Radius of an Open Ball inside the Intersection of an Open, Dense Subset and the Ball Around a Point in the...
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Let $I subseteq mathbf{R} $ be dense and open.
Because $I$ is dense, for $x in mathbf{R}$,
$$exists yin I:y=I cap B_{epsilon}(x)$$
Because $I$ and $B_{epsilon}(x)$ are open, $I cap B_{epsilon}(x)$ is also open and thus $exists delta, text{as a function of y, }delta(y) text{, such that}$:
$$B_{delta}(y) subseteq I cap B_{epsilon}(x) tag{*}$$
Question: Is there a way to construct $y$ so that we can have an upper bound for $delta (y)$ that makes (*) holds? In other words, can we find a $y' in I$ such that:
$$delta (y') < alpha to B_{delta(y')}(y')subseteq I cap B_{epsilon}(x)$$
general-topology
asked Nov 29 at 4:02
A Slow Learner
403212
|
show 5 more comments
up vote
0
down vote
favorite
Let $I subseteq mathbf{R} $ be dense and open.
Because $I$ is dense, for $x in mathbf{R}$,
$$exists yin I:y=I cap B_{epsilon}(x)$$
Because $I$ and $B_{epsilon}(x)$ are open, $I cap B_{epsilon}(x)$ is also open and thus $exists delta, text{as a function of y, }delta(y) text{, such that}$:
$$B_{delta}(y) subseteq I cap B_{epsilon}(x) tag{*}$$
Question: Is there a way to construct $y$ so that we can have an upper bound for $delta (y)$ that makes (*) holds? In other words, can we find a $y' in I$ such that:
$$delta (y') < alpha to B_{delta(y')}(y')subseteq I cap B_{epsilon}(x)$$
general-topology
asked Nov 29 at 4:02
A Slow Learner
403212
We can think of diameter of $I cap B_{epsilon} left( x right)$ to find an upper bound?
– Aniruddha Deshmukh
Nov 29 at 4:07
Wouldn't the interesting question be to find a lower bound on $delta(y)$?
– user25959
Nov 29 at 4:12
@AniruddhaDeshmukh The only thing about the diameter $d$ of $Icap B$ that I can come up with is that $d<epsilon$. What do you think and how are you going to proceed with this?
– A Slow Learner
Nov 29 at 4:39
@user25959 I am not sure about this. Obviously, one of the lower bounds is $0$. But I don't think there exists a largest lower bound.
– A Slow Learner
Nov 29 at 4:45
@ASlowLearner I think an upper bound for your function $delta left( y right)$ should be half of the diameter. Although I do not know if we can find a tighter bound.
– Aniruddha Deshmukh
Nov 29 at 6:00
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $I subseteq mathbf{R} $ be dense and open.
Because $I$ is dense, for $x in mathbf{R}$,
$$exists yin I:y=I cap B_{epsilon}(x)$$
Because $I$ and $B_{epsilon}(x)$ are open, $I cap B_{epsilon}(x)$ is also open and thus $exists delta, text{as a function of y, }delta(y) text{, such that}$:
$$B_{delta}(y) subseteq I cap B_{epsilon}(x) tag{*}$$
Question: Is there a way to construct $y$ so that we can have an upper bound for $delta (y)$ that makes (*) holds? In other words, can we find a $y' in I$ such that:
$$delta (y') < alpha to B_{delta(y')}(y')subseteq I cap B_{epsilon}(x)$$
general-topology
asked Nov 29 at 4:02
A Slow Learner
403212
Let $I subseteq mathbf{R} $ be dense and open.
