Prove the set $leftvecxinmathbbR^n : 0 < x_i < 1, 1leq i leq nright$ is open. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that any subset of $mathbbR^2$ which is $D$-open is also $D_1$-open.Prove that a set is openShow a set is open using open ballsHelp with proving triangle law to prove these distance functions are metric spacesProve this set is open in $mathbbR^n$Prove that $g(x,y) = left((x,y), (x,y)right)$ is continuous.Properties of $d(x,y) = vert x_1 - y_1vert^frac12 + vert x_2-y_2vert^frac12$Prove that the set $A=(x,y)inmathbbR^2: x^2 + y^2 > 4, y < 6$ is open in $mathbbR^2$Prove that $N((0,0);1)$ is an open set in $mathbbRtimesmathbbR$ with metric $d((x_1,x_2),(y_1,y_2))=|x_1-y_1|+|x_2-y_2|$.Determine whether the following subsets of $mathbbR^2$ are open with respect to the metric $d$.
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Prove the set $left{vecxinmathbbR^n : 0
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that any subset of $mathbbR^2$ which is $D$-open is also $D_1$-open.Prove that a set is openShow a set is open using open ballsHelp with proving triangle law to prove these distance functions are metric spacesProve this set is open in $mathbbR^n$Prove that $g(x,y) = left((x,y), (x,y)right)$ is continuous.Properties of $d(x,y) = vert x_1 - y_1vert^frac12 + vert x_2-y_2vert^frac12$Prove that the set $A=(x,y)inmathbbR^2: x^2 + y^2 > 4, y < 6$ is open in $mathbbR^2$Prove that $N((0,0);1)$ is an open set in $mathbbRtimesmathbbR$ with metric $d((x_1,x_2),(y_1,y_2))=|x_1-y_1|+|x_2-y_2|$.Determine whether the following subsets of $mathbbR^2$ are open with respect to the metric $d$.
$begingroup$
I'm having a lot of difficulty with proving the set $S = leftvecxinmathbbR^2 : 0 < x_1,x_2 < 1right$ is open. I need to use the definition of an open set to prove this. At this point I mostly have "scratch work."
For any $vecain S$, $Vert vecaVert < 2$ since each component of the vector is between $0$ and $1$.
Let $vecy,vecain S$ and let $r = 2-Vertveca-vecyVert>0$. If $vecx in B_r(veca)$, then
$$Vertvecx-vecyVert leq Vertvecx-vecaVert + Vertveca-vecyVert <2.$$
Thus $|x_1-y_1|-|x_2-y_2| leq Vert(x_1-y_1,x_2-y_2)Vert<2$.
I have a hunch this is not the radius I should be using for the ball, but I'm not sure how to properly go about this problem.
real-analysis general-topology
edited Mar 31 at 8:29
Ingix
5,232259
asked Mar 31 at 8:24
moofasamoofasa
82
$endgroup$
|
show 2 more comments
$begingroup$
I'm having a lot of difficulty with proving the set $S = leftvecxinmathbbR^2 : 0 < x_1,x_2 < 1right$ is open. I need to use the definition of an open set to prove this. At this point I mostly have "scratch work."
For any $vecain S$, $Vert vecaVert < 2$ since each component of the vector is between $0$ and $1$.
Let $vecy,vecain S$ and let $r = 2-Vertveca-vecyVert>0$. If $vecx in B_r(veca)$, then
$$Vertvecx-vecyVert leq Vertvecx-vecaVert + Vertveca-vecyVert <2.$$
Thus $|x_1-y_1|-|x_2-y_2| leq Vert(x_1-y_1,x_2-y_2)Vert<2$.
I have a hunch this is not the radius I should be using for the ball, but I'm not sure how to properly go about this problem.
real-analysis general-topology
edited Mar 31 at 8:29
Ingix
5,232259
asked Mar 31 at 8:24
moofasamoofasa
82
$endgroup$
3
$begingroup$
what metric are you using?
$endgroup$
– Riquelme
Mar 31 at 8:25
$begingroup$
Euclidean metric
$endgroup$
– moofasa
Mar 31 at 8:29
1
$begingroup$
All norms are equivalent on finite dimensional vector spaces, and as you want each component of the vector to have mod less than 1, use the sup (max) norm, then the norm of an arbitrary vector in your set would be $c$ for some $c < 1$ then a ball of radius $r<1-c$ should work?
