Is $0$ a cluster point of the set ${0}$?












3












$begingroup$


I just read the theorem




A subset of R is closed if and only if it contains all of its cluster points.




And I found that the set {0} is closed (Is that correct?), then the set {0} should contains all its cluster points, right? But I just checked the definition of cluster point in the textbook (it appears at the beginning of the limit of function) and found that 0 is not the cluster point of the set {0}



I want to know where the false is ?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you think that the set is not closed? If yes then can you give a cluster point which is not in the set? The definition says that A set is closed iff it contains all its cluster point. If 0 is not the cluster point then why do you even worry about it?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 6:51










  • $begingroup$
    I got it ! I just thought 0 "should" be a cluster point of the set .Then the set does not have a cluster point, right? Thank you. But wait a minute, the set {0} contains the element 0 !
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:06








  • 3




    $begingroup$
    Post the definition of "cluster point" in your book.
    $endgroup$
    – xbh
    Dec 15 '18 at 7:12










  • $begingroup$
    I think it appears at the answer provided by Anthony Ter .
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:19










  • $begingroup$
    @YangGao The set {0} contains the element 0! Yes, It contains. So what?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 9:21
















3












$begingroup$


I just read the theorem




A subset of R is closed if and only if it contains all of its cluster points.




And I found that the set {0} is closed (Is that correct?), then the set {0} should contains all its cluster points, right? But I just checked the definition of cluster point in the textbook (it appears at the beginning of the limit of function) and found that 0 is not the cluster point of the set {0}



I want to know where the false is ?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you think that the set is not closed? If yes then can you give a cluster point which is not in the set? The definition says that A set is closed iff it contains all its cluster point. If 0 is not the cluster point then why do you even worry about it?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 6:51










  • $begingroup$
    I got it ! I just thought 0 "should" be a cluster point of the set .Then the set does not have a cluster point, right? Thank you. But wait a minute, the set {0} contains the element 0 !
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:06








  • 3




    $begingroup$
    Post the definition of "cluster point" in your book.
    $endgroup$
    – xbh
    Dec 15 '18 at 7:12










  • $begingroup$
    I think it appears at the answer provided by Anthony Ter .
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:19










  • $begingroup$
    @YangGao The set {0} contains the element 0! Yes, It contains. So what?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 9:21














3












3








3


2



$begingroup$


I just read the theorem




A subset of R is closed if and only if it contains all of its cluster points.




And I found that the set {0} is closed (Is that correct?), then the set {0} should contains all its cluster points, right? But I just checked the definition of cluster point in the textbook (it appears at the beginning of the limit of function) and found that 0 is not the cluster point of the set {0}



I want to know where the false is ?










share|cite|improve this question











$endgroup$




I just read the theorem




A subset of R is closed if and only if it contains all of its cluster points.




And I found that the set {0} is closed (Is that correct?), then the set {0} should contains all its cluster points, right? But I just checked the definition of cluster point in the textbook (it appears at the beginning of the limit of function) and found that 0 is not the cluster point of the set {0}



I want to know where the false is ?







real-analysis real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 7:13









anomaly

17.8k42666




17.8k42666










asked Dec 15 '18 at 6:48









Yang GaoYang Gao

186




186












  • $begingroup$
    Do you think that the set is not closed? If yes then can you give a cluster point which is not in the set? The definition says that A set is closed iff it contains all its cluster point. If 0 is not the cluster point then why do you even worry about it?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 6:51










  • $begingroup$
    I got it ! I just thought 0 "should" be a cluster point of the set .Then the set does not have a cluster point, right? Thank you. But wait a minute, the set {0} contains the element 0 !
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:06








  • 3




    $begingroup$
    Post the definition of "cluster point" in your book.
    $endgroup$
    – xbh
    Dec 15 '18 at 7:12










  • $begingroup$
    I think it appears at the answer provided by Anthony Ter .
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:19










  • $begingroup$
    @YangGao The set {0} contains the element 0! Yes, It contains. So what?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 9:21


















  • $begingroup$
    Do you think that the set is not closed? If yes then can you give a cluster point which is not in the set? The definition says that A set is closed iff it contains all its cluster point. If 0 is not the cluster point then why do you even worry about it?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 6:51