Because $I$ is dense, for $x in mathbf{R}$,
$$exists yin I:y=I cap B_{epsilon}(x)$$
Because $I$ and $B_{epsilon}(x)$ are open, $I cap B_{epsilon}(x)$ is also open and thus $exists delta, text{as a function of y, }delta(y) text{, such that}$:
$$B_{delta}(y) subseteq I cap B_{epsilon}(x) tag{*}$$
Question: Is there a way to construct $y$ so that we can have an upper bound for $delta (y)$ that makes (*) holds? In other words, can we find a $y' in I$ such that:
$$delta (y') < alpha to B_{delta(y')}(y')subseteq I cap B_{epsilon}(x)$$
general-topology
general-topology
asked Nov 29 at 4:02
A Slow Learner
403212
asked Nov 29 at 4:02
A Slow Learner
403212
edited Nov 29 at 4:07
asked Nov 29 at 4:02
A Slow Learner
403212
asked Nov 29 at 4:02
A Slow Learner
403212
asked Nov 29 at 4:02
A Slow Learner
403212
403212
We can think of diameter of $I cap B_{epsilon} left( x right)$ to find an upper bound?
– Aniruddha Deshmukh
Nov 29 at 4:07
Wouldn't the interesting question be to find a lower bound on $delta(y)$?
– user25959
Nov 29 at 4:12
@AniruddhaDeshmukh The only thing about the diameter $d$ of $Icap B$ that I can come up with is that $d<epsilon$. What do you think and how are you going to proceed with this?
– A Slow Learner
Nov 29 at 4:39
@user25959 I am not sure about this. Obviously, one of the lower bounds is $0$. But I don't think there exists a largest lower bound.
– A Slow Learner
Nov 29 at 4:45
@ASlowLearner I think an upper bound for your function $delta left( y right)$ should be half of the diameter. Although I do not know if we can find a tighter bound.
– Aniruddha Deshmukh
Nov 29 at 6:00
|
show 5 more comments
We can think of diameter of $I cap B_{epsilon} left( x right)$ to find an upper bound?
– Aniruddha Deshmukh
Nov 29 at 4:07
Wouldn't the interesting question be to find a lower bound on $delta(y)$?
– user25959
Nov 29 at 4:12
@AniruddhaDeshmukh The only thing about the diameter $d$ of $Icap B$ that I can come up with is that $d<epsilon$. What do you think and how are you going to proceed with this?
– A Slow Learner
Nov 29 at 4:39
@user25959 I am not sure about this. Obviously, one of the lower bounds is $0$. But I don't think there exists a largest lower bound.
– A Slow Learner
Nov 29 at 4:45
@ASlowLearner I think an upper bound for your function $delta left( y right)$ should be half of the diameter. Although I do not know if we can find a tighter bound.
– Aniruddha Deshmukh
Nov 29 at 6:00
We can think of diameter of $I cap B_{epsilon} left( x right)$ to find an upper bound?
– Aniruddha Deshmukh
Nov 29 at 4:07
We can think of diameter of $I cap B_{epsilon} left( x right)$ to find an upper bound?
– Aniruddha Deshmukh
Nov 29 at 4:07
Wouldn't the interesting question be to find a lower bound on $delta(y)$?
– user25959
Nov 29 at 4:12
Wouldn't the interesting question be to find a lower bound on $delta(y)$?
– user25959
Nov 29 at 4:12
@AniruddhaDeshmukh The only thing about the diameter $d$ of $Icap B$ that I can come up with is that $d<epsilon$. What do you think and how are you going to proceed with this?
– A Slow Learner
Nov 29 at 4:39
@AniruddhaDeshmukh The only thing about the diameter $d$ of $Icap B$ that I can come up with is that $d<epsilon$. What do you think and how are you going to proceed with this?
– A Slow Learner
Nov 29 at 4:39
@user25959 I am not sure about this. Obviously, one of the lower bounds is $0$. But I don't think there exists a largest lower bound.
– A Slow Learner
Nov 29 at 4:45
@user25959 I am not sure about this. Obviously, one of the lower bounds is $0$. But I don't think there exists a largest lower bound.
