Geometrical interpretation of disjoint union
$begingroup$
Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.
algebraic-topology
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
$endgroup$
add a comment |
$begingroup$
Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.
algebraic-topology
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
$endgroup$
$begingroup$
What makes you think there is one? The disjoint union is an operation on sets.
$endgroup$
– John Douma
Jan 4 at 4:13
2
$begingroup$
You mean disjoint union of topological spaces?
$endgroup$
– Randall
Jan 4 at 4:21
$begingroup$
Are you perhaps looking for a visual representation of disjoint union, e.g. this?
$endgroup$
– Alexis
Jan 4 at 4:44
$begingroup$
@Randall yes. Do u have any example?
$endgroup$
– Prince Thomas
Jan 4 at 5:28
add a comment |
$begingroup$
Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.
algebraic-topology
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
$endgroup$
Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.
algebraic-topology
algebraic-topology
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
asked Jan 4 at 3:40
Prince ThomasPrince Thomas
625311
625311
$begingroup$
What makes you think there is one? The disjoint union is an operation on sets.
$endgroup$
– John Douma
Jan 4 at 4:13
2
$begingroup$
You mean disjoint union of topological spaces?
$endgroup$
– Randall
Jan 4 at 4:21
$begingroup$
Are you perhaps looking for a visual representation of disjoint union, e.g. this?
$endgroup$
– Alexis
Jan 4 at 4:44
$begingroup$
@Randall yes. Do u have any example?
$endgroup$
– Prince Thomas
Jan 4 at 5:28
add a comment |
$begingroup$
What makes you think there is one? The disjoint union is an operation on sets.
$endgroup$
– John Douma
Jan 4 at 4:13
2
$begingroup$
You mean disjoint union of topological spaces?
$endgroup$
– Randall
Jan 4 at 4:21
$begingroup$
Are you perhaps looking for a visual representation of disjoint union, e.g. this?
$endgroup$
– Alexis
Jan 4 at 4:44
$begingroup$
@Randall yes. Do u have any example?
$endgroup$
– Prince Thomas
Jan 4 at 5:28
$begingroup$
What makes you think there is one? The disjoint union is an operation on sets.
$endgroup$
– John Douma
Jan 4 at 4:13
$begingroup$
What makes you think there is one? The disjoint union is an operation on sets.
$endgroup$
– John Douma
Jan 4 at 4:13
2
2
$begingroup$
You mean disjoint union of topological spaces?
$endgroup$
– Randall
Jan 4 at 4:21
$begingroup$
You mean disjoint union of topological spaces?
$endgroup$
– Randall
Jan 4 at 4:21
$begingroup$
Are you perhaps looking for a visual representation of disjoint union, e.g. this?
$endgroup$
– Alexis
Jan 4 at 4:44
$begingroup$
Are you perhaps looking for a visual representation of disjoint union, e.g. this?
$endgroup$
– Alexis
Jan 4 at 4:44
$begingroup$
@Randall yes. Do u have any example?
$endgroup$
– Prince Thomas
Jan 4 at 5:28
$begingroup$
@Randall yes. Do u have any example?
$endgroup$
– Prince Thomas
Jan 4 at 5:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.
So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.
I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.
So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.
I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.
So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.
I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
add a comment |
$begingroup$
Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.
So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.
I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
$endgroup$
Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.
So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.
I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
answered Jan 4 at 18:26
Ronnie BrownRonnie Brown
12.1k12939
12.1k12939
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$begingroup$
What makes you think there is one? The disjoint union is an operation on sets.
$endgroup$
– John Douma
Jan 4 at 4:13
2
$begingroup$
You mean disjoint union of topological spaces?
$endgroup$
– Randall
Jan 4 at 4:21
$begingroup$
Are you perhaps looking for a visual representation of disjoint union, e.g. this?
$endgroup$
– Alexis
Jan 4 at 4:44
$begingroup$
@Randall yes. Do u have any example?
$endgroup$
– Prince Thomas
Jan 4 at 5:28