Interior or border point of a set
$begingroup$
I am working on a classical real analysis problem as follow:
Find $int (S)$ and $bd (S)$ if $S = { frac{1}{n} | n in mathbb N }= { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$.
The answers are respectively $ int (S) = emptyset$ and $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$. And here are my textbook's definition of interior point and border point:
Let $S subseteq mathbb R$. A point $x$ in $mathbb R$ is an interior point of S if there exists a neighborhood $N$ of $x$ such that $N subseteq S$. If for every neighborhood of $N$ of $x$, $N cap S neq emptyset$ and $N cap S^c neq emptyset$, then $x$ is a border point of $S$.
And then there is also this:
Every point $x in S$ is either interior or border point of $S$.
To my inexperienced mind, the answer to $int (S) = emptyset $ is understandable because while neighborhood $N (x | epsilon >0)$ is an open set (per textbook), but $S = { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$ is incomplete, meaning that there is gap between any two consecutive elements of $S$. Therefore by technicality $N (x | epsilon )$ does not exist inside $S$. (Am I correct here?)
The answer to $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$ is understandably the direct consequence of $int (S) = emptyset $, i.e., if none of the elements of $S$ is interior point, then all of them must be border points, plus the zero.
But what troubles me is this: What happens if I instead answer $int(S)$ first and then $bd (S)$ second? Am I going to get the same answers?
(i) For ${0}$, I was tempted to conclude that zero is indeed a border point because $N (0 | epsilon) cap S neq emptyset$ and $N (0 | epsilon) cap S^c neq emptyset$ for all $epsilon >0$. But had I just concluded that $N (x | epsilon)$ does not exist in $S$?
(ii) How about for the rest of the points, i.e., $frac{1}{n}$? How do I conclude $N (frac{1}{n} | epsilon ) cap S neq emptyset$ and $N (frac{1}{n} | epsilon ) cap S^c neq emptyset$ for all possible $epsilon >0$?
I think I must have missed something here. Any help would therefore be very much appreciated. Thank you for your time.
real-analysis
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
$endgroup$
add a comment |
$begingroup$
I am working on a classical real analysis problem as follow:
Find $int (S)$ and $bd (S)$ if $S = { frac{1}{n} | n in mathbb N }= { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$.
The answers are respectively $ int (S) = emptyset$ and $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$. And here are my textbook's definition of interior point and border point:
Let $S subseteq mathbb R$. A point $x$ in $mathbb R$ is an interior point of S if there exists a neighborhood $N$ of $x$ such that $N subseteq S$. If for every neighborhood of $N$ of $x$, $N cap S neq emptyset$ and $N cap S^c neq emptyset$, then $x$ is a border point of $S$.
And then there is also this:
Every point $x in S$ is either interior or border point of $S$.
To my inexperienced mind, the answer to $int (S) = emptyset $ is understandable because while neighborhood $N (x | epsilon >0)$ is an open set (per textbook), but $S = { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$ is incomplete, meaning that there is gap between any two consecutive elements of $S$. Therefore by technicality $N (x | epsilon )$ does not exist inside $S$. (Am I correct here?)
The answer to $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$ is understandably the direct consequence of $int (S) = emptyset $, i.e., if none of the elements of $S$ is interior point, then all of them must be border points, plus the zero.
But what troubles me is this: What happens if I instead answer $int(S)$ first and then $bd (S)$ second? Am I going to get the same answers?
(i) For ${0}$, I was tempted to conclude that zero is indeed a border point because $N (0 | epsilon) cap S neq emptyset$ and $N (0 | epsilon) cap S^c neq emptyset$ for all $epsilon >0$. But had I just concluded that $N (x | epsilon)$ does not exist in $S$?
(ii) How about for the rest of the points, i.e., $frac{1}{n}$? How do I conclude $N (frac{1}{n} | epsilon ) cap S neq emptyset$ and $N (frac{1}{n} | epsilon ) cap S^c neq emptyset$ for all possible $epsilon >0$?