$endgroup$
– Displayname
Mar 31 at 8:31
$begingroup$
Your attempted argument is confusing because you use $a,y,x$ without telling what they are supposed to mean. For example, it seems that $a$ is any point in $S$ and you want to prove that some ball around it is still in $S$. But I can't figure out what you need $y$ for, and how it is determined.
$endgroup$
– Ingix
Mar 31 at 8:33
$begingroup$
y and a are both in S, as stated. I was playing around with the triangle inequality to see if I could get somewhere useful. I tried googling what the sup (max) norm is but I'm still a bit confused. For reference, this is for a real analysis course covering the basics about topology of R^n.
$endgroup$
– moofasa
Mar 31 at 8:46
|
show 2 more comments
$begingroup$
I'm having a lot of difficulty with proving the set $S = leftvecxinmathbbR^2 : 0 < x_1,x_2 < 1right$ is open. I need to use the definition of an open set to prove this. At this point I mostly have "scratch work."
For any $vecain S$, $Vert vecaVert < 2$ since each component of the vector is between $0$ and $1$.
Let $vecy,vecain S$ and let $r = 2-Vertveca-vecyVert>0$. If $vecx in B_r(veca)$, then
$$Vertvecx-vecyVert leq Vertvecx-vecaVert + Vertveca-vecyVert <2.$$
Thus $|x_1-y_1|-|x_2-y_2| leq Vert(x_1-y_1,x_2-y_2)Vert<2$.
I have a hunch this is not the radius I should be using for the ball, but I'm not sure how to properly go about this problem.
real-analysis general-topology
edited Mar 31 at 8:29
Ingix
5,232259
asked Mar 31 at 8:24
moofasamoofasa
82
$endgroup$
I'm having a lot of difficulty with proving the set $S = leftvecxinmathbbR^2 : 0 < x_1,x_2 < 1right$ is open. I need to use the definition of an open set to prove this. At this point I mostly have "scratch work."
For any $vecain S$, $Vert vecaVert < 2$ since each component of the vector is between $0$ and $1$.
Let $vecy,vecain S$ and let $r = 2-Vertveca-vecyVert>0$. If $vecx in B_r(veca)$, then
$$Vertvecx-vecyVert leq Vertvecx-vecaVert + Vertveca-vecyVert <2.$$
Thus $|x_1-y_1|-|x_2-y_2| leq Vert(x_1-y_1,x_2-y_2)Vert<2$.
I have a hunch this is not the radius I should be using for the ball, but I'm not sure how to properly go about this problem.
real-analysis general-topology
real-analysis general-topology
edited Mar 31 at 8:29
Ingix
5,232259
asked Mar 31 at 8:24
moofasamoofasa
82
edited Mar 31 at 8:29
Ingix
5,232259
asked Mar 31 at 8:24
moofasamoofasa
82
edited Mar 31 at 8:29
Ingix
5,232259
edited Mar 31 at 8:29
Ingix
5,232259
edited Mar 31 at 8:29
Ingix
5,232259
5,232259
asked Mar 31 at 8:24
moofasamoofasa
82
asked Mar 31 at 8:24
moofasamoofasa
82
asked Mar 31 at 8:24
moofasamoofasa
82
82
3
$begingroup$
what metric are you using?
$endgroup$
– Riquelme
Mar 31 at 8:25
$begingroup$
Euclidean metric
$endgroup$
– moofasa
Mar 31 at 8:29
1
$begingroup$
All norms are equivalent on finite dimensional vector spaces, and as you want each component of the vector to have mod less than 1, use the sup (max) norm, then the norm of an arbitrary vector in your set would be $c$ for some $c < 1$ then a ball of radius $r<1-c$ should work?
$endgroup$
– Displayname
Mar 31 at 8:31
$begingroup$
Your attempted argument is confusing because you use $a,y,x$ without telling what they are supposed to mean. For example, it seems that $a$ is any point in $S$ and you want to prove that some ball around it is still in $S$. But I can't figure out what you need $y$ for, and how it is determined.
