Show that a subset of $mathbbR^2$ is open
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Show that the subset of $mathbbR^2$ given by $(x_1, x_2)in mathbbR^2:x_1>x_2$ is open.
Can anyone give me some help, or a proof for it. The book I'm using defines a subset $S$ of $E$ as open if for each $pin S$, $S$ contains some open ball of center $p$.
Any input is appreciated.
general-topology
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
add a comment |Â
up vote
2
down vote
favorite
Show that the subset of $mathbbR^2$ given by $(x_1, x_2)in mathbbR^2:x_1>x_2$ is open.
Can anyone give me some help, or a proof for it. The book I'm using defines a subset $S$ of $E$ as open if for each $pin S$, $S$ contains some open ball of center $p$.
Any input is appreciated.
general-topology
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
What is $E$?
– wj32
Nov 23 '12 at 0:10
I guess that $E^2$ stands for the Euclidean plane.
– Giuseppe Negro
Nov 23 '12 at 0:17
1
What about trying to build such an open ball for every point in your subset ? Given a point $p$, what is its distance to the boundary of the subset ?
– beauby
Nov 23 '12 at 0:22
1
I highly recommend drawing a picture.
– icurays1
Nov 23 '12 at 0:36
Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.
– Pax Kivimae
Nov 23 '12 at 0:51
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Show that the subset of $mathbbR^2$ given by $(x_1, x_2)in mathbbR^2:x_1>x_2$ is open.
Can anyone give me some help, or a proof for it. The book I'm using defines a subset $S$ of $E$ as open if for each $pin S$, $S$ contains some open ball of center $p$.
Any input is appreciated.
general-topology
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
Show that the subset of $mathbbR^2$ given by $(x_1, x_2)in mathbbR^2:x_1>x_2$ is open.
Can anyone give me some help, or a proof for it. The book I'm using defines a subset $S$ of $E$ as open if for each $pin S$, $S$ contains some open ball of center $p$.
Any input is appreciated.
general-topology
general-topology
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
edited Jul 21 at 4:18
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
asked Nov 23 '12 at 0:08
Pax Kivimae
3,34131132
3,34131132
What is $E$?
– wj32
Nov 23 '12 at 0:10
I guess that $E^2$ stands for the Euclidean plane.
– Giuseppe Negro
Nov 23 '12 at 0:17
1
What about trying to build such an open ball for every point in your subset ? Given a point $p$, what is its distance to the boundary of the subset ?
– beauby
Nov 23 '12 at 0:22
1
I highly recommend drawing a picture.
– icurays1
Nov 23 '12 at 0:36
Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.
– Pax Kivimae
Nov 23 '12 at 0:51
add a comment |Â
What is $E$?
– wj32
Nov 23 '12 at 0:10
I guess that $E^2$ stands for the Euclidean plane.
– Giuseppe Negro
Nov 23 '12 at 0:17
1
What about trying to build such an open ball for every point in your subset ? Given a point $p$, what is its distance to the boundary of the subset ?
– beauby
Nov 23 '12 at 0:22
1
I highly recommend drawing a picture.
– icurays1
Nov 23 '12 at 0:36
Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.
– Pax Kivimae
Nov 23 '12 at 0:51
What is $E$?
– wj32
Nov 23 '12 at 0:10
What is $E$?
– wj32
Nov 23 '12 at 0:10
I guess that $E^2$ stands for the Euclidean plane.
– Giuseppe Negro
Nov 23 '12 at 0:17
I guess that $E^2$ stands for the Euclidean plane.
– Giuseppe Negro
Nov 23 '12 at 0:17
1
1
What about trying to build such an open ball for every point in your subset ? Given a point $p$, what is its distance to the boundary of the subset ?
– beauby
Nov 23 '12 at 0:22
What about trying to build such an open ball for every point in your subset ? Given a point $p$, what is its distance to the boundary of the subset ?
– beauby
Nov 23 '12 at 0:22
1
1
I highly recommend drawing a picture.
– icurays1
Nov 23 '12 at 0:36
I highly recommend drawing a picture.
– icurays1
Nov 23 '12 at 0:36
Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.
– Pax Kivimae
Nov 23 '12 at 0:51
Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.
– Pax Kivimae
Nov 23 '12 at 0:51
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Let's call this subset $S$, and let $(x_1,x_2) in S$. Suppose that $x_1 - x_2 = r > 0$. Can see why the open ball of radius $r/10$ around $(x_1,x_2)$ should remain in $S$?
Suppose that there is a point $(y_1,y_2)$ in this ball. The triangle inequality tells us
$$ r = |x_1 -x_2| leq |x_1 - y_1| + |y_1 - y_2| + |y_2 - x_2|$$
We know that $|x_1 - y_1|$ and $|y_2 - x_2|$ can be at most $r/10$, so their sum is at most $r/5$. This means that $|y_1 - y_2|$ is at least $4r/5 > 0$, and so $(y_1,y_2) in S$.
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
add a comment |Â
up vote
3
down vote
Define $f:E^2tomathbb R$ by $f(x_1,x_2)=x_1-x_2$. Your set is then $f^-1 (0,infty)$ which is open because $(0,infty)$ is open and $f$ is continuous.
answered Nov 24 '12 at 10:09
nonpop
1,708614
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let's call this subset $S$, and let $(x_1,x_2) in S$. Suppose that $x_1 - x_2 = r > 0$. Can see why the open ball of radius $r/10$ around $(x_1,x_2)$ should remain in $S$?
