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Proof that the triangle inequality holds in the following metric?
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Let's say we have a finite metric space $(S,d)$. I need to prove that $P(S)$ (the power set of $S$) is a metric space with the metric $bar d:P(S)times P(S)to[0,infty)$ defined as
$$bar d(X,Y)=sum_{x,yin Xcup Y}d(x,y)-sum_{x,yin Xcap Y}d(x,y)$$
Non-negativity, identity of discernibles and symmetry are almost obvious, but what about the triangle inequality? I'm not able to prove that. I cannot find a value between $bar d(X,Z)$ and $bar d(X,Y)+bar d(Y,Z)$ that sends me in the right track. Thanks.
metric-spaces
asked 22 hours ago
Garmekain
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Let's say we have a finite metric space $(S,d)$. I need to prove that $P(S)$ (the power set of $S$) is a metric space with the metric $bar d:P(S)times P(S)to[0,infty)$ defined as
$$bar d(X,Y)=sum_{x,yin Xcup Y}d(x,y)-sum_{x,yin Xcap Y}d(x,y)$$
Non-negativity, identity of discernibles and symmetry are almost obvious, but what about the triangle inequality? I'm not able to prove that. I cannot find a value between $bar d(X,Z)$ and $bar d(X,Y)+bar d(Y,Z)$ that sends me in the right track. Thanks.
metric-spaces
asked 22 hours ago
Garmekain
1,243719
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1
down vote
favorite
up vote
1
down vote
favorite
Let's say we have a finite metric space $(S,d)$. I need to prove that $P(S)$ (the power set of $S$) is a metric space with the metric $bar d:P(S)times P(S)to[0,infty)$ defined as
$$bar d(X,Y)=sum_{x,yin Xcup Y}d(x,y)-sum_{x,yin Xcap Y}d(x,y)$$
Non-negativity, identity of discernibles and symmetry are almost obvious, but what about the triangle inequality? I'm not able to prove that. I cannot find a value between $bar d(X,Z)$ and $bar d(X,Y)+bar d(Y,Z)$ that sends me in the right track. Thanks.
metric-spaces
asked 22 hours ago
Garmekain
1,243719
Let's say we have a finite metric space $(S,d)$. I need to prove that $P(S)$ (the power set of $S$) is a metric space with the metric $bar d:P(S)times P(S)to[0,infty)$ defined as
$$bar d(X,Y)=sum_{x,yin Xcup Y}d(x,y)-sum_{x,yin Xcap Y}d(x,y)$$
Non-negativity, identity of discernibles and symmetry are almost obvious, but what about the triangle inequality? I'm not able to prove that. I cannot find a value between $bar d(X,Z)$ and $bar d(X,Y)+bar d(Y,Z)$ that sends me in the right track. Thanks.
metric-spaces
metric-spaces
asked 22 hours ago
Garmekain
1,243719
asked 22 hours ago
Garmekain
1,243719
edited 21 hours ago
asked 22 hours ago
Garmekain
1,243719
asked 22 hours ago
Garmekain
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asked 22 hours ago
Garmekain
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1,243719
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2 Answers
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There is a quite nice visual answer to this question.
Suppose we treat the venn diagram of $A$, $B$ and $C$-in OP's example, $X$, $Y$ and $Z$ respectively- as an unordered graph $G$ where each vertex is a portion of the venn diagram with no intersection with the rest, and an edge between two portions corresponds to the sum of the distances between the points of each set, so $d(a,b)=sum_{x,yin acup b}d(x,y)$ for any distinct nodes $a,bin V(G)$-where $V(G)$ is the set of vertices of $G$- because $acap b=emptyset$ for all $a,bin V(G)$. We will enumerate the nodes like so
This will be useful later.
