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How do we go from $d(x,x') le d(x,y) +d(y,y') +d(y',x')$ to $d(x,x') ge d(y,y') -d(x,y)-d(y',x')$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to prove that there is a unique geodisic segment that is pependicular to two other geodesics?Spacing nodes by moving the shortest distance possible.closure of the unit ballFind length of a side from given mesurementsIs there a metric on $X = (-1,1)$ such that the completion of $X$ is just one additional point?Show that if $(A,d)$ is complete then $A$ is closed.How to solve the following mixed exponential inequality? (from Spivak's Calculus)Finding the distance of a point from a lineSolving an inequality with multiple casesHow do you find $x$ such as $y leq sqrt2x-x^2$?
0
$begingroup$
How do we go from $d(x,x') le d(x,y) +d(y,y') +d(y',x')$ to $d(x,x') ge d(y,y') -d(x,y)-d(y',x')$?
I am not sure how to proceed here.
geometry inequality metric-spaces
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
$endgroup$
add a comment |
0
$begingroup$
How do we go from $d(x,x') le d(x,y) +d(y,y') +d(y',x')$ to $d(x,x') ge d(y,y') -d(x,y)-d(y',x')$?
I am not sure how to proceed here.
geometry inequality metric-spaces
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
$endgroup$
add a comment |
0
0
0
$begingroup$
How do we go from $d(x,x') le d(x,y) +d(y,y') +d(y',x')$ to $d(x,x') ge d(y,y') -d(x,y)-d(y',x')$?
I am not sure how to proceed here.
geometry inequality metric-spaces
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
$endgroup$
How do we go from $d(x,x') le d(x,y) +d(y,y') +d(y',x')$ to $d(x,x') ge d(y,y') -d(x,y)-d(y',x')$?
I am not sure how to proceed here.
geometry inequality metric-spaces
geometry inequality metric-spaces
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
asked Mar 25 at 20:10
Al JebrAl Jebr
4,42543479
4,42543479
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
Subtract $d(x,y)$ and $d(y',x')$ from both sides of your original inequality, and then relabel all $x,x'$ as $y,y'$ and vice versa.
answered Mar 25 at 20:14
jawheelejawheele
51639
$endgroup$
1
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
Subtract $d(x,y)$ and $d(y',x')$ from both sides of your original inequality, and then relabel all $x,x'$ as $y,y'$ and vice versa.
answered Mar 25 at 20:14
jawheelejawheele
51639
$endgroup$
1
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
add a comment |
3
$begingroup$
Subtract $d(x,y)$ and $d(y',x')$ from both sides of your original inequality, and then relabel all $x,x'$ as $y,y'$ and vice versa.
answered Mar 25 at 20:14
jawheelejawheele
51639
$endgroup$
1
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
add a comment |
3
3
3
$begingroup$
Subtract $d(x,y)$ and $d(y',x')$ from both sides of your original inequality, and then relabel all $x,x'$ as $y,y'$ and vice versa.
answered Mar 25 at 20:14
jawheelejawheele
51639
$endgroup$
Subtract $d(x,y)$ and $d(y',x')$ from both sides of your original inequality, and then relabel all $x,x'$ as $y,y'$ and vice versa.
answered Mar 25 at 20:14
jawheelejawheele
51639
edited Mar 25 at 20:21
answered Mar 25 at 20:14
jawheelejawheele
51639
answered Mar 25 at 20:14
jawheelejawheele
51639
answered Mar 25 at 20:14
jawheelejawheele
51639
51639
1
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
add a comment |
1
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
1
1
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
That procedure gets $d(x,,x^prime)ge d(y,,y^prime)-d(y,,x)-d(x^prime,,y^prime)$, so we also need to use $d(a,,b)=d(b,,a)$, which no doubt the original exercise allows us to assume.
$endgroup$
– J.G.
Mar 25 at 20:17
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
Right! I had taken the symmetry as a given, assuming this is in the context of a metric space. I agree that warranted clarification, as that context was not explicitly stated.
$endgroup$
– jawheele
Mar 25 at 20:20
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
@jawheele What do you mean relabel all? How can we just relabel things?
$endgroup$
– Al Jebr
Mar 25 at 20:22
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
I mean replace $x$ by $y$, $x'$ by $y'$, $y$ by $x$, and $y'$ by $x'$. Assuming your first inequality is supposed to hold for all choices of $x,x',y,y'$ in whatever set you're working in, this is valid to do (you're just applying the first inequality to a different combination of points). If it makes the validity clearer, you might make the substitutions before the subtractions.
$endgroup$
– jawheele
Mar 25 at 20:25
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
$begingroup$
@jawheele Thanks. It did turn out to be clearer when I made the substitutions first. One could also start by considering the triangle inequality starting with $d(y,y')$.
$endgroup$
– Al Jebr
Mar 31 at 22:18
add a comment |
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