Suppose that there is a point $(y_1,y_2)$ in this ball. The triangle inequality tells us
$$ r = |x_1 -x_2| leq |x_1 - y_1| + |y_1 - y_2| + |y_2 - x_2|$$
We know that $|x_1 - y_1|$ and $|y_2 - x_2|$ can be at most $r/10$, so their sum is at most $r/5$. This means that $|y_1 - y_2|$ is at least $4r/5 > 0$, and so $(y_1,y_2) in S$.
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
add a comment |Â
up vote
4
down vote
accepted
Let's call this subset $S$, and let $(x_1,x_2) in S$. Suppose that $x_1 - x_2 = r > 0$. Can see why the open ball of radius $r/10$ around $(x_1,x_2)$ should remain in $S$?
Suppose that there is a point $(y_1,y_2)$ in this ball. The triangle inequality tells us
$$ r = |x_1 -x_2| leq |x_1 - y_1| + |y_1 - y_2| + |y_2 - x_2|$$
We know that $|x_1 - y_1|$ and $|y_2 - x_2|$ can be at most $r/10$, so their sum is at most $r/5$. This means that $|y_1 - y_2|$ is at least $4r/5 > 0$, and so $(y_1,y_2) in S$.
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let's call this subset $S$, and let $(x_1,x_2) in S$. Suppose that $x_1 - x_2 = r > 0$. Can see why the open ball of radius $r/10$ around $(x_1,x_2)$ should remain in $S$?
Suppose that there is a point $(y_1,y_2)$ in this ball. The triangle inequality tells us
$$ r = |x_1 -x_2| leq |x_1 - y_1| + |y_1 - y_2| + |y_2 - x_2|$$
We know that $|x_1 - y_1|$ and $|y_2 - x_2|$ can be at most $r/10$, so their sum is at most $r/5$. This means that $|y_1 - y_2|$ is at least $4r/5 > 0$, and so $(y_1,y_2) in S$.
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
Let's call this subset $S$, and let $(x_1,x_2) in S$. Suppose that $x_1 - x_2 = r > 0$. Can see why the open ball of radius $r/10$ around $(x_1,x_2)$ should remain in $S$?
Suppose that there is a point $(y_1,y_2)$ in this ball. The triangle inequality tells us
$$ r = |x_1 -x_2| leq |x_1 - y_1| + |y_1 - y_2| + |y_2 - x_2|$$
We know that $|x_1 - y_1|$ and $|y_2 - x_2|$ can be at most $r/10$, so their sum is at most $r/5$. This means that $|y_1 - y_2|$ is at least $4r/5 > 0$, and so $(y_1,y_2) in S$.
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
edited Nov 23 '12 at 1:28
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
answered Nov 23 '12 at 0:51
Elchanan Solomon
21.4k44174
21.4k44174
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
add a comment |Â
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
$|y_1-y_2|>0$? I'm sorry I don't see anything else
– Pax Kivimae
Nov 23 '12 at 1:26
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
I've tried to clarify it. Does that help?
– Elchanan Solomon
Nov 23 '12 at 1:28
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
Yes, thank you.
– Pax Kivimae
Nov 23 '12 at 1:49
add a comment |Â
up vote
3
down vote
Define $f:E^2tomathbb R$ by $f(x_1,x_2)=x_1-x_2$. Your set is then $f^-1 (0,infty)$ which is open because $(0,infty)$ is open and $f$ is continuous.
answered Nov 24 '12 at 10:09
nonpop
1,708614
add a comment |Â
up vote
3
down vote
Define $f:E^2tomathbb R$ by $f(x_1,x_2)=x_1-x_2$. Your set is then $f^-1 (0,infty)$ which is open because $(0,infty)$ is open and $f$ is continuous.
answered Nov 24 '12 at 10:09
nonpop
1,708614
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Define $f:E^2tomathbb R$ by $f(x_1,x_2)=x_1-x_2$. Your set is then $f^-1 (0,infty)$ which is open because $(0,infty)$ is open and $f$ is continuous.
answered Nov 24 '12 at 10:09
nonpop
1,708614
Define $f:E^2tomathbb R$ by $f(x_1,x_2)=x_1-x_2$. Your set is then $f^-1 (0,infty)$ which is open because $(0,infty)$ is open and $f$ is continuous.
answered Nov 24 '12 at 10:09
nonpop
1,708614
answered Nov 24 '12 at 10:09
nonpop
1,708614
answered Nov 24 '12 at 10:09
nonpop
1,708614
answered Nov 24 '12 at 10:09
nonpop
1,708614
1,708614
add a comment |Â
add a comment |Â
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What is $E$?
– wj32
Nov 23 '12 at 0:10
I guess that $E^2$ stands for the Euclidean plane.
– Giuseppe Negro
Nov 23 '12 at 0:17
1
What about trying to build such an open ball for every point in your subset ? Given a point $p$, what is its distance to the boundary of the subset ?
– beauby
Nov 23 '12 at 0:22
1
I highly recommend drawing a picture.
– icurays1
Nov 23 '12 at 0:36
Sorry, $E^2$ is the Euclidean plane with the standard distance function. I see that is true, I just can't find a way to show it.
– Pax Kivimae
Nov 23 '12 at 0:51