– A Slow Learner
Nov 29 at 4:45
@ASlowLearner I think an upper bound for your function $delta left( y right)$ should be half of the diameter. Although I do not know if we can find a tighter bound.
– Aniruddha Deshmukh
Nov 29 at 6:00
@ASlowLearner I think an upper bound for your function $delta left( y right)$ should be half of the diameter. Although I do not know if we can find a tighter bound.
– Aniruddha Deshmukh
Nov 29 at 6:00
|
show 5 more comments
1 Answer
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Let $left( X, d right)$ be any metric space with the $d$ as the distance function. Consider your dense open subset $I$ and any point $x in X$. Now for a given epsilon, consider $I cap B_{epsilon} left( x right)$. Now, if $y$ belongs to the intersection, in particular, $y$ belongs to $I$.
Therefore, $exists delta_1 > 0$ such that $B_{delta_1} left( y right) subseteq I$.
Also, $y$ is in $B_{epsilon} left( x right)$. Therefore, $exists delta_2 > 0$ such that $B_{delta_2} left( y right) subseteq B_{epsilon} left( x right)$. Now, if you choose $delta = min leftlbrace delta_1, epsilon - d left( x, y right) rightrbrace$, then $B_{delta} left( y right) subseteq I cap B_{epsilon} left( x right)$ which will do your work.
I have attached an image below for better understanding. In the image, $y$ is shown with $2$ open balls, one is relative to $I$ and other is relative to $B_{epsilon} left( x right)$. Once you choose the lesser radius, you are done with what you want.
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
|
show 1 more comment
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1 Answer
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Let $left( X, d right)$ be any metric space with the $d$ as the distance function. Consider your dense open subset $I$ and any point $x in X$. Now for a given epsilon, consider $I cap B_{epsilon} left( x right)$. Now, if $y$ belongs to the intersection, in particular, $y$ belongs to $I$.
Therefore, $exists delta_1 > 0$ such that $B_{delta_1} left( y right) subseteq I$.
Also, $y$ is in $B_{epsilon} left( x right)$. Therefore, $exists delta_2 > 0$ such that $B_{delta_2} left( y right) subseteq B_{epsilon} left( x right)$. Now, if you choose $delta = min leftlbrace delta_1, epsilon - d left( x, y right) rightrbrace$, then $B_{delta} left( y right) subseteq I cap B_{epsilon} left( x right)$ which will do your work.
I have attached an image below for better understanding. In the image, $y$ is shown with $2$ open balls, one is relative to $I$ and other is relative to $B_{epsilon} left( x right)$. Once you choose the lesser radius, you are done with what you want.
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
|
show 1 more comment
up vote
0
down vote
Let $left( X, d right)$ be any metric space with the $d$ as the distance function. Consider your dense open subset $I$ and any point $x in X$. Now for a given epsilon, consider $I cap B_{epsilon} left( x right)$. Now, if $y$ belongs to the intersection, in particular, $y$ belongs to $I$.
Therefore, $exists delta_1 > 0$ such that $B_{delta_1} left( y right) subseteq I$.
Also, $y$ is in $B_{epsilon} left( x right)$. Therefore, $exists delta_2 > 0$ such that $B_{delta_2} left( y right) subseteq B_{epsilon} left( x right)$. Now, if you choose $delta = min leftlbrace delta_1, epsilon - d left( x, y right) rightrbrace$, then $B_{delta} left( y right) subseteq I cap B_{epsilon} left( x right)$ which will do your work.
I have attached an image below for better understanding. In the image, $y$ is shown with $2$ open balls, one is relative to $I$ and other is relative to $B_{epsilon} left( x right)$. Once you choose the lesser radius, you are done with what you want.
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
|
show 1 more comment
up vote
0
down vote
up vote
0
down vote
Let $left( X, d right)$ be any metric space with the $d$ as the distance function. Consider your dense open subset $I$ and any point $x in X$. Now for a given epsilon, consider $I cap B_{epsilon} left( x right)$. Now, if $y$ belongs to the intersection, in particular, $y$ belongs to $I$.