I think I must have missed something here. Any help would therefore be very much appreciated. Thank you for your time.
real-analysis
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
$endgroup$
3
$begingroup$
${emptyset}$ is different than $emptyset$. If you mean that none of the elements of $S$ is an interior point, then you mean to say that $int(S)=emptyset$. Saying that $int(S)={emptyset}$ would imply that there is an element called $emptyset$ and this element is an interior point of $S$, just like how if you say $int(S)={1}$ that would mean that $1$ is an interior element of $S$, however $emptyset$ is not an element of $S$.
$endgroup$
– JMoravitz
Dec 15 '18 at 15:05
$begingroup$
Oops! My bad! I will make correction. Thank you for your time.
$endgroup$
– Amanda.M
Dec 15 '18 at 15:08
add a comment |
$begingroup$
I am working on a classical real analysis problem as follow:
Find $int (S)$ and $bd (S)$ if $S = { frac{1}{n} | n in mathbb N }= { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$.
The answers are respectively $ int (S) = emptyset$ and $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$. And here are my textbook's definition of interior point and border point:
Let $S subseteq mathbb R$. A point $x$ in $mathbb R$ is an interior point of S if there exists a neighborhood $N$ of $x$ such that $N subseteq S$. If for every neighborhood of $N$ of $x$, $N cap S neq emptyset$ and $N cap S^c neq emptyset$, then $x$ is a border point of $S$.
And then there is also this:
Every point $x in S$ is either interior or border point of $S$.
To my inexperienced mind, the answer to $int (S) = emptyset $ is understandable because while neighborhood $N (x | epsilon >0)$ is an open set (per textbook), but $S = { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$ is incomplete, meaning that there is gap between any two consecutive elements of $S$. Therefore by technicality $N (x | epsilon )$ does not exist inside $S$. (Am I correct here?)
The answer to $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$ is understandably the direct consequence of $int (S) = emptyset $, i.e., if none of the elements of $S$ is interior point, then all of them must be border points, plus the zero.
But what troubles me is this: What happens if I instead answer $int(S)$ first and then $bd (S)$ second? Am I going to get the same answers?
(i) For ${0}$, I was tempted to conclude that zero is indeed a border point because $N (0 | epsilon) cap S neq emptyset$ and $N (0 | epsilon) cap S^c neq emptyset$ for all $epsilon >0$. But had I just concluded that $N (x | epsilon)$ does not exist in $S$?
(ii) How about for the rest of the points, i.e., $frac{1}{n}$? How do I conclude $N (frac{1}{n} | epsilon ) cap S neq emptyset$ and $N (frac{1}{n} | epsilon ) cap S^c neq emptyset$ for all possible $epsilon >0$?
I think I must have missed something here. Any help would therefore be very much appreciated. Thank you for your time.
real-analysis
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
$endgroup$
I am working on a classical real analysis problem as follow:
Find $int (S)$ and $bd (S)$ if $S = { frac{1}{n} | n in mathbb N }= { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$.
The answers are respectively $ int (S) = emptyset$ and $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$. And here are my textbook's definition of interior point and border point:
Let $S subseteq mathbb R$. A point $x$ in $mathbb R$ is an interior point of S if there exists a neighborhood $N$ of $x$ such that $N subseteq S$. If for every neighborhood of $N$ of $x$, $N cap S neq emptyset$ and $N cap S^c neq emptyset$, then $x$ is a border point of $S$.
And then there is also this:
Every point $x in S$ is either interior or border point of $S$.
To my inexperienced mind, the answer to $int (S) = emptyset $ is understandable because while neighborhood $N (x | epsilon >0)$ is an open set (per textbook), but $S = { frac{1}{1}, frac{1}{2}, frac{1}{3}, ... }$ is incomplete, meaning that there is gap between any two consecutive elements of $S$. Therefore by technicality $N (x | epsilon )$ does not exist inside $S$. (Am I correct here?)
The answer to $bd (S) = {0} cup { frac{1}{n} | n in mathbb N }$ is understandably the direct consequence of $int (S) = emptyset $, i.e., if none of the elements of $S$ is interior point, then all of them must be border points, plus the zero.
But what troubles me is this: What happens if I instead answer $int(S)$ first and then $bd (S)$ second? Am I going to get the same answers?