$endgroup$
– Ingix
Mar 31 at 8:33
$begingroup$
y and a are both in S, as stated. I was playing around with the triangle inequality to see if I could get somewhere useful. I tried googling what the sup (max) norm is but I'm still a bit confused. For reference, this is for a real analysis course covering the basics about topology of R^n.
$endgroup$
– moofasa
Mar 31 at 8:46
|
show 2 more comments
3
$begingroup$
what metric are you using?
$endgroup$
– Riquelme
Mar 31 at 8:25
$begingroup$
Euclidean metric
$endgroup$
– moofasa
Mar 31 at 8:29
1
$begingroup$
All norms are equivalent on finite dimensional vector spaces, and as you want each component of the vector to have mod less than 1, use the sup (max) norm, then the norm of an arbitrary vector in your set would be $c$ for some $c < 1$ then a ball of radius $r<1-c$ should work?
$endgroup$
– Displayname
Mar 31 at 8:31
$begingroup$
Your attempted argument is confusing because you use $a,y,x$ without telling what they are supposed to mean. For example, it seems that $a$ is any point in $S$ and you want to prove that some ball around it is still in $S$. But I can't figure out what you need $y$ for, and how it is determined.
$endgroup$
– Ingix
Mar 31 at 8:33
$begingroup$
y and a are both in S, as stated. I was playing around with the triangle inequality to see if I could get somewhere useful. I tried googling what the sup (max) norm is but I'm still a bit confused. For reference, this is for a real analysis course covering the basics about topology of R^n.
$endgroup$
– moofasa
Mar 31 at 8:46
3
3
$begingroup$
what metric are you using?
$endgroup$
– Riquelme
Mar 31 at 8:25
$begingroup$
what metric are you using?
$endgroup$
– Riquelme
Mar 31 at 8:25
$begingroup$
Euclidean metric
$endgroup$
– moofasa
Mar 31 at 8:29
$begingroup$
Euclidean metric
$endgroup$
– moofasa
Mar 31 at 8:29
1
1
$begingroup$
All norms are equivalent on finite dimensional vector spaces, and as you want each component of the vector to have mod less than 1, use the sup (max) norm, then the norm of an arbitrary vector in your set would be $c$ for some $c < 1$ then a ball of radius $r<1-c$ should work?
$endgroup$
– Displayname
Mar 31 at 8:31
$begingroup$
All norms are equivalent on finite dimensional vector spaces, and as you want each component of the vector to have mod less than 1, use the sup (max) norm, then the norm of an arbitrary vector in your set would be $c$ for some $c < 1$ then a ball of radius $r<1-c$ should work?
$endgroup$
– Displayname
Mar 31 at 8:31
$begingroup$
Your attempted argument is confusing because you use $a,y,x$ without telling what they are supposed to mean. For example, it seems that $a$ is any point in $S$ and you want to prove that some ball around it is still in $S$. But I can't figure out what you need $y$ for, and how it is determined.
$endgroup$
– Ingix
Mar 31 at 8:33
$begingroup$
Your attempted argument is confusing because you use $a,y,x$ without telling what they are supposed to mean. For example, it seems that $a$ is any point in $S$ and you want to prove that some ball around it is still in $S$. But I can't figure out what you need $y$ for, and how it is determined.
$endgroup$
– Ingix
Mar 31 at 8:33
$begingroup$
y and a are both in S, as stated. I was playing around with the triangle inequality to see if I could get somewhere useful. I tried googling what the sup (max) norm is but I'm still a bit confused. For reference, this is for a real analysis course covering the basics about topology of R^n.
$endgroup$
– moofasa
Mar 31 at 8:46
$begingroup$
y and a are both in S, as stated. I was playing around with the triangle inequality to see if I could get somewhere useful. I tried googling what the sup (max) norm is but I'm still a bit confused. For reference, this is for a real analysis course covering the basics about topology of R^n.
$endgroup$
– moofasa
Mar 31 at 8:46
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Assume (a,b) in S = [0,1]^2.
Let r > 0 be less than a, 1-a, b, and 1-b.