Therefore, $exists delta_1 > 0$ such that $B_{delta_1} left( y right) subseteq I$.
Also, $y$ is in $B_{epsilon} left( x right)$. Therefore, $exists delta_2 > 0$ such that $B_{delta_2} left( y right) subseteq B_{epsilon} left( x right)$. Now, if you choose $delta = min leftlbrace delta_1, epsilon - d left( x, y right) rightrbrace$, then $B_{delta} left( y right) subseteq I cap B_{epsilon} left( x right)$ which will do your work.
I have attached an image below for better understanding. In the image, $y$ is shown with $2$ open balls, one is relative to $I$ and other is relative to $B_{epsilon} left( x right)$. Once you choose the lesser radius, you are done with what you want.
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
Let $left( X, d right)$ be any metric space with the $d$ as the distance function. Consider your dense open subset $I$ and any point $x in X$. Now for a given epsilon, consider $I cap B_{epsilon} left( x right)$. Now, if $y$ belongs to the intersection, in particular, $y$ belongs to $I$.
Therefore, $exists delta_1 > 0$ such that $B_{delta_1} left( y right) subseteq I$.
Also, $y$ is in $B_{epsilon} left( x right)$. Therefore, $exists delta_2 > 0$ such that $B_{delta_2} left( y right) subseteq B_{epsilon} left( x right)$. Now, if you choose $delta = min leftlbrace delta_1, epsilon - d left( x, y right) rightrbrace$, then $B_{delta} left( y right) subseteq I cap B_{epsilon} left( x right)$ which will do your work.
I have attached an image below for better understanding. In the image, $y$ is shown with $2$ open balls, one is relative to $I$ and other is relative to $B_{epsilon} left( x right)$. Once you choose the lesser radius, you are done with what you want.
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
answered Nov 29 at 7:18
Aniruddha Deshmukh
811418
811418
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
|
show 1 more comment
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
I am not sure but is there a typo in the $6^{th}$ line?
– A Slow Learner
Nov 29 at 7:33
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Where do you think exactly is the error?
– Aniruddha Deshmukh
Nov 29 at 7:52
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Shouldn't it be $delta = min leftlbrace delta_1, delta_2 rightrbrace$ instead?
– A Slow Learner
Nov 29 at 7:56
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Yes, that will also work. Even what I have written will work, since $epsilon - d left( x, y right)$ is the distance of the point $y$ from the boundary of $B_{epsilon} left( x right)$. Now, if you see carefully, the $delta$ I have proposed guarantees that the open ball lies within $I$ (since $delta leq delta_1$ and also it is less than or equal to the distance of $y$ from the boundary of $B_{epsilon} left( x right)$. Hence, the open ball will be completely inside the intersection.
– Aniruddha Deshmukh
Nov 29 at 7:57
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
Ok so I am confused now. Did you mean both will work?
– A Slow Learner
Nov 29 at 7:58
|
show 1 more comment
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We can think of diameter of $I cap B_{epsilon} left( x right)$ to find an upper bound?
– Aniruddha Deshmukh
Nov 29 at 4:07
Wouldn't the interesting question be to find a lower bound on $delta(y)$?
– user25959
Nov 29 at 4:12
@AniruddhaDeshmukh The only thing about the diameter $d$ of $Icap B$ that I can come up with is that $d<epsilon$. What do you think and how are you going to proceed with this?
– A Slow Learner
Nov 29 at 4:39
@user25959 I am not sure about this. Obviously, one of the lower bounds is $0$. But I don't think there exists a largest lower bound.
– A Slow Learner
Nov 29 at 4:45
@ASlowLearner I think an upper bound for your function $delta left( y right)$ should be half of the diameter. Although I do not know if we can find a tighter bound.
– Aniruddha Deshmukh
Nov 29 at 6:00