For example, assuming wether the definition of $d$ is over $S$ or $P(S)$(where $P(S)$ is the power set of $S$), is given by context, so we don't need the bar to distinguish them, and $^c$ is the complement of a set, then $d(Acap B^ccap C^c,Acap Bcap C^c)$ could be defined with the following graph:
meaning that
$$d(Acap B^ccap C^c,Acap Bcap C^c)=sum_{x,yin Acap B^ccap C^c}d(x,y)+sum_{x,yin Acap Bcap C^c}d(x,y)+{sum_{xin Acap B^ccap C^c}sum_{yin Acap Bcap C^c}d(x,y)}+{sum_{xin Acap Bcap C^c}sum_{yin Acap B^ccap C^c}d(x,y)}$$
Knowing that edges correspond to sums, and that adding an edge just adds to the sum, then $d(A,B)$ could be defined with the following one:
and $d(B,C)$ with
Knowing that unioning the graphs of $d(A,B)$ and $d(B,C)$ gives a number less than their sum, because we are only counting once even the sum appears more than once, we will prove that the union of these graphs is indeed larger than the graph of $d(A,C)$. If we union those graphs together we get the graph
comparing with the graph of $d(A,C)$, which is
we can see that the only edge of the graph of $d(A,C)$ which is not contained in the graph of the union is the edge between the nodes $3$ and $5$, so if there is a way to compensate that edge in the graph of the union we are done, and there is.
We have a lot of unused edges going from the node $1$, and two of those edges are from $3$ to $1$, and from $1$ to $5$. This means that for every sum of the graph of $d(x,z)$ where $xintext{Region 5}$ and $zintext{Region 3}$ there is a sum in the graph of the union where for some $yintext{Region 1}$ we have that $d(x,z)leq d(x,y)+d(y,z)$, meaning that we have compensated that missing edge on the graph of the union, meaning that the union of the graphs $d(A,B)$ and $d(B,C)$ are indeed bigger than $d(A,C)$, so $d(A,B)+d(B,C)$ must be bigger than $d(A,C)$, proving that OP's $bar d$ is in fact a metric.
answered 13 hours ago
Garmekain
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The 8 sets
$$
A_{xyz}=Xcap Y cap Z,,,
A_{xyz'}=Xcap Y cap Z^c,,,
A_{xy'z}=Xcap Y^c cap Z,,,
A_{xy'z'}=Xcap Y^c cap Z^c,\
A_{x'yz}=X^ccap Y cap Z,,,
A_{x'yz'}=Xcap Y cap Z^c,,,
A_{x'y'z}=X^ccap Y^c cap Z,,,
A_{x'y'z'}=X^ccap Y^c cap Z^c,
$$
are mutually disjoint, and their union is $Xcup Ycup Z$.
To see this draw the Venn diagram.
Then
$$
overline{d}(X,Y)=sum_{x,yin Xcup Y}-
sum_{x,yin Xcap Y}=sum_{x,yin A_{xy'z'}}+sum_{x,yin A_{xy'z}}+
sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{x'yz}}
$$
and
$$
overline{d}(Y,Z)=sum_{x,yin Ycup Z}-
sum_{x,yin Ycap Z}=sum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}
+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}
$$
and hence
$$
overline{d}(X,Y)+overline{d}(Y,Z)gesum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}=overline{d}(X,Z)
$$
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
1
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 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
1
down vote
There is a quite nice visual answer to this question.
Suppose we treat the venn diagram of $A$, $B$ and $C$-in OP's example, $X$, $Y$ and $Z$ respectively- as an unordered graph $G$ where each vertex is a portion of the venn diagram with no intersection with the rest, and an edge between two portions corresponds to the sum of the distances between the points of each set, so $d(a,b)=sum_{x,yin acup b}d(x,y)$ for any distinct nodes $a,bin V(G)$-where $V(G)$ is the set of vertices of $G$- because $acap b=emptyset$ for all $a,bin V(G)$. We will enumerate the nodes like so
This will be useful later.