Show B((a,b),r) is a subset of S. (Euclidean metric)
Thus, as every point of S is inside an open subset S, S is open.
Alternatively, show S is the union of all the above described balls and conclude that as S is a union of open sets, it is open.
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
$endgroup$
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
add a comment |
$begingroup$
Consider $(a,b) in (0,1)×(0,1)$;
$d_x:= min (1-a,a)$; $d_y:= min (1-b,b)$;
$r:=min (d_x,d_y).$
$B_r(a,b)= $
sqrt(x-a)^2 +(y-b)^2<r$.
$(x,y) in B_r(a,b)$ implies
$|x-a| < r le d_x$, and $|y-b| <r le d_y$, hence
$B_r(a,b) subset (0,1)×(0,1)$.
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
Hint
Show that if $x=(x_1,x_2,cdots,x_n)in S$, then by defining $$y=(y_1,y_2,cdots , y_n)\|y_i-x_i|<epsilon$$$y$ will also belong to $S$ for sufficiently small $epsilon>0$.
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
$endgroup$
– moofasa
Apr 1 at 1:16
$begingroup$
Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume (a,b) in S = [0,1]^2.
Let r > 0 be less than a, 1-a, b, and 1-b.
Show B((a,b),r) is a subset of S. (Euclidean metric)
Thus, as every point of S is inside an open subset S, S is open.
Alternatively, show S is the union of all the above described balls and conclude that as S is a union of open sets, it is open.
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
$endgroup$
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
add a comment |
$begingroup$
Assume (a,b) in S = [0,1]^2.
Let r > 0 be less than a, 1-a, b, and 1-b.
Show B((a,b),r) is a subset of S. (Euclidean metric)
Thus, as every point of S is inside an open subset S, S is open.
Alternatively, show S is the union of all the above described balls and conclude that as S is a union of open sets, it is open.
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
$endgroup$
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
add a comment |
$begingroup$
Assume (a,b) in S = [0,1]^2.
Let r > 0 be less than a, 1-a, b, and 1-b.
Show B((a,b),r) is a subset of S. (Euclidean metric)
Thus, as every point of S is inside an open subset S, S is open.
Alternatively, show S is the union of all the above described balls and conclude that as S is a union of open sets, it is open.
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
$endgroup$
Assume (a,b) in S = [0,1]^2.
Let r > 0 be less than a, 1-a, b, and 1-b.
Show B((a,b),r) is a subset of S. (Euclidean metric)
Thus, as every point of S is inside an open subset S, S is open.
Alternatively, show S is the union of all the above described balls and conclude that as S is a union of open sets, it is open.
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
answered Mar 31 at 9:09
William ElliotWilliam Elliot
9,1812820
9,1812820
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
add a comment |
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
$begingroup$
I'm not seeing how to relate these balls to showing that (x,y) in B((a,b),r) => 0<x<1 and 0<y<1.
$endgroup$
– moofasa
Mar 31 at 9:48
add a comment |
$begingroup$
Consider $(a,b) in (0,1)×(0,1)$;
$d_x:= min (1-a,a)$; $d_y:= min (1-b,b)$;
$r:=min (d_x,d_y).$
$B_r(a,b)= $
sqrt(x-a)^2 +(y-b)^2<r$.
$(x,y) in B_r(a,b)$ implies
$|x-a| < r le d_x$, and $|y-b| <r le d_y$, hence
$B_r(a,b) subset (0,1)×(0,1)$.
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
Consider $(a,b) in (0,1)×(0,1)$;
$d_x:= min (1-a,a)$; $d_y:= min (1-b,b)$;
$r:=min (d_x,d_y).$
$B_r(a,b)= $
sqrt(x-a)^2 +(y-b)^2<r$.
$(x,y) in B_r(a,b)$ implies
$|x-a| < r le d_x$, and $|y-b| <r le d_y$, hence
$B_r(a,b) subset (0,1)×(0,1)$.
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
Consider $(a,b) in (0,1)×(0,1)$;
$d_x:= min (1-a,a)$; $d_y:= min (1-b,b)$;
$r:=min (d_x,d_y).$
$B_r(a,b)= $
sqrt(x-a)^2 +(y-b)^2<r$.