(i) For ${0}$, I was tempted to conclude that zero is indeed a border point because $N (0 | epsilon) cap S neq emptyset$ and $N (0 | epsilon) cap S^c neq emptyset$ for all $epsilon >0$. But had I just concluded that $N (x | epsilon)$ does not exist in $S$?
(ii) How about for the rest of the points, i.e., $frac{1}{n}$? How do I conclude $N (frac{1}{n} | epsilon ) cap S neq emptyset$ and $N (frac{1}{n} | epsilon ) cap S^c neq emptyset$ for all possible $epsilon >0$?
I think I must have missed something here. Any help would therefore be very much appreciated. Thank you for your time.
real-analysis
real-analysis
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
edited Dec 15 '18 at 15:15
Amanda.M
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
asked Dec 15 '18 at 15:02
Amanda.MAmanda.M
1,62411435
1,62411435
3
$begingroup$
${emptyset}$ is different than $emptyset$. If you mean that none of the elements of $S$ is an interior point, then you mean to say that $int(S)=emptyset$. Saying that $int(S)={emptyset}$ would imply that there is an element called $emptyset$ and this element is an interior point of $S$, just like how if you say $int(S)={1}$ that would mean that $1$ is an interior element of $S$, however $emptyset$ is not an element of $S$.
$endgroup$
– JMoravitz
Dec 15 '18 at 15:05
$begingroup$
Oops! My bad! I will make correction. Thank you for your time.
$endgroup$
– Amanda.M
Dec 15 '18 at 15:08
add a comment |
3
$begingroup$
${emptyset}$ is different than $emptyset$. If you mean that none of the elements of $S$ is an interior point, then you mean to say that $int(S)=emptyset$. Saying that $int(S)={emptyset}$ would imply that there is an element called $emptyset$ and this element is an interior point of $S$, just like how if you say $int(S)={1}$ that would mean that $1$ is an interior element of $S$, however $emptyset$ is not an element of $S$.
$endgroup$
– JMoravitz
Dec 15 '18 at 15:05
$begingroup$
Oops! My bad! I will make correction. Thank you for your time.
$endgroup$
– Amanda.M
Dec 15 '18 at 15:08
3
3
$begingroup$
${emptyset}$ is different than $emptyset$. If you mean that none of the elements of $S$ is an interior point, then you mean to say that $int(S)=emptyset$. Saying that $int(S)={emptyset}$ would imply that there is an element called $emptyset$ and this element is an interior point of $S$, just like how if you say $int(S)={1}$ that would mean that $1$ is an interior element of $S$, however $emptyset$ is not an element of $S$.
$endgroup$
– JMoravitz
Dec 15 '18 at 15:05
$begingroup$
${emptyset}$ is different than $emptyset$. If you mean that none of the elements of $S$ is an interior point, then you mean to say that $int(S)=emptyset$. Saying that $int(S)={emptyset}$ would imply that there is an element called $emptyset$ and this element is an interior point of $S$, just like how if you say $int(S)={1}$ that would mean that $1$ is an interior element of $S$, however $emptyset$ is not an element of $S$.
$endgroup$
– JMoravitz
Dec 15 '18 at 15:05
$begingroup$
Oops! My bad! I will make correction. Thank you for your time.
$endgroup$
– Amanda.M
Dec 15 '18 at 15:08
$begingroup$
Oops! My bad! I will make correction. Thank you for your time.
$endgroup$
– Amanda.M
Dec 15 '18 at 15:08
add a comment |
1 Answer
1
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$begingroup$
Personally, I like the following definitions:
A point $p$ is an interior point of $X$ if $exists r>0$ such that $B_r(p) subset X$. The interior of $X$ is the set of all interior points.
A point $q$ is a boundary point of $X$ if $forall r>0$ $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$.
(With this, open sets have none of their boundary points, closed sets have all of their boundary points, giving at least some dichotomy to open-closed. Can we call sets that have some of their boundary points "ajar"? :) )
Your sequence has no interior points because any small neighborhood around $frac{1}{n}$ will contain points that are not in the sequence. Therefore, they are all boundary points. However, you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point. All other points in $mathbb{R}$ are interior points of the complement, so not boundary points of the sequence.
