23:07
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So: the same question I asked becomes: Why state codimension ${} ge 2$ instead of codimension ${}ne 1$?
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– GEdgar
Jan 30 at 23:12
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No particular reason. They mean the same thing.
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– Matheus Andrade
Jan 30 at 23:14