For example, assuming wether the definition of $d$ is over $S$ or $P(S)$(where $P(S)$ is the power set of $S$), is given by context, so we don't need the bar to distinguish them, and $^c$ is the complement of a set, then $d(Acap B^ccap C^c,Acap Bcap C^c)$ could be defined with the following graph:
meaning that
$$d(Acap B^ccap C^c,Acap Bcap C^c)=sum_{x,yin Acap B^ccap C^c}d(x,y)+sum_{x,yin Acap Bcap C^c}d(x,y)+{sum_{xin Acap B^ccap C^c}sum_{yin Acap Bcap C^c}d(x,y)}+{sum_{xin Acap Bcap C^c}sum_{yin Acap B^ccap C^c}d(x,y)}$$
Knowing that edges correspond to sums, and that adding an edge just adds to the sum, then $d(A,B)$ could be defined with the following one:
and $d(B,C)$ with
Knowing that unioning the graphs of $d(A,B)$ and $d(B,C)$ gives a number less than their sum, because we are only counting once even the sum appears more than once, we will prove that the union of these graphs is indeed larger than the graph of $d(A,C)$. If we union those graphs together we get the graph
comparing with the graph of $d(A,C)$, which is
we can see that the only edge of the graph of $d(A,C)$ which is not contained in the graph of the union is the edge between the nodes $3$ and $5$, so if there is a way to compensate that edge in the graph of the union we are done, and there is.
We have a lot of unused edges going from the node $1$, and two of those edges are from $3$ to $1$, and from $1$ to $5$. This means that for every sum of the graph of $d(x,z)$ where $xintext{Region 5}$ and $zintext{Region 3}$ there is a sum in the graph of the union where for some $yintext{Region 1}$ we have that $d(x,z)leq d(x,y)+d(y,z)$, meaning that we have compensated that missing edge on the graph of the union, meaning that the union of the graphs $d(A,B)$ and $d(B,C)$ are indeed bigger than $d(A,C)$, so $d(A,B)+d(B,C)$ must be bigger than $d(A,C)$, proving that OP's $bar d$ is in fact a metric.
answered 13 hours ago
Garmekain
1,243719
add a comment |
up vote
1
down vote
There is a quite nice visual answer to this question.
Suppose we treat the venn diagram of $A$, $B$ and $C$-in OP's example, $X$, $Y$ and $Z$ respectively- as an unordered graph $G$ where each vertex is a portion of the venn diagram with no intersection with the rest, and an edge between two portions corresponds to the sum of the distances between the points of each set, so $d(a,b)=sum_{x,yin acup b}d(x,y)$ for any distinct nodes $a,bin V(G)$-where $V(G)$ is the set of vertices of $G$- because $acap b=emptyset$ for all $a,bin V(G)$. We will enumerate the nodes like so
This will be useful later.
For example, assuming wether the definition of $d$ is over $S$ or $P(S)$(where $P(S)$ is the power set of $S$), is given by context, so we don't need the bar to distinguish them, and $^c$ is the complement of a set, then $d(Acap B^ccap C^c,Acap Bcap C^c)$ could be defined with the following graph:
meaning that
$$d(Acap B^ccap C^c,Acap Bcap C^c)=sum_{x,yin Acap B^ccap C^c}d(x,y)+sum_{x,yin Acap Bcap C^c}d(x,y)+{sum_{xin Acap B^ccap C^c}sum_{yin Acap Bcap C^c}d(x,y)}+{sum_{xin Acap Bcap C^c}sum_{yin Acap B^ccap C^c}d(x,y)}$$
Knowing that edges correspond to sums, and that adding an edge just adds to the sum, then $d(A,B)$ could be defined with the following one:
and $d(B,C)$ with
Knowing that unioning the graphs of $d(A,B)$ and $d(B,C)$ gives a number less than their sum, because we are only counting once even the sum appears more than once, we will prove that the union of these graphs is indeed larger than the graph of $d(A,C)$. If we union those graphs together we get the graph
comparing with the graph of $d(A,C)$, which is
we can see that the only edge of the graph of $d(A,C)$ which is not contained in the graph of the union is the edge between the nodes $3$ and $5$, so if there is a way to compensate that edge in the graph of the union we are done, and there is.
