Axioms of a metric space [duplicate]
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This question already has an answer here:
Metric Space Axioms
2 answers
$d$ :$X×X$ $to$ $mathbb R$
is a metric space iff it satisfies the four conditions -
$d(x,y)$ $ge$ $0$
$d(x,y)=0$ iff $x=y$
$d(x,y)$=$d(y,x)$ for all $x,y in$ $X$
$d(x,y)$ $le$ $d(x,z)$ +$d(z,y)$ for all $x,y,z$ $in$ $X$
These are the very familiar axioms of a metric space.
But from axiom $4$ we can deduce the axiom $3$ but still why in many books, statement $3$ is used as an axiom(although it is not an axiom) ?
Please suggest some edit.
metric-spaces
edited Jan 18 at 9:00
red_trumpet
1,063319
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
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marked as duplicate by José Carlos Santos, Misha Lavrov, max_zorn, Cesareo, Shailesh Jan 27 at 3:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 2 more comments
$begingroup$
This question already has an answer here:
Metric Space Axioms
2 answers
$d$ :$X×X$ $to$ $mathbb R$
is a metric space iff it satisfies the four conditions -
$d(x,y)$ $ge$ $0$
$d(x,y)=0$ iff $x=y$
$d(x,y)$=$d(y,x)$ for all $x,y in$ $X$
$d(x,y)$ $le$ $d(x,z)$ +$d(z,y)$ for all $x,y,z$ $in$ $X$
These are the very familiar axioms of a metric space.
But from axiom $4$ we can deduce the axiom $3$ but still why in many books, statement $3$ is used as an axiom(although it is not an axiom) ?
Please suggest some edit.
metric-spaces
edited Jan 18 at 9:00
red_trumpet
1,063319
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
$endgroup$
marked as duplicate by José Carlos Santos, Misha Lavrov, max_zorn, Cesareo, Shailesh Jan 27 at 3:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
I think 4) uses $leq$ because of the case of $x=y=z$.
$endgroup$
– twnly
Jan 18 at 8:24
$begingroup$
Sorry but I am not too much familiar with this..writing in stack exchange
$endgroup$
– Supriyo Banerjee
Jan 18 at 8:26
1
$begingroup$
How do we deduce 3) from 4)? Also note, that in my analysis 2 book (Forster), property 1): $d(x,y) geq 0$ is deduces from 2)-4).
$endgroup$
– red_trumpet
Jan 18 at 8:29
2
$begingroup$
In the second axiom, it should say "iff $x=y$", shouldn't it?
$endgroup$
– Miksu
Jan 18 at 8:34
1
$begingroup$
It should be $ge$ in 1) and in the triangle inequality 4) it has to be $le$
$endgroup$
– ohno
Jan 18 at 8:49
|
show 2 more comments
$begingroup$
This question already has an answer here:
Metric Space Axioms
2 answers
$d$ :$X×X$ $to$ $mathbb R$
is a metric space iff it satisfies the four conditions -
$d(x,y)$ $ge$ $0$
$d(x,y)=0$ iff $x=y$
$d(x,y)$=$d(y,x)$ for all $x,y in$ $X$
$d(x,y)$ $le$ $d(x,z)$ +$d(z,y)$ for all $x,y,z$ $in$ $X$
These are the very familiar axioms of a metric space.
But from axiom $4$ we can deduce the axiom $3$ but still why in many books, statement $3$ is used as an axiom(although it is not an axiom) ?
Please suggest some edit.
metric-spaces
edited Jan 18 at 9:00
red_trumpet
1,063319
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
$endgroup$
This question already has an answer here:
Metric Space Axioms
2 answers
$d$ :$X×X$ $to$ $mathbb R$
is a metric space iff it satisfies the four conditions -
$d(x,y)$ $ge$ $0$
$d(x,y)=0$ iff $x=y$
$d(x,y)$=$d(y,x)$ for all $x,y in$ $X$
$d(x,y)$ $le$ $d(x,z)$ +$d(z,y)$ for all $x,y,z$ $in$ $X$
These are the very familiar axioms of a metric space.
But from axiom $4$ we can deduce the axiom $3$ but still why in many books, statement $3$ is used as an axiom(although it is not an axiom) ?
Please suggest some edit.