$(x,y) in B_r(a,b)$ implies
$|x-a| < r le d_x$, and $|y-b| <r le d_y$, hence
$B_r(a,b) subset (0,1)×(0,1)$.
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
$endgroup$
Consider $(a,b) in (0,1)×(0,1)$;
$d_x:= min (1-a,a)$; $d_y:= min (1-b,b)$;
$r:=min (d_x,d_y).$
$B_r(a,b)= $
sqrt(x-a)^2 +(y-b)^2<r$.
$(x,y) in B_r(a,b)$ implies
$|x-a| < r le d_x$, and $|y-b| <r le d_y$, hence
$B_r(a,b) subset (0,1)×(0,1)$.
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
answered Mar 31 at 10:35
Peter SzilasPeter Szilas
12k2822
12k2822
add a comment |
add a comment |
$begingroup$
Hint
Show that if $x=(x_1,x_2,cdots,x_n)in S$, then by defining $$y=(y_1,y_2,cdots , y_n)\|y_i-x_i|<epsilon$$$y$ will also belong to $S$ for sufficiently small $epsilon>0$.
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
$endgroup$
– moofasa
Apr 1 at 1:16
$begingroup$
Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
add a comment |
$begingroup$
Hint
Show that if $x=(x_1,x_2,cdots,x_n)in S$, then by defining $$y=(y_1,y_2,cdots , y_n)\|y_i-x_i|<epsilon$$$y$ will also belong to $S$ for sufficiently small $epsilon>0$.
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
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$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
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– moofasa
Apr 1 at 1:16
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Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
add a comment |
$begingroup$
Hint
Show that if $x=(x_1,x_2,cdots,x_n)in S$, then by defining $$y=(y_1,y_2,cdots , y_n)\|y_i-x_i|<epsilon$$$y$ will also belong to $S$ for sufficiently small $epsilon>0$.
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Show that if $x=(x_1,x_2,cdots,x_n)in S$, then by defining $$y=(y_1,y_2,cdots , y_n)\|y_i-x_i|<epsilon$$$y$ will also belong to $S$ for sufficiently small $epsilon>0$.
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 31 at 12:08
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
$endgroup$
– moofasa
Apr 1 at 1:16
$begingroup$
Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
add a comment |
$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
$endgroup$
– moofasa
Apr 1 at 1:16
$begingroup$
Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
$endgroup$
– moofasa
Apr 1 at 1:16
$begingroup$
Maybe I am overthinking this, but I don't see how showing that the distance is smaller than epsilon argues that $vecy$ has components that are each between 0 and 1. Could you elaborate?
$endgroup$
– moofasa
Apr 1 at 1:16
$begingroup$
Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
$begingroup$
Sure. If any $x_i$ falls within $(0,1)$, then by definition $$x_i-epsilon<y_i<x_i+epsilon$$and you can choose $epsilon>0$ sufficiently small such that $$0<x_i-epsilon<y_i<x_i+epsilon<1$$this is because from $$0<u<1$$one can choose some $epsilon>0$ such that$$epsilon<u<1-epsilon$$
$endgroup$
– Mostafa Ayaz
Apr 1 at 7:53
add a comment |
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what metric are you using?
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– Riquelme
Mar 31 at 8:25
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Euclidean metric
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– moofasa
Mar 31 at 8:29
1
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All norms are equivalent on finite dimensional vector spaces, and as you want each component of the vector to have mod less than 1, use the sup (max) norm, then the norm of an arbitrary vector in your set would be $c$ for some $c < 1$ then a ball of radius $r<1-c$ should work?
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– Displayname
Mar 31 at 8:31
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Your attempted argument is confusing because you use $a,y,x$ without telling what they are supposed to mean. For example, it seems that $a$ is any point in $S$ and you want to prove that some ball around it is still in $S$. But I can't figure out what you need $y$ for, and how it is determined.
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– Ingix
Mar 31 at 8:33
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y and a are both in S, as stated. I was playing around with the triangle inequality to see if I could get somewhere useful. I tried googling what the sup (max) norm is but I'm still a bit confused. For reference, this is for a real analysis course covering the basics about topology of R^n.
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– moofasa
Mar 31 at 8:46