Let $S = [0, 1] × [0, 1]$. Which of the following statements are true? [closed]
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If $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ are arbitrary points in the plane, define
the metric
$d(P, Q) = max{|x_1 − x_2|, |y_1 − y_2|}.$
Let $P = (2,frac{1}{2})$. Let $S = [0, 1] × [0, 1]$.
Which of the following statements are
true?
$a.$ There does not exist any point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$b.$ There exists a unique point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$c.$ There exist infinitely many points $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}$
I thinks option $b)$ is correct
Any hints/solution
thanks u
metric-spaces
asked Nov 22 at 22:18
Messi fifa
50111
closed as too broad by amWhy, GEdgar, Leucippus, Cesareo, Chinnapparaj R Nov 23 at 2:19
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
If $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ are arbitrary points in the plane, define
the metric
$d(P, Q) = max{|x_1 − x_2|, |y_1 − y_2|}.$
Let $P = (2,frac{1}{2})$. Let $S = [0, 1] × [0, 1]$.
Which of the following statements are
true?
$a.$ There does not exist any point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$b.$ There exists a unique point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$c.$ There exist infinitely many points $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}$
I thinks option $b)$ is correct
Any hints/solution
thanks u
metric-spaces
asked Nov 22 at 22:18
Messi fifa
50111
closed as too broad by amWhy, GEdgar, Leucippus, Cesareo, Chinnapparaj R Nov 23 at 2:19
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
min{d(P,X)|X∊S} is absolutely = to 1, and there is only one point in S that has that minimum distance ... so (b) is the answer. This isn't really any theorem as such ... I think it's just an exercise in interpreting set-notation correctly.
– AmbretteOrrisey
Nov 23 at 7:28
The point being (1,1/2).
– AmbretteOrrisey
Nov 23 at 7:29
Ah no ... alright ... yes if you're not used to set-theoretical notation & not as well-versed it it as you ought to be: it's all the points on the right-hand vertical edge, isn't it! It's coz my mind keeps defaulting to Euclidean distance, and it's difficult to stop it from doing so!
– AmbretteOrrisey
Nov 23 at 7:33
So (c) then ... that's why we are set these exercises, I suppose!
– AmbretteOrrisey
Nov 23 at 7:35
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
If $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ are arbitrary points in the plane, define
the metric
$d(P, Q) = max{|x_1 − x_2|, |y_1 − y_2|}.$
Let $P = (2,frac{1}{2})$. Let $S = [0, 1] × [0, 1]$.
Which of the following statements are
true?
$a.$ There does not exist any point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$b.$ There exists a unique point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$c.$ There exist infinitely many points $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}$
I thinks option $b)$ is correct
Any hints/solution
thanks u
metric-spaces
asked Nov 22 at 22:18
Messi fifa
50111
If $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ are arbitrary points in the plane, define
the metric
$d(P, Q) = max{|x_1 − x_2|, |y_1 − y_2|}.$
Let $P = (2,frac{1}{2})$. Let $S = [0, 1] × [0, 1]$.
Which of the following statements are
true?
$a.$ There does not exist any point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$b.$ There exists a unique point $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}.$
$c.$ There exist infinitely many points $Q ∈ S$ such that
$d(P, Q) = min{d(P, X) | X ∈ S}$
I thinks option $b)$ is correct
Any hints/solution
thanks u
metric-spaces
metric-spaces
asked Nov 22 at 22:18
Messi fifa
50111
asked Nov 22 at 22:18
Messi fifa
50111
asked Nov 22 at 22:18
Messi fifa
50111
asked Nov 22 at 22:18
Messi fifa
50111
asked Nov 22 at 22:18
Messi fifa
50111
50111
closed as too broad by amWhy, GEdgar, Leucippus, Cesareo, Chinnapparaj R Nov 23 at 2:19
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by amWhy, GEdgar, Leucippus, Cesareo, Chinnapparaj R Nov 23 at 2:19
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
min{d(P,X)|X∊S} is absolutely = to 1, and there is only one point in S that has that minimum distance ... so (b) is the answer. This isn't really any theorem as such ... I think it's just an exercise in interpreting set-notation correctly.
– AmbretteOrrisey
Nov 23 at 7:28
The point being (1,1/2).
– AmbretteOrrisey
Nov 23 at 7:29
Ah no ... alright ... yes if you're not used to set-theoretical notation & not as well-versed it it as you ought to be: it's all the points on the right-hand vertical edge, isn't it! It's coz my mind keeps defaulting to Euclidean distance, and it's difficult to stop it from doing so!
– AmbretteOrrisey
Nov 23 at 7:33
So (c) then ... that's why we are set these exercises, I suppose!
– AmbretteOrrisey
Nov 23 at 7:35
add a comment |
min{d(P,X)|X∊S} is absolutely = to 1, and there is only one point in S that has that minimum distance ... so (b) is the answer. This isn't really any theorem as such ... I think it's just an exercise in interpreting set-notation correctly.
– AmbretteOrrisey
Nov 23 at 7:28
The point being (1,1/2).