We have a lot of unused edges going from the node $1$, and two of those edges are from $3$ to $1$, and from $1$ to $5$. This means that for every sum of the graph of $d(x,z)$ where $xintext{Region 5}$ and $zintext{Region 3}$ there is a sum in the graph of the union where for some $yintext{Region 1}$ we have that $d(x,z)leq d(x,y)+d(y,z)$, meaning that we have compensated that missing edge on the graph of the union, meaning that the union of the graphs $d(A,B)$ and $d(B,C)$ are indeed bigger than $d(A,C)$, so $d(A,B)+d(B,C)$ must be bigger than $d(A,C)$, proving that OP's $bar d$ is in fact a metric.
answered 13 hours ago
Garmekain
1,243719
add a comment |
up vote
1
down vote
up vote
1
down vote
There is a quite nice visual answer to this question.
Suppose we treat the venn diagram of $A$, $B$ and $C$-in OP's example, $X$, $Y$ and $Z$ respectively- as an unordered graph $G$ where each vertex is a portion of the venn diagram with no intersection with the rest, and an edge between two portions corresponds to the sum of the distances between the points of each set, so $d(a,b)=sum_{x,yin acup b}d(x,y)$ for any distinct nodes $a,bin V(G)$-where $V(G)$ is the set of vertices of $G$- because $acap b=emptyset$ for all $a,bin V(G)$. We will enumerate the nodes like so
This will be useful later.
For example, assuming wether the definition of $d$ is over $S$ or $P(S)$(where $P(S)$ is the power set of $S$), is given by context, so we don't need the bar to distinguish them, and $^c$ is the complement of a set, then $d(Acap B^ccap C^c,Acap Bcap C^c)$ could be defined with the following graph:
meaning that
$$d(Acap B^ccap C^c,Acap Bcap C^c)=sum_{x,yin Acap B^ccap C^c}d(x,y)+sum_{x,yin Acap Bcap C^c}d(x,y)+{sum_{xin Acap B^ccap C^c}sum_{yin Acap Bcap C^c}d(x,y)}+{sum_{xin Acap Bcap C^c}sum_{yin Acap B^ccap C^c}d(x,y)}$$
Knowing that edges correspond to sums, and that adding an edge just adds to the sum, then $d(A,B)$ could be defined with the following one:
and $d(B,C)$ with
Knowing that unioning the graphs of $d(A,B)$ and $d(B,C)$ gives a number less than their sum, because we are only counting once even the sum appears more than once, we will prove that the union of these graphs is indeed larger than the graph of $d(A,C)$. If we union those graphs together we get the graph
comparing with the graph of $d(A,C)$, which is
we can see that the only edge of the graph of $d(A,C)$ which is not contained in the graph of the union is the edge between the nodes $3$ and $5$, so if there is a way to compensate that edge in the graph of the union we are done, and there is.
We have a lot of unused edges going from the node $1$, and two of those edges are from $3$ to $1$, and from $1$ to $5$. This means that for every sum of the graph of $d(x,z)$ where $xintext{Region 5}$ and $zintext{Region 3}$ there is a sum in the graph of the union where for some $yintext{Region 1}$ we have that $d(x,z)leq d(x,y)+d(y,z)$, meaning that we have compensated that missing edge on the graph of the union, meaning that the union of the graphs $d(A,B)$ and $d(B,C)$ are indeed bigger than $d(A,C)$, so $d(A,B)+d(B,C)$ must be bigger than $d(A,C)$, proving that OP's $bar d$ is in fact a metric.
answered 13 hours ago
Garmekain
1,243719
There is a quite nice visual answer to this question.
Suppose we treat the venn diagram of $A$, $B$ and $C$-in OP's example, $X$, $Y$ and $Z$ respectively- as an unordered graph $G$ where each vertex is a portion of the venn diagram with no intersection with the rest, and an edge between two portions corresponds to the sum of the distances between the points of each set, so $d(a,b)=sum_{x,yin acup b}d(x,y)$ for any distinct nodes $a,bin V(G)$-where $V(G)$ is the set of vertices of $G$- because $acap b=emptyset$ for all $a,bin V(G)$. We will enumerate the nodes like so
This will be useful later.
