Show that the union of convex sets does not have to be convex.
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2
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The following is an example that I've come up with:
Suppose that $pin A$ and $qin B$ so that $p,q in Acup B$, where $A$ and $B$ are two mutually disjoint, convex, unit circles centered at $x=0,2$ in $mathbb{R^2}$, respectively. Also let $p:=(frac{1}{2},0)$ and $q:= (frac{3}{2},0)$. The set of points satisfying $lambda p + (1-lambda)q$ for $0 < lambda < 1$ forms a line between $p$ and $q$. But for $lambda = frac{1}{2}$, we have that $z = frac{1}{2}p + (1-frac{1}{2})q = frac{1}{2}(p+q)=(1,0)$, which is not in $Acup B$.
I was wondering if there's a simpler example that shows that the union of two convex sets does not have to be convex?
analysis
asked 5 hours ago
K.M
648312
add a comment |
up vote
2
down vote
favorite
The following is an example that I've come up with:
Suppose that $pin A$ and $qin B$ so that $p,q in Acup B$, where $A$ and $B$ are two mutually disjoint, convex, unit circles centered at $x=0,2$ in $mathbb{R^2}$, respectively. Also let $p:=(frac{1}{2},0)$ and $q:= (frac{3}{2},0)$. The set of points satisfying $lambda p + (1-lambda)q$ for $0 < lambda < 1$ forms a line between $p$ and $q$. But for $lambda = frac{1}{2}$, we have that $z = frac{1}{2}p + (1-frac{1}{2})q = frac{1}{2}(p+q)=(1,0)$, which is not in $Acup B$.
I was wondering if there's a simpler example that shows that the union of two convex sets does not have to be convex?
analysis
asked 5 hours ago
K.M
648312
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The following is an example that I've come up with:
Suppose that $pin A$ and $qin B$ so that $p,q in Acup B$, where $A$ and $B$ are two mutually disjoint, convex, unit circles centered at $x=0,2$ in $mathbb{R^2}$, respectively. Also let $p:=(frac{1}{2},0)$ and $q:= (frac{3}{2},0)$. The set of points satisfying $lambda p + (1-lambda)q$ for $0 < lambda < 1$ forms a line between $p$ and $q$. But for $lambda = frac{1}{2}$, we have that $z = frac{1}{2}p + (1-frac{1}{2})q = frac{1}{2}(p+q)=(1,0)$, which is not in $Acup B$.
I was wondering if there's a simpler example that shows that the union of two convex sets does not have to be convex?
analysis
asked 5 hours ago
K.M
648312
The following is an example that I've come up with:
Suppose that $pin A$ and $qin B$ so that $p,q in Acup B$, where $A$ and $B$ are two mutually disjoint, convex, unit circles centered at $x=0,2$ in $mathbb{R^2}$, respectively. Also let $p:=(frac{1}{2},0)$ and $q:= (frac{3}{2},0)$. The set of points satisfying $lambda p + (1-lambda)q$ for $0 < lambda < 1$ forms a line between $p$ and $q$. But for $lambda = frac{1}{2}$, we have that $z = frac{1}{2}p + (1-frac{1}{2})q = frac{1}{2}(p+q)=(1,0)$, which is not in $Acup B$.
I was wondering if there's a simpler example that shows that the union of two convex sets does not have to be convex?
analysis
analysis
asked 5 hours ago
K.M
648312
asked 5 hours ago
K.M
648312
asked 5 hours ago
K.M
648312
asked 5 hours ago
K.M
648312
asked 5 hours ago
K.M
648312
648312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
$(0,1) cup (2,3)$ is a simpler example. $frac {0.5+2.5} 2$ does not belong to this union.
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
add a comment |
up vote
3
down vote
Even easier: two points in the plane.
answered 5 hours ago
ncmathsadist
42k259101
Or in the line.
– Martin Argerami
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$(0,1) cup (2,3)$ is a simpler example. $frac {0.5+2.5} 2$ does not belong to this union.
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
add a comment |
up vote
4
down vote
accepted
$(0,1) cup (2,3)$ is a simpler example. $frac {0.5+2.5} 2$ does not belong to this union.
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$(0,1) cup (2,3)$ is a simpler example. $frac {0.5+2.5} 2$ does not belong to this union.
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
$(0,1) cup (2,3)$ is a simpler example. $frac {0.5+2.5} 2$ does not belong to this union.
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
answered 5 hours ago
Kavi Rama Murthy
45.1k31852
45.1k31852
add a comment |
add a comment |
up vote
3
down vote
Even easier: two points in the plane.
answered 5 hours ago
ncmathsadist
42k259101
Or in the line.
– Martin Argerami
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
add a comment |
up vote
3
down vote
Even easier: two points in the plane.
answered 5 hours ago
ncmathsadist
42k259101
Or in the line.
– Martin Argerami
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Even easier: two points in the plane.
answered 5 hours ago
ncmathsadist
42k259101
Even easier: two points in the plane.
answered 5 hours ago
ncmathsadist
42k259101
answered 5 hours ago
ncmathsadist
42k259101
answered 5 hours ago
ncmathsadist
42k259101
answered 5 hours ago
ncmathsadist
42k259101
42k259101
Or in the line.
– Martin Argerami
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
add a comment |
Or in the line.
– Martin Argerami
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
Or in the line.
– Martin Argerami
4 hours ago
Or in the line.
– Martin Argerami
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
They "puncture" this conjecture oh-so-prettily.
– ncmathsadist
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
when you say two points in the plane, do you mean that each point is a trivial convex set?
– K.M
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
Verily. A point is about as convex as you can get.
– ncmathsadist
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
@ncmathsadist: wouldn't this be considered more of a counterexample?
– K.M
4 hours ago
add a comment |
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