Mixing
Let X = (0,1) ∪ {2} and let d7: X x X->[0,∞+) be a function defined
$begingroup$
In topology,Let $X = (0,1) ∪ {2}$ and let $d_7: X times Xto [0,∞)$ be a function defined formulas:
$$
d_7(x,y)=cases{|x-y| & if $x ∈ (0,1)$ and $y ∈ (0,1)$\7 & if $x=2$ and $y ∈(0,1)$\ 7& if $x ∈(0,1)$ and $y=2$\0& if $x=2$ and $y=2$}
$$
Please check if $d_7$ is a distance function on $X$.
Does anybody explain and solve this question ?
metric-spaces
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
$endgroup$
add a comment |
$begingroup$
In topology,Let $X = (0,1) ∪ {2}$ and let $d_7: X times Xto [0,∞)$ be a function defined formulas:
$$
d_7(x,y)=cases{|x-y| & if $x ∈ (0,1)$ and $y ∈ (0,1)$\7 & if $x=2$ and $y ∈(0,1)$\ 7& if $x ∈(0,1)$ and $y=2$\0& if $x=2$ and $y=2$}
$$
Please check if $d_7$ is a distance function on $X$.
Does anybody explain and solve this question ?
metric-spaces
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
$endgroup$
$begingroup$
Hi, and welcome to MSE! Please provide some context so that users know how best to help you with this. If you show some of your own thoughts, even (especially) ones resulting in dead ends, this will help users steer you in the right direction, and properly address your question at your level of understanding the subject. Cheers!
$endgroup$
– Adam Hrankowski
Dec 23 '18 at 3:10
add a comment |
$begingroup$
In topology,Let $X = (0,1) ∪ {2}$ and let $d_7: X times Xto [0,∞)$ be a function defined formulas:
$$
d_7(x,y)=cases{|x-y| & if $x ∈ (0,1)$ and $y ∈ (0,1)$\7 & if $x=2$ and $y ∈(0,1)$\ 7& if $x ∈(0,1)$ and $y=2$\0& if $x=2$ and $y=2$}
$$
Please check if $d_7$ is a distance function on $X$.
Does anybody explain and solve this question ?
metric-spaces
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
$endgroup$
In topology,Let $X = (0,1) ∪ {2}$ and let $d_7: X times Xto [0,∞)$ be a function defined formulas:
$$
d_7(x,y)=cases{|x-y| & if $x ∈ (0,1)$ and $y ∈ (0,1)$\7 & if $x=2$ and $y ∈(0,1)$\ 7& if $x ∈(0,1)$ and $y=2$\0& if $x=2$ and $y=2$}
$$
Please check if $d_7$ is a distance function on $X$.
Does anybody explain and solve this question ?
metric-spaces
metric-spaces
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
edited Dec 22 '18 at 3:29
Alex Ravsky
40.8k32282
40.8k32282
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
asked Dec 18 '18 at 19:21
Arda Batuhan DemirArda Batuhan Demir
11
11
$begingroup$
Hi, and welcome to MSE! Please provide some context so that users know how best to help you with this. If you show some of your own thoughts, even (especially) ones resulting in dead ends, this will help users steer you in the right direction, and properly address your question at your level of understanding the subject. Cheers!
$endgroup$
– Adam Hrankowski
Dec 23 '18 at 3:10
add a comment |
$begingroup$
Hi, and welcome to MSE! Please provide some context so that users know how best to help you with this. If you show some of your own thoughts, even (especially) ones resulting in dead ends, this will help users steer you in the right direction, and properly address your question at your level of understanding the subject. Cheers!
$endgroup$
– Adam Hrankowski
Dec 23 '18 at 3:10
$begingroup$
Hi, and welcome to MSE! Please provide some context so that users know how best to help you with this. If you show some of your own thoughts, even (especially) ones resulting in dead ends, this will help users steer you in the right direction, and properly address your question at your level of understanding the subject. Cheers!
$endgroup$
– Adam Hrankowski
Dec 23 '18 at 3:10
$begingroup$
Hi, and welcome to MSE! Please provide some context so that users know how best to help you with this. If you show some of your own thoughts, even (especially) ones resulting in dead ends, this will help users steer you in the right direction, and properly address your question at your level of understanding the subject. Cheers!
$endgroup$
– Adam Hrankowski
Dec 23 '18 at 3:10
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Yes, $d_7$ is a distance function, because for each $x,y,zin X$ we have $d_7(x,y)=0$ iff $x=y$, $d_7(x,y)=d_7(y,x)$, and $d_7(x,z)le d_7(x,y)+d_7 (y,z)$. The first two properties are easy to check while looking at the definition on $d_7$. The last inequality clearly holds when $x,y,zin (0,1)$, because then it is the triangle inequality for the standard distance on $Bbb R$. If $d_7(x,y)=7$ or $d_7(y,z)=7$ then the inequality holds, because anyway $d_7(x,z)le 7$. The remaining case is $x=y=z=2$, in which the inequality holds trivially.