For example, assuming wether the definition of $d$ is over $S$ or $P(S)$(where $P(S)$ is the power set of $S$), is given by context, so we don't need the bar to distinguish them, and $^c$ is the complement of a set, then $d(Acap B^ccap C^c,Acap Bcap C^c)$ could be defined with the following graph:
meaning that
$$d(Acap B^ccap C^c,Acap Bcap C^c)=sum_{x,yin Acap B^ccap C^c}d(x,y)+sum_{x,yin Acap Bcap C^c}d(x,y)+{sum_{xin Acap B^ccap C^c}sum_{yin Acap Bcap C^c}d(x,y)}+{sum_{xin Acap Bcap C^c}sum_{yin Acap B^ccap C^c}d(x,y)}$$
Knowing that edges correspond to sums, and that adding an edge just adds to the sum, then $d(A,B)$ could be defined with the following one:
and $d(B,C)$ with
Knowing that unioning the graphs of $d(A,B)$ and $d(B,C)$ gives a number less than their sum, because we are only counting once even the sum appears more than once, we will prove that the union of these graphs is indeed larger than the graph of $d(A,C)$. If we union those graphs together we get the graph
comparing with the graph of $d(A,C)$, which is
we can see that the only edge of the graph of $d(A,C)$ which is not contained in the graph of the union is the edge between the nodes $3$ and $5$, so if there is a way to compensate that edge in the graph of the union we are done, and there is.
We have a lot of unused edges going from the node $1$, and two of those edges are from $3$ to $1$, and from $1$ to $5$. This means that for every sum of the graph of $d(x,z)$ where $xintext{Region 5}$ and $zintext{Region 3}$ there is a sum in the graph of the union where for some $yintext{Region 1}$ we have that $d(x,z)leq d(x,y)+d(y,z)$, meaning that we have compensated that missing edge on the graph of the union, meaning that the union of the graphs $d(A,B)$ and $d(B,C)$ are indeed bigger than $d(A,C)$, so $d(A,B)+d(B,C)$ must be bigger than $d(A,C)$, proving that OP's $bar d$ is in fact a metric.
answered 13 hours ago
Garmekain
1,243719
edited 11 hours ago
answered 13 hours ago
Garmekain
1,243719
answered 13 hours ago
Garmekain
1,243719
answered 13 hours ago
Garmekain
1,243719
1,243719
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add a comment |
up vote
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The 8 sets
$$
A_{xyz}=Xcap Y cap Z,,,
A_{xyz'}=Xcap Y cap Z^c,,,
A_{xy'z}=Xcap Y^c cap Z,,,
A_{xy'z'}=Xcap Y^c cap Z^c,\
A_{x'yz}=X^ccap Y cap Z,,,
A_{x'yz'}=Xcap Y cap Z^c,,,
A_{x'y'z}=X^ccap Y^c cap Z,,,
A_{x'y'z'}=X^ccap Y^c cap Z^c,
$$
are mutually disjoint, and their union is $Xcup Ycup Z$.
To see this draw the Venn diagram.
Then
$$
overline{d}(X,Y)=sum_{x,yin Xcup Y}-
sum_{x,yin Xcap Y}=sum_{x,yin A_{xy'z'}}+sum_{x,yin A_{xy'z}}+
sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{x'yz}}
$$
and
$$
overline{d}(Y,Z)=sum_{x,yin Ycup Z}-
sum_{x,yin Ycap Z}=sum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}
+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}
$$
and hence
$$
overline{d}(X,Y)+overline{d}(Y,Z)gesum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}=overline{d}(X,Z)
$$
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
1
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 hours ago
add a comment |
up vote
0
down vote
The 8 sets
$$
A_{xyz}=Xcap Y cap Z,,,
A_{xyz'}=Xcap Y cap Z^c,,,
A_{xy'z}=Xcap Y^c cap Z,,,
A_{xy'z'}=Xcap Y^c cap Z^c,\
A_{x'yz}=X^ccap Y cap Z,,,
A_{x'yz'}=Xcap Y cap Z^c,,,
A_{x'y'z}=X^ccap Y^c cap Z,,,
A_{x'y'z'}=X^ccap Y^c cap Z^c,
$$
are mutually disjoint, and their union is $Xcup Ycup Z$.