  • $begingroup$
    I got it ! I just thought 0 "should" be a cluster point of the set .Then the set does not have a cluster point, right? Thank you. But wait a minute, the set {0} contains the element 0 !
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:06








  • 3




    $begingroup$
    Post the definition of "cluster point" in your book.
    $endgroup$
    – xbh
    Dec 15 '18 at 7:12










  • $begingroup$
    I think it appears at the answer provided by Anthony Ter .
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:19










  • $begingroup$
    @YangGao The set {0} contains the element 0! Yes, It contains. So what?
    $endgroup$
    – Rakesh Bhatt
    Dec 15 '18 at 9:21
















$begingroup$
Do you think that the set is not closed? If yes then can you give a cluster point which is not in the set? The definition says that A set is closed iff it contains all its cluster point. If 0 is not the cluster point then why do you even worry about it?
$endgroup$
– Rakesh Bhatt
Dec 15 '18 at 6:51




$begingroup$
Do you think that the set is not closed? If yes then can you give a cluster point which is not in the set? The definition says that A set is closed iff it contains all its cluster point. If 0 is not the cluster point then why do you even worry about it?
$endgroup$
– Rakesh Bhatt
Dec 15 '18 at 6:51












$begingroup$
I got it ! I just thought 0 "should" be a cluster point of the set .Then the set does not have a cluster point, right? Thank you. But wait a minute, the set {0} contains the element 0 !
$endgroup$
– Yang Gao
Dec 15 '18 at 7:06






$begingroup$
I got it ! I just thought 0 "should" be a cluster point of the set .Then the set does not have a cluster point, right? Thank you. But wait a minute, the set {0} contains the element 0 !
$endgroup$
– Yang Gao
Dec 15 '18 at 7:06






3




3




$begingroup$
Post the definition of "cluster point" in your book.
$endgroup$
– xbh
Dec 15 '18 at 7:12




$begingroup$
Post the definition of "cluster point" in your book.
$endgroup$
– xbh
Dec 15 '18 at 7:12












$begingroup$
I think it appears at the answer provided by Anthony Ter .
$endgroup$
– Yang Gao
Dec 15 '18 at 7:19




$begingroup$
I think it appears at the answer provided by Anthony Ter .
$endgroup$
– Yang Gao
Dec 15 '18 at 7:19












$begingroup$
@YangGao The set {0} contains the element 0! Yes, It contains. So what?
$endgroup$
– Rakesh Bhatt
Dec 15 '18 at 9:21




$begingroup$
@YangGao The set {0} contains the element 0! Yes, It contains. So what?
$endgroup$
– Rakesh Bhatt
Dec 15 '18 at 9:21










1 Answer
1






active

oldest

votes


















2












$begingroup$

If $0$ was a cluster point of ${0}$ then for every $delta > 0$ there would exist an $x neq 0, x in N_delta(0)$ such that $x in {0}$. clearly no such $x$ exists, for any $delta$. Thus $0$ is not a cluster point of ${0}$.



So is ${0 }$ closed? Does it contain all of its limit points? Well, $emptyset subseteq {0}$, so ${0}$ is closed.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:17










  • $begingroup$
    Now the empty set is the only cluster point of the set, right? Just a little "weird".
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:25






  • 1




    $begingroup$
    No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:27








  • 1




    $begingroup$
    So the set does not have a cluster point ? wow
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:31










  • $begingroup$
    Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:35












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

If $0$ was a cluster point of ${0}$ then for every $delta > 0$ there would exist an $x neq 0, x in N_delta(0)$ such that $x in {0}$. clearly no such $x$ exists, for any $delta$. Thus $0$ is not a cluster point of ${0}$.