We have to show that $d_7(x,y)+d_7 (y,z)ge d_7(x,z)$ for each $x,y,zin X$. The following cases are possible.
1) $x,y,zin (0,1)$. Then $d_7(x,z)=|x-z|$, $d_7(x,y)=|x-y|$, and $d_7(y,z)=|y-z|$. Then $d_7(x,z)le d_7(x,y)+d_7 (y,z)$ because $|a|+|b|ge |a+b| $ for all real $a$ and $b$, so $$|x-y|+|y-z|ge |(x-y)+(y-z)|=|x-z|.$$
2) $xin (0,1)$, $y=2$ or $yin (0,1)$, $x=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(x,y)=7ge d_7(x,z)$.
3) $zin (0,1)$, $y=2$ or $yin (0,1)$, $z=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(y,z)=7ge d_7(x,z)$.
4) $x=y=z=2$. Then $d_7(x,y)+d_7(y,z)=0+0=0=d_7(x,z)$.
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
$endgroup$
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
1
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
add a comment |
Your Answer
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$begingroup$
Yes, $d_7$ is a distance function, because for each $x,y,zin X$ we have $d_7(x,y)=0$ iff $x=y$, $d_7(x,y)=d_7(y,x)$, and $d_7(x,z)le d_7(x,y)+d_7 (y,z)$. The first two properties are easy to check while looking at the definition on $d_7$. The last inequality clearly holds when $x,y,zin (0,1)$, because then it is the triangle inequality for the standard distance on $Bbb R$. If $d_7(x,y)=7$ or $d_7(y,z)=7$ then the inequality holds, because anyway $d_7(x,z)le 7$. The remaining case is $x=y=z=2$, in which the inequality holds trivially.
We have to show that $d_7(x,y)+d_7 (y,z)ge d_7(x,z)$ for each $x,y,zin X$. The following cases are possible.
1) $x,y,zin (0,1)$. Then $d_7(x,z)=|x-z|$, $d_7(x,y)=|x-y|$, and $d_7(y,z)=|y-z|$. Then $d_7(x,z)le d_7(x,y)+d_7 (y,z)$ because $|a|+|b|ge |a+b| $ for all real $a$ and $b$, so $$|x-y|+|y-z|ge |(x-y)+(y-z)|=|x-z|.$$
2) $xin (0,1)$, $y=2$ or $yin (0,1)$, $x=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(x,y)=7ge d_7(x,z)$.
3) $zin (0,1)$, $y=2$ or $yin (0,1)$, $z=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(y,z)=7ge d_7(x,z)$.
4) $x=y=z=2$. Then $d_7(x,y)+d_7(y,z)=0+0=0=d_7(x,z)$.
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
$endgroup$
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
1
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
add a comment |
$begingroup$
Yes, $d_7$ is a distance function, because for each $x,y,zin X$ we have $d_7(x,y)=0$ iff $x=y$, $d_7(x,y)=d_7(y,x)$, and $d_7(x,z)le d_7(x,y)+d_7 (y,z)$. The first two properties are easy to check while looking at the definition on $d_7$. The last inequality clearly holds when $x,y,zin (0,1)$, because then it is the triangle inequality for the standard distance on $Bbb R$. If $d_7(x,y)=7$ or $d_7(y,z)=7$ then the inequality holds, because anyway $d_7(x,z)le 7$. The remaining case is $x=y=z=2$, in which the inequality holds trivially.
We have to show that $d_7(x,y)+d_7 (y,z)ge d_7(x,z)$ for each $x,y,zin X$. The following cases are possible.
1) $x,y,zin (0,1)$. Then $d_7(x,z)=|x-z|$, $d_7(x,y)=|x-y|$, and $d_7(y,z)=|y-z|$. Then $d_7(x,z)le d_7(x,y)+d_7 (y,z)$ because $|a|+|b|ge |a+b| $ for all real $a$ and $b$, so $$|x-y|+|y-z|ge |(x-y)+(y-z)|=|x-z|.$$
2) $xin (0,1)$, $y=2$ or $yin (0,1)$, $x=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(x,y)=7ge d_7(x,z)$.
3) $zin (0,1)$, $y=2$ or $yin (0,1)$, $z=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(y,z)=7ge d_7(x,z)$.
4) $x=y=z=2$. Then $d_7(x,y)+d_7(y,z)=0+0=0=d_7(x,z)$.