– AmbretteOrrisey
Nov 23 at 7:29
Ah no ... alright ... yes if you're not used to set-theoretical notation & not as well-versed it it as you ought to be: it's all the points on the right-hand vertical edge, isn't it! It's coz my mind keeps defaulting to Euclidean distance, and it's difficult to stop it from doing so!
– AmbretteOrrisey
Nov 23 at 7:33
So (c) then ... that's why we are set these exercises, I suppose!
– AmbretteOrrisey
Nov 23 at 7:35
min{d(P,X)|X∊S} is absolutely = to 1, and there is only one point in S that has that minimum distance ... so (b) is the answer. This isn't really any theorem as such ... I think it's just an exercise in interpreting set-notation correctly.
– AmbretteOrrisey
Nov 23 at 7:28
min{d(P,X)|X∊S} is absolutely = to 1, and there is only one point in S that has that minimum distance ... so (b) is the answer. This isn't really any theorem as such ... I think it's just an exercise in interpreting set-notation correctly.
– AmbretteOrrisey
Nov 23 at 7:28
The point being (1,1/2).
– AmbretteOrrisey
Nov 23 at 7:29
The point being (1,1/2).
– AmbretteOrrisey
Nov 23 at 7:29
Ah no ... alright ... yes if you're not used to set-theoretical notation & not as well-versed it it as you ought to be: it's all the points on the right-hand vertical edge, isn't it! It's coz my mind keeps defaulting to Euclidean distance, and it's difficult to stop it from doing so!
– AmbretteOrrisey
Nov 23 at 7:33
Ah no ... alright ... yes if you're not used to set-theoretical notation & not as well-versed it it as you ought to be: it's all the points on the right-hand vertical edge, isn't it! It's coz my mind keeps defaulting to Euclidean distance, and it's difficult to stop it from doing so!
– AmbretteOrrisey
Nov 23 at 7:33
So (c) then ... that's why we are set these exercises, I suppose!
– AmbretteOrrisey
Nov 23 at 7:35
So (c) then ... that's why we are set these exercises, I suppose!
– AmbretteOrrisey
Nov 23 at 7:35
add a comment |
1 Answer
1
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up vote
2
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accepted
a. is false: $Q = left(1,frac{1}{2}right)$ is such a point (the minimum being 1).
b. is false: $Q = left(1,1right)$ is such a point (and I have no idea what $f$ is supposed to have to do with anything.
c. is true: $Q = left(1,xright)$ is such a point for every $x in [frac{-1}{2},1]$.
answered Nov 22 at 22:25
user3482749
2,086414
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
a. is false: $Q = left(1,frac{1}{2}right)$ is such a point (the minimum being 1).
b. is false: $Q = left(1,1right)$ is such a point (and I have no idea what $f$ is supposed to have to do with anything.
c. is true: $Q = left(1,xright)$ is such a point for every $x in [frac{-1}{2},1]$.
answered Nov 22 at 22:25
user3482749
2,086414
add a comment |
up vote
2
down vote
accepted
a. is false: $Q = left(1,frac{1}{2}right)$ is such a point (the minimum being 1).
b. is false: $Q = left(1,1right)$ is such a point (and I have no idea what $f$ is supposed to have to do with anything.
c. is true: $Q = left(1,xright)$ is such a point for every $x in [frac{-1}{2},1]$.
answered Nov 22 at 22:25
user3482749
2,086414
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
a. is false: $Q = left(1,frac{1}{2}right)$ is such a point (the minimum being 1).
b. is false: $Q = left(1,1right)$ is such a point (and I have no idea what $f$ is supposed to have to do with anything.
c. is true: $Q = left(1,xright)$ is such a point for every $x in [frac{-1}{2},1]$.
answered Nov 22 at 22:25
user3482749
2,086414
a. is false: $Q = left(1,frac{1}{2}right)$ is such a point (the minimum being 1).
b. is false: $Q = left(1,1right)$ is such a point (and I have no idea what $f$ is supposed to have to do with anything.
c. is true: $Q = left(1,xright)$ is such a point for every $x in [frac{-1}{2},1]$.
answered Nov 22 at 22:25
user3482749
2,086414
answered Nov 22 at 22:25
user3482749
2,086414
answered Nov 22 at 22:25
user3482749
2,086414
answered Nov 22 at 22:25
user3482749
2,086414
2,086414
add a comment |
add a comment |
min{d(P,X)|X∊S} is absolutely = to 1, and there is only one point in S that has that minimum distance ... so (b) is the answer. This isn't really any theorem as such ... I think it's just an exercise in interpreting set-notation correctly.
– AmbretteOrrisey
Nov 23 at 7:28
The point being (1,1/2).
– AmbretteOrrisey
Nov 23 at 7:29
Ah no ... alright ... yes if you're not used to set-theoretical notation & not as well-versed it it as you ought to be: it's all the points on the right-hand vertical edge, isn't it! It's coz my mind keeps defaulting to Euclidean distance, and it's difficult to stop it from doing so!
– AmbretteOrrisey
Nov 23 at 7:33
So (c) then ... that's why we are set these exercises, I suppose!
– AmbretteOrrisey
Nov 23 at 7:35