So is ${0 }$ closed? Does it contain all of its limit points? Well, $emptyset subseteq {0}$, so ${0}$ is closed.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:17










  • $begingroup$
    Now the empty set is the only cluster point of the set, right? Just a little "weird".
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:25






  • 1




    $begingroup$
    No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:27








  • 1




    $begingroup$
    So the set does not have a cluster point ? wow
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:31










  • $begingroup$
    Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:35
















2












$begingroup$

If $0$ was a cluster point of ${0}$ then for every $delta > 0$ there would exist an $x neq 0, x in N_delta(0)$ such that $x in {0}$. clearly no such $x$ exists, for any $delta$. Thus $0$ is not a cluster point of ${0}$.



So is ${0 }$ closed? Does it contain all of its limit points? Well, $emptyset subseteq {0}$, so ${0}$ is closed.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:17










  • $begingroup$
    Now the empty set is the only cluster point of the set, right? Just a little "weird".
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:25






  • 1




    $begingroup$
    No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:27








  • 1




    $begingroup$
    So the set does not have a cluster point ? wow
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:31










  • $begingroup$
    Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:35














2












2








2





$begingroup$

If $0$ was a cluster point of ${0}$ then for every $delta > 0$ there would exist an $x neq 0, x in N_delta(0)$ such that $x in {0}$. clearly no such $x$ exists, for any $delta$. Thus $0$ is not a cluster point of ${0}$.



So is ${0 }$ closed? Does it contain all of its limit points? Well, $emptyset subseteq {0}$, so ${0}$ is closed.






share|cite|improve this answer











$endgroup$



If $0$ was a cluster point of ${0}$ then for every $delta > 0$ there would exist an $x neq 0, x in N_delta(0)$ such that $x in {0}$. clearly no such $x$ exists, for any $delta$. Thus $0$ is not a cluster point of ${0}$.



So is ${0 }$ closed? Does it contain all of its limit points? Well, $emptyset subseteq {0}$, so ${0}$ is closed.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 15 '18 at 7:28

























answered Dec 15 '18 at 7:04









Anthony TerAnthony Ter

39616




39616












  • $begingroup$
    Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:17










  • $begingroup$
    Now the empty set is the only cluster point of the set, right? Just a little "weird".
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:25






  • 1




    $begingroup$
    No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:27








  • 1




    $begingroup$
    So the set does not have a cluster point ? wow
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:31










  • $begingroup$
    Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:35


















  • $begingroup$
    Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:17










  • $begingroup$
    Now the empty set is the only cluster point of the set, right? Just a little "weird".
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:25






  • 1




    $begingroup$
    No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:27








  • 1




    $begingroup$
    So the set does not have a cluster point ? wow
    $endgroup$
    – Yang Gao
    Dec 15 '18 at 7:31










  • $begingroup$
    Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
    $endgroup$
    – Anthony Ter
    Dec 15 '18 at 7:35
















$begingroup$
Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
$endgroup$
– Yang Gao
Dec 15 '18 at 7:17




$begingroup$
Thanks. Now I really get it . The closed set contains all of its cluster point but not all of its elements! Thank you!
$endgroup$
– Yang Gao
Dec 15 '18 at 7:17












$begingroup$
Now the empty set is the only cluster point of the set, right? Just a little "weird".
$endgroup$
– Yang Gao
Dec 15 '18 at 7:25




$begingroup$
Now the empty set is the only cluster point of the set, right? Just a little "weird".
$endgroup$
– Yang Gao
Dec 15 '18 at 7:25




1




1




$begingroup$
No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
$endgroup$
– Anthony Ter
Dec 15 '18 at 7:27






$begingroup$
No,the empty set isn't a point. The empty set represents the set of limit points, and I was trying to show that that set is contained in ${0}$. I should have used $subseteq$, not $in$. I'll edit it now.
$endgroup$
– Anthony Ter
Dec 15 '18 at 7:27






1




1




$begingroup$
So the set does not have a cluster point ? wow
$endgroup$
– Yang Gao
Dec 15 '18 at 7:31




$begingroup$
So the set does not have a cluster point ? wow
$endgroup$
– Yang Gao
Dec 15 '18 at 7:31












$begingroup$
Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
$endgroup$
– Anthony Ter
Dec 15 '18 at 7:35




$begingroup$
Yep, in fact, no finite subset of $mathbb{R}$ has a cluster point.
$endgroup$
– Anthony Ter
Dec 15 '18 at 7:35


















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