This question already has an answer here:
Metric Space Axioms
2 answers
metric-spaces
metric-spaces
edited Jan 18 at 9:00
red_trumpet
1,063319
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
edited Jan 18 at 9:00
red_trumpet
1,063319
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
edited Jan 18 at 9:00
red_trumpet
1,063319
edited Jan 18 at 9:00
red_trumpet
1,063319
edited Jan 18 at 9:00
red_trumpet
1,063319
1,063319
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
asked Jan 18 at 8:21
Supriyo BanerjeeSupriyo Banerjee
1056
1056
marked as duplicate by José Carlos Santos, Misha Lavrov, max_zorn, Cesareo, Shailesh Jan 27 at 3:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, Misha Lavrov, max_zorn, Cesareo, Shailesh Jan 27 at 3:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
I think 4) uses $leq$ because of the case of $x=y=z$.
$endgroup$
– twnly
Jan 18 at 8:24
$begingroup$
Sorry but I am not too much familiar with this..writing in stack exchange
$endgroup$
– Supriyo Banerjee
Jan 18 at 8:26
1
$begingroup$
How do we deduce 3) from 4)? Also note, that in my analysis 2 book (Forster), property 1): $d(x,y) geq 0$ is deduces from 2)-4).
$endgroup$
– red_trumpet
Jan 18 at 8:29
2
$begingroup$
In the second axiom, it should say "iff $x=y$", shouldn't it?
$endgroup$
– Miksu
Jan 18 at 8:34
1
$begingroup$
It should be $ge$ in 1) and in the triangle inequality 4) it has to be $le$
$endgroup$
– ohno
Jan 18 at 8:49
|
show 2 more comments
$begingroup$
I think 4) uses $leq$ because of the case of $x=y=z$.
$endgroup$
– twnly
Jan 18 at 8:24
$begingroup$
Sorry but I am not too much familiar with this..writing in stack exchange
$endgroup$
– Supriyo Banerjee
Jan 18 at 8:26
1
$begingroup$
How do we deduce 3) from 4)? Also note, that in my analysis 2 book (Forster), property 1): $d(x,y) geq 0$ is deduces from 2)-4).
$endgroup$
– red_trumpet
Jan 18 at 8:29
2
$begingroup$
In the second axiom, it should say "iff $x=y$", shouldn't it?
$endgroup$
– Miksu
Jan 18 at 8:34
1
$begingroup$
It should be $ge$ in 1) and in the triangle inequality 4) it has to be $le$
$endgroup$
– ohno
Jan 18 at 8:49
$begingroup$
I think 4) uses $leq$ because of the case of $x=y=z$.
$endgroup$
– twnly
Jan 18 at 8:24
$begingroup$
I think 4) uses $leq$ because of the case of $x=y=z$.
$endgroup$
– twnly
Jan 18 at 8:24
$begingroup$
Sorry but I am not too much familiar with this..writing in stack exchange
$endgroup$
– Supriyo Banerjee
Jan 18 at 8:26
$begingroup$
Sorry but I am not too much familiar with this..writing in stack exchange
$endgroup$
– Supriyo Banerjee
Jan 18 at 8:26
1
1
$begingroup$
How do we deduce 3) from 4)? Also note, that in my analysis 2 book (Forster), property 1): $d(x,y) geq 0$ is deduces from 2)-4).
$endgroup$
– red_trumpet
Jan 18 at 8:29
$begingroup$
How do we deduce 3) from 4)? Also note, that in my analysis 2 book (Forster), property 1): $d(x,y) geq 0$ is deduces from 2)-4).
$endgroup$
– red_trumpet
Jan 18 at 8:29
2
2
$begingroup$
In the second axiom, it should say "iff $x=y$", shouldn't it?
$endgroup$
– Miksu
Jan 18 at 8:34
$begingroup$
In the second axiom, it should say "iff $x=y$", shouldn't it?
$endgroup$
– Miksu
Jan 18 at 8:34
1
1
$begingroup$
It should be $ge$ in 1) and in the triangle inequality 4) it has to be $le$
$endgroup$
– ohno
Jan 18 at 8:49
$begingroup$
It should be $ge$ in 1) and in the triangle inequality 4) it has to be $le$
$endgroup$
– ohno
Jan 18 at 8:49
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I think this requires some sort of meta-answer since it is true that at least one of the axioms is redundant, but so the question becomes why they include the axiom in textbooks on metric spaces.
