Mixing
Limit point alternative definition
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I have a question, why isn't an alternative definition of a $p in X$ being a limit point of a set $E$ : $forall r>0 N_r(p) cap E neq emptyset$, why does it have to be the punctured neighborhood?
Note that $E$ is a subset of $X$ and $X$ is a metric space.
Note that the definition of a limit point of a set E is a point $p in X$ such that every neighborhood contains a point $q$ such that $qneq p$ and $ qin E$.
real-analysis general-topology analysis proof-verification definition
edited Jan 14 at 7:00
Henno Brandsma
109k347115
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
$endgroup$
add a comment |
$begingroup$
I have a question, why isn't an alternative definition of a $p in X$ being a limit point of a set $E$ : $forall r>0 N_r(p) cap E neq emptyset$, why does it have to be the punctured neighborhood?
Note that $E$ is a subset of $X$ and $X$ is a metric space.
Note that the definition of a limit point of a set E is a point $p in X$ such that every neighborhood contains a point $q$ such that $qneq p$ and $ qin E$.
real-analysis general-topology analysis proof-verification definition
edited Jan 14 at 7:00
Henno Brandsma
109k347115
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
$endgroup$
$begingroup$
Then the limit point is same as a point of the closure of the set. $xin Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty
$endgroup$
– Ameryr
Jan 14 at 3:40
$begingroup$
Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure.
$endgroup$
– Matt A Pelto
Jan 14 at 4:16
add a comment |
$begingroup$
I have a question, why isn't an alternative definition of a $p in X$ being a limit point of a set $E$ : $forall r>0 N_r(p) cap E neq emptyset$, why does it have to be the punctured neighborhood?
Note that $E$ is a subset of $X$ and $X$ is a metric space.
Note that the definition of a limit point of a set E is a point $p in X$ such that every neighborhood contains a point $q$ such that $qneq p$ and $ qin E$.
real-analysis general-topology analysis proof-verification definition
edited Jan 14 at 7:00
Henno Brandsma
109k347115
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
$endgroup$
I have a question, why isn't an alternative definition of a $p in X$ being a limit point of a set $E$ : $forall r>0 N_r(p) cap E neq emptyset$, why does it have to be the punctured neighborhood?
Note that $E$ is a subset of $X$ and $X$ is a metric space.
Note that the definition of a limit point of a set E is a point $p in X$ such that every neighborhood contains a point $q$ such that $qneq p$ and $ qin E$.
real-analysis general-topology analysis proof-verification definition
real-analysis general-topology analysis proof-verification definition
edited Jan 14 at 7:00
Henno Brandsma
109k347115
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
edited Jan 14 at 7:00
Henno Brandsma
109k347115
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
edited Jan 14 at 7:00
Henno Brandsma
109k347115
edited Jan 14 at 7:00
Henno Brandsma
109k347115
edited Jan 14 at 7:00
Henno Brandsma
109k347115
109k347115
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
asked Jan 14 at 3:28
mathsssislifemathsssislife
408
408
$begingroup$
Then the limit point is same as a point of the closure of the set. $xin Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty
$endgroup$
– Ameryr
Jan 14 at 3:40
$begingroup$
Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure.
$endgroup$
– Matt A Pelto
Jan 14 at 4:16
add a comment |
$begingroup$
Then the limit point is same as a point of the closure of the set. $xin Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty
$endgroup$
– Ameryr
Jan 14 at 3:40
$begingroup$
Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure.
$endgroup$
– Matt A Pelto
Jan 14 at 4:16
$begingroup$
Then the limit point is same as a point of the closure of the set. $xin Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty
$endgroup$
– Ameryr
Jan 14 at 3:40
$begingroup$
Then the limit point is same as a point of the closure of the set. $xin Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty
$endgroup$
– Ameryr
Jan 14 at 3:40
$begingroup$
Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure.
$endgroup$
– Matt A Pelto
Jan 14 at 4:16
$begingroup$
Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure.
$endgroup$
– Matt A Pelto
Jan 14 at 4:16
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The definition of a limit point uses punctured neighborhoods to intentionally exclude isolated points of $E$ from being limit points. To see this just consider any closed set $E$ with an isolated point $p$. And notice that if we didn't use punctured neighborhoods, then the isolated point $p$ would be considered a limit point of $E$ (hypothetically).
We typically say $p in text {Cl}(E)$ if and only if $p$ is an adherent point of $E$. An adherent point of $E$ is either a limit point of $E$ or an isolated point of $E$.
To summarize:
$p$ is a limit point of $E$ means there is a sequence of points ${p_n}_{n=1}^infty$ with $p_n in E$ and $p_n neq p$ for every $n in mathbb N$ and such that $lim_{n to infty} p_n = p$. Notice how an isolated point of $E$ can't possibly be the limit of such a sequence.
$p$ is an adherent point of $E$ means there is no open neighborhood $U_p$ such that $U_p cap E=emptyset $. More informally, $p$ sticks to $E$ in the sense that we cannot find a open set containing $p$ that is disjoint from $E$.