To see this draw the Venn diagram.
Then
$$
overline{d}(X,Y)=sum_{x,yin Xcup Y}-
sum_{x,yin Xcap Y}=sum_{x,yin A_{xy'z'}}+sum_{x,yin A_{xy'z}}+
sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{x'yz}}
$$
and
$$
overline{d}(Y,Z)=sum_{x,yin Ycup Z}-
sum_{x,yin Ycap Z}=sum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}
+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}
$$
and hence
$$
overline{d}(X,Y)+overline{d}(Y,Z)gesum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}=overline{d}(X,Z)
$$
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
1
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
The 8 sets
$$
A_{xyz}=Xcap Y cap Z,,,
A_{xyz'}=Xcap Y cap Z^c,,,
A_{xy'z}=Xcap Y^c cap Z,,,
A_{xy'z'}=Xcap Y^c cap Z^c,\
A_{x'yz}=X^ccap Y cap Z,,,
A_{x'yz'}=Xcap Y cap Z^c,,,
A_{x'y'z}=X^ccap Y^c cap Z,,,
A_{x'y'z'}=X^ccap Y^c cap Z^c,
$$
are mutually disjoint, and their union is $Xcup Ycup Z$.
To see this draw the Venn diagram.
Then
$$
overline{d}(X,Y)=sum_{x,yin Xcup Y}-
sum_{x,yin Xcap Y}=sum_{x,yin A_{xy'z'}}+sum_{x,yin A_{xy'z}}+
sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{x'yz}}
$$
and
$$
overline{d}(Y,Z)=sum_{x,yin Ycup Z}-
sum_{x,yin Ycap Z}=sum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}
+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}
$$
and hence
$$
overline{d}(X,Y)+overline{d}(Y,Z)gesum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}=overline{d}(X,Z)
$$
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
The 8 sets
$$
A_{xyz}=Xcap Y cap Z,,,
A_{xyz'}=Xcap Y cap Z^c,,,
A_{xy'z}=Xcap Y^c cap Z,,,
A_{xy'z'}=Xcap Y^c cap Z^c,\
A_{x'yz}=X^ccap Y cap Z,,,
A_{x'yz'}=Xcap Y cap Z^c,,,
A_{x'y'z}=X^ccap Y^c cap Z,,,
A_{x'y'z'}=X^ccap Y^c cap Z^c,
$$
are mutually disjoint, and their union is $Xcup Ycup Z$.
To see this draw the Venn diagram.
Then
$$
overline{d}(X,Y)=sum_{x,yin Xcup Y}-
sum_{x,yin Xcap Y}=sum_{x,yin A_{xy'z'}}+sum_{x,yin A_{xy'z}}+
sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{x'yz}}
$$
and
$$
overline{d}(Y,Z)=sum_{x,yin Ycup Z}-
sum_{x,yin Ycap Z}=sum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}
+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}
$$
and hence
$$
overline{d}(X,Y)+overline{d}(Y,Z)gesum_{x,yin A_{xyz'}}+sum_{x,yin A_{x'yz'}}+sum_{x,yin A_{xy'z}}+sum_{x,yin A_{x'y'z}}=overline{d}(X,Z)
$$
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
answered 21 hours ago
Yiorgos S. Smyrlis
61.4k1383161
61.4k1383161
1
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 hours ago
add a comment |
1
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 hours ago
1
1
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
there is a mistake in some of the equations: you forget terms of the form $d(x,y)$ where $x,y$ are not in the same $A_{dots}$-set.
– supinf
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 hours ago
Drawing a venn diagram helped me out. Thanks.
– Garmekain
19 hours ago
add a comment |
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