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
$endgroup$
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
1
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
add a comment |
$begingroup$
Yes, $d_7$ is a distance function, because for each $x,y,zin X$ we have $d_7(x,y)=0$ iff $x=y$, $d_7(x,y)=d_7(y,x)$, and $d_7(x,z)le d_7(x,y)+d_7 (y,z)$. The first two properties are easy to check while looking at the definition on $d_7$. The last inequality clearly holds when $x,y,zin (0,1)$, because then it is the triangle inequality for the standard distance on $Bbb R$. If $d_7(x,y)=7$ or $d_7(y,z)=7$ then the inequality holds, because anyway $d_7(x,z)le 7$. The remaining case is $x=y=z=2$, in which the inequality holds trivially.
We have to show that $d_7(x,y)+d_7 (y,z)ge d_7(x,z)$ for each $x,y,zin X$. The following cases are possible.
1) $x,y,zin (0,1)$. Then $d_7(x,z)=|x-z|$, $d_7(x,y)=|x-y|$, and $d_7(y,z)=|y-z|$. Then $d_7(x,z)le d_7(x,y)+d_7 (y,z)$ because $|a|+|b|ge |a+b| $ for all real $a$ and $b$, so $$|x-y|+|y-z|ge |(x-y)+(y-z)|=|x-z|.$$
2) $xin (0,1)$, $y=2$ or $yin (0,1)$, $x=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(x,y)=7ge d_7(x,z)$.
3) $zin (0,1)$, $y=2$ or $yin (0,1)$, $z=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(y,z)=7ge d_7(x,z)$.
4) $x=y=z=2$. Then $d_7(x,y)+d_7(y,z)=0+0=0=d_7(x,z)$.
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
$endgroup$
Yes, $d_7$ is a distance function, because for each $x,y,zin X$ we have $d_7(x,y)=0$ iff $x=y$, $d_7(x,y)=d_7(y,x)$, and $d_7(x,z)le d_7(x,y)+d_7 (y,z)$. The first two properties are easy to check while looking at the definition on $d_7$. The last inequality clearly holds when $x,y,zin (0,1)$, because then it is the triangle inequality for the standard distance on $Bbb R$. If $d_7(x,y)=7$ or $d_7(y,z)=7$ then the inequality holds, because anyway $d_7(x,z)le 7$. The remaining case is $x=y=z=2$, in which the inequality holds trivially.
We have to show that $d_7(x,y)+d_7 (y,z)ge d_7(x,z)$ for each $x,y,zin X$. The following cases are possible.
1) $x,y,zin (0,1)$. Then $d_7(x,z)=|x-z|$, $d_7(x,y)=|x-y|$, and $d_7(y,z)=|y-z|$. Then $d_7(x,z)le d_7(x,y)+d_7 (y,z)$ because $|a|+|b|ge |a+b| $ for all real $a$ and $b$, so $$|x-y|+|y-z|ge |(x-y)+(y-z)|=|x-z|.$$
2) $xin (0,1)$, $y=2$ or $yin (0,1)$, $x=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(x,y)=7ge d_7(x,z)$.
3) $zin (0,1)$, $y=2$ or $yin (0,1)$, $z=2$. Then $d_7(x,y)+d_7(y,z)ge d_7(y,z)=7ge d_7(x,z)$.
4) $x=y=z=2$. Then $d_7(x,y)+d_7(y,z)=0+0=0=d_7(x,z)$.
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
edited Jan 10 at 21:48
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
answered Dec 22 '18 at 3:29
Alex RavskyAlex Ravsky
40.8k32282
40.8k32282
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
1
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
add a comment |
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
1
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
Could you giving more then clearly details by formally ? and writing all possible cases....This is so significant to me like as e.g 1) x,y,z ∈ (0,1) 2) d7(y,z)=7 for clearly and formally solutions ? Thank you so much!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 16:46
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
$begingroup$
@ArdaBatuhanDemir Done.
$endgroup$
– Alex Ravsky
Jan 10 at 21:48
1
1
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Many thanks for your feedback!
$endgroup$
– Arda Batuhan Demir
Jan 10 at 22:10
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
$begingroup$
Can you look up this question ? math.stackexchange.com/questions/3067642/…
$endgroup$
– Arda Batuhan Demir
Jan 11 at 15:47
add a comment |
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Hi, and welcome to MSE! Please provide some context so that users know how best to help you with this. If you show some of your own thoughts, even (especially) ones resulting in dead ends, this will help users steer you in the right direction, and properly address your question at your level of understanding the subject. Cheers!
$endgroup$
– Adam Hrankowski
Dec 23 '18 at 3:10