I hope this helps to clarify things.
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
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1 Answer
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1 Answer
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$begingroup$
The definition of a limit point uses punctured neighborhoods to intentionally exclude isolated points of $E$ from being limit points. To see this just consider any closed set $E$ with an isolated point $p$. And notice that if we didn't use punctured neighborhoods, then the isolated point $p$ would be considered a limit point of $E$ (hypothetically).
We typically say $p in text {Cl}(E)$ if and only if $p$ is an adherent point of $E$. An adherent point of $E$ is either a limit point of $E$ or an isolated point of $E$.
To summarize:
$p$ is a limit point of $E$ means there is a sequence of points ${p_n}_{n=1}^infty$ with $p_n in E$ and $p_n neq p$ for every $n in mathbb N$ and such that $lim_{n to infty} p_n = p$. Notice how an isolated point of $E$ can't possibly be the limit of such a sequence.
$p$ is an adherent point of $E$ means there is no open neighborhood $U_p$ such that $U_p cap E=emptyset $. More informally, $p$ sticks to $E$ in the sense that we cannot find a open set containing $p$ that is disjoint from $E$.
I hope this helps to clarify things.
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
add a comment |
$begingroup$
The definition of a limit point uses punctured neighborhoods to intentionally exclude isolated points of $E$ from being limit points. To see this just consider any closed set $E$ with an isolated point $p$. And notice that if we didn't use punctured neighborhoods, then the isolated point $p$ would be considered a limit point of $E$ (hypothetically).
We typically say $p in text {Cl}(E)$ if and only if $p$ is an adherent point of $E$. An adherent point of $E$ is either a limit point of $E$ or an isolated point of $E$.
To summarize:
$p$ is a limit point of $E$ means there is a sequence of points ${p_n}_{n=1}^infty$ with $p_n in E$ and $p_n neq p$ for every $n in mathbb N$ and such that $lim_{n to infty} p_n = p$. Notice how an isolated point of $E$ can't possibly be the limit of such a sequence.
$p$ is an adherent point of $E$ means there is no open neighborhood $U_p$ such that $U_p cap E=emptyset $. More informally, $p$ sticks to $E$ in the sense that we cannot find a open set containing $p$ that is disjoint from $E$.
I hope this helps to clarify things.
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
add a comment |
$begingroup$
The definition of a limit point uses punctured neighborhoods to intentionally exclude isolated points of $E$ from being limit points. To see this just consider any closed set $E$ with an isolated point $p$. And notice that if we didn't use punctured neighborhoods, then the isolated point $p$ would be considered a limit point of $E$ (hypothetically).
We typically say $p in text {Cl}(E)$ if and only if $p$ is an adherent point of $E$. An adherent point of $E$ is either a limit point of $E$ or an isolated point of $E$.
To summarize:
$p$ is a limit point of $E$ means there is a sequence of points ${p_n}_{n=1}^infty$ with $p_n in E$ and $p_n neq p$ for every $n in mathbb N$ and such that $lim_{n to infty} p_n = p$. Notice how an isolated point of $E$ can't possibly be the limit of such a sequence.
$p$ is an adherent point of $E$ means there is no open neighborhood $U_p$ such that $U_p cap E=emptyset $. More informally, $p$ sticks to $E$ in the sense that we cannot find a open set containing $p$ that is disjoint from $E$.
I hope this helps to clarify things.
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
$endgroup$
The definition of a limit point uses punctured neighborhoods to intentionally exclude isolated points of $E$ from being limit points. To see this just consider any closed set $E$ with an isolated point $p$. And notice that if we didn't use punctured neighborhoods, then the isolated point $p$ would be considered a limit point of $E$ (hypothetically).
We typically say $p in text {Cl}(E)$ if and only if $p$ is an adherent point of $E$. An adherent point of $E$ is either a limit point of $E$ or an isolated point of $E$.
To summarize:
$p$ is a limit point of $E$ means there is a sequence of points ${p_n}_{n=1}^infty$ with $p_n in E$ and $p_n neq p$ for every $n in mathbb N$ and such that $lim_{n to infty} p_n = p$. Notice how an isolated point of $E$ can't possibly be the limit of such a sequence.
$p$ is an adherent point of $E$ means there is no open neighborhood $U_p$ such that $U_p cap E=emptyset $. More informally, $p$ sticks to $E$ in the sense that we cannot find a open set containing $p$ that is disjoint from $E$.
I hope this helps to clarify things.
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
edited Jan 14 at 7:34
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
answered Jan 14 at 4:14
Matt A PeltoMatt A Pelto
2,602621
2,602621
add a comment |
add a comment |
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$begingroup$
Then the limit point is same as a point of the closure of the set. $xin Cl(A)$ if for every neighborhood containing $x$ the intersection between it and the set is non empty
$endgroup$
– Ameryr
Jan 14 at 3:40
$begingroup$
Limit points are in the closure of a set but unless a nonstandard definition of limit point is being used, then the set of all limit points won't necessarily equal the closure.
$endgroup$
– Matt A Pelto
Jan 14 at 4:16