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
$endgroup$
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
add a comment |
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$begingroup$
Personally, I like the following definitions:
A point $p$ is an interior point of $X$ if $exists r>0$ such that $B_r(p) subset X$. The interior of $X$ is the set of all interior points.
A point $q$ is a boundary point of $X$ if $forall r>0$ $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$.
(With this, open sets have none of their boundary points, closed sets have all of their boundary points, giving at least some dichotomy to open-closed. Can we call sets that have some of their boundary points "ajar"? :) )
Your sequence has no interior points because any small neighborhood around $frac{1}{n}$ will contain points that are not in the sequence. Therefore, they are all boundary points. However, you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point. All other points in $mathbb{R}$ are interior points of the complement, so not boundary points of the sequence.
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
$endgroup$
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
add a comment |
$begingroup$
Personally, I like the following definitions:
A point $p$ is an interior point of $X$ if $exists r>0$ such that $B_r(p) subset X$. The interior of $X$ is the set of all interior points.
A point $q$ is a boundary point of $X$ if $forall r>0$ $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$.
(With this, open sets have none of their boundary points, closed sets have all of their boundary points, giving at least some dichotomy to open-closed. Can we call sets that have some of their boundary points "ajar"? :) )
Your sequence has no interior points because any small neighborhood around $frac{1}{n}$ will contain points that are not in the sequence. Therefore, they are all boundary points. However, you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point. All other points in $mathbb{R}$ are interior points of the complement, so not boundary points of the sequence.
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
$endgroup$
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
add a comment |
$begingroup$
Personally, I like the following definitions:
A point $p$ is an interior point of $X$ if $exists r>0$ such that $B_r(p) subset X$. The interior of $X$ is the set of all interior points.
A point $q$ is a boundary point of $X$ if $forall r>0$ $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$.
(With this, open sets have none of their boundary points, closed sets have all of their boundary points, giving at least some dichotomy to open-closed. Can we call sets that have some of their boundary points "ajar"? :) )
Your sequence has no interior points because any small neighborhood around $frac{1}{n}$ will contain points that are not in the sequence. Therefore, they are all boundary points. However, you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point. All other points in $mathbb{R}$ are interior points of the complement, so not boundary points of the sequence.
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
$endgroup$
Personally, I like the following definitions:
A point $p$ is an interior point of $X$ if $exists r>0$ such that $B_r(p) subset X$. The interior of $X$ is the set of all interior points.
A point $q$ is a boundary point of $X$ if $forall r>0$ $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$.
(With this, open sets have none of their boundary points, closed sets have all of their boundary points, giving at least some dichotomy to open-closed. Can we call sets that have some of their boundary points "ajar"? :) )
Your sequence has no interior points because any small neighborhood around $frac{1}{n}$ will contain points that are not in the sequence. Therefore, they are all boundary points. However, you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point. All other points in $mathbb{R}$ are interior points of the complement, so not boundary points of the sequence.
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
answered Dec 15 '18 at 15:19
MatthiasMatthias
2287
2287
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
add a comment |
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
$begingroup$
Thanks for your speedy response. Referring to your line "... you can also show that every neighborhood of $0$ has infinitely many points $frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point." I would like to point out that definition of border point involves "$forall r >0$" which has the same meaning as "all possible value of $r$". Sufficiently large value of $r$ will satisfy $B_r(q) cap X neq emptyset$ and $B_r(q) cap X^C neq emptyset$, but I doubt that a sufficiently small value will also do. Thank you again for your quick response.
$endgroup$
– Amanda.M
Dec 16 '18 at 12:46
add a comment |
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
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${emptyset}$ is different than $emptyset$. If you mean that none of the elements of $S$ is an interior point, then you mean to say that $int(S)=emptyset$. Saying that $int(S)={emptyset}$ would imply that there is an element called $emptyset$ and this element is an interior point of $S$, just like how if you say $int(S)={1}$ that would mean that $1$ is an interior element of $S$, however $emptyset$ is not an element of $S$.
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– JMoravitz
Dec 15 '18 at 15:05
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Oops! My bad! I will make correction. Thank you for your time.
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– Amanda.M
Dec 15 '18 at 15:08