The reason is that when coming up with definitions for certain mathematical objects, the minimality of the axioms in the definition (measured in how many properties the mathematical object has to satisfy) is not that important. Especially considering that it is a textbook, there are pedagogical reasons to explicitly write out axioms 1-4).
So in the metric space case, just writing out the axiom 3) or 1), whichever is redundant, just lets the reader focus on those properties, which are semantically a bit different, even though they can be derived from the other axioms.
answered Jan 18 at 8:42
twnlytwnly
1,2341214
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think this requires some sort of meta-answer since it is true that at least one of the axioms is redundant, but so the question becomes why they include the axiom in textbooks on metric spaces.
The reason is that when coming up with definitions for certain mathematical objects, the minimality of the axioms in the definition (measured in how many properties the mathematical object has to satisfy) is not that important. Especially considering that it is a textbook, there are pedagogical reasons to explicitly write out axioms 1-4).
So in the metric space case, just writing out the axiom 3) or 1), whichever is redundant, just lets the reader focus on those properties, which are semantically a bit different, even though they can be derived from the other axioms.
answered Jan 18 at 8:42
twnlytwnly
1,2341214
$endgroup$
add a comment |
$begingroup$
I think this requires some sort of meta-answer since it is true that at least one of the axioms is redundant, but so the question becomes why they include the axiom in textbooks on metric spaces.
The reason is that when coming up with definitions for certain mathematical objects, the minimality of the axioms in the definition (measured in how many properties the mathematical object has to satisfy) is not that important. Especially considering that it is a textbook, there are pedagogical reasons to explicitly write out axioms 1-4).
So in the metric space case, just writing out the axiom 3) or 1), whichever is redundant, just lets the reader focus on those properties, which are semantically a bit different, even though they can be derived from the other axioms.
answered Jan 18 at 8:42
twnlytwnly
1,2341214
$endgroup$
add a comment |
$begingroup$
I think this requires some sort of meta-answer since it is true that at least one of the axioms is redundant, but so the question becomes why they include the axiom in textbooks on metric spaces.
The reason is that when coming up with definitions for certain mathematical objects, the minimality of the axioms in the definition (measured in how many properties the mathematical object has to satisfy) is not that important. Especially considering that it is a textbook, there are pedagogical reasons to explicitly write out axioms 1-4).
So in the metric space case, just writing out the axiom 3) or 1), whichever is redundant, just lets the reader focus on those properties, which are semantically a bit different, even though they can be derived from the other axioms.
answered Jan 18 at 8:42
twnlytwnly
1,2341214
$endgroup$
I think this requires some sort of meta-answer since it is true that at least one of the axioms is redundant, but so the question becomes why they include the axiom in textbooks on metric spaces.
The reason is that when coming up with definitions for certain mathematical objects, the minimality of the axioms in the definition (measured in how many properties the mathematical object has to satisfy) is not that important. Especially considering that it is a textbook, there are pedagogical reasons to explicitly write out axioms 1-4).
So in the metric space case, just writing out the axiom 3) or 1), whichever is redundant, just lets the reader focus on those properties, which are semantically a bit different, even though they can be derived from the other axioms.
answered Jan 18 at 8:42
twnlytwnly
1,2341214
edited Jan 18 at 8:48
answered Jan 18 at 8:42
twnlytwnly
1,2341214
answered Jan 18 at 8:42
twnlytwnly
1,2341214
answered Jan 18 at 8:42
twnlytwnly
1,2341214
1,2341214
add a comment |
add a comment |
$begingroup$
I think 4) uses $leq$ because of the case of $x=y=z$.
$endgroup$
– twnly
Jan 18 at 8:24
$begingroup$
Sorry but I am not too much familiar with this..writing in stack exchange
$endgroup$
– Supriyo Banerjee
Jan 18 at 8:26
1
$begingroup$
How do we deduce 3) from 4)? Also note, that in my analysis 2 book (Forster), property 1): $d(x,y) geq 0$ is deduces from 2)-4).
$endgroup$
– red_trumpet
Jan 18 at 8:29
2
$begingroup$
In the second axiom, it should say "iff $x=y$", shouldn't it?
$endgroup$
– Miksu
Jan 18 at 8:34
1
$begingroup$
It should be $ge$ in 1) and in the triangle inequality 4) it has to be $le$
$endgroup$
– ohno
Jan 18 at 8:49