Mixing
Question Regarding Isolation and Limit Points
$begingroup$
I have been self studying real analysis and came across a question regarding isolated vs interior points. I am reading that an isolated point p of a set S is an element of S such that p not a limit point of S; however, a limit point of S is a point such that any neighborhood of p contains a point q not equal to p in S. My question is, wouldn't any point in a set have a neighborhood that contains a different point in that set, and therefore be a limit point?
Thank You
real-analysis general-topology
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
$endgroup$
add a comment |
$begingroup$
I have been self studying real analysis and came across a question regarding isolated vs interior points. I am reading that an isolated point p of a set S is an element of S such that p not a limit point of S; however, a limit point of S is a point such that any neighborhood of p contains a point q not equal to p in S. My question is, wouldn't any point in a set have a neighborhood that contains a different point in that set, and therefore be a limit point?
Thank You
real-analysis general-topology
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
$endgroup$
$begingroup$
No. Consider the integers with the subspace topology from $mathbb{R}$. For $a in mathbb{Z}$, ${a}$ is an open neighbourhood of $a$ that has empty intersection with $mathbb{Z} setminus {a}$.
$endgroup$
– greelious
Jan 10 at 23:06
add a comment |
$begingroup$
I have been self studying real analysis and came across a question regarding isolated vs interior points. I am reading that an isolated point p of a set S is an element of S such that p not a limit point of S; however, a limit point of S is a point such that any neighborhood of p contains a point q not equal to p in S. My question is, wouldn't any point in a set have a neighborhood that contains a different point in that set, and therefore be a limit point?
Thank You
real-analysis general-topology
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
$endgroup$
I have been self studying real analysis and came across a question regarding isolated vs interior points. I am reading that an isolated point p of a set S is an element of S such that p not a limit point of S; however, a limit point of S is a point such that any neighborhood of p contains a point q not equal to p in S. My question is, wouldn't any point in a set have a neighborhood that contains a different point in that set, and therefore be a limit point?
Thank You
real-analysis general-topology
real-analysis general-topology
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
asked Jan 10 at 23:00
Zachary CarterZachary Carter
517
517
$begingroup$
No. Consider the integers with the subspace topology from $mathbb{R}$. For $a in mathbb{Z}$, ${a}$ is an open neighbourhood of $a$ that has empty intersection with $mathbb{Z} setminus {a}$.
$endgroup$
– greelious
Jan 10 at 23:06
add a comment |
$begingroup$
No. Consider the integers with the subspace topology from $mathbb{R}$. For $a in mathbb{Z}$, ${a}$ is an open neighbourhood of $a$ that has empty intersection with $mathbb{Z} setminus {a}$.
$endgroup$
– greelious
Jan 10 at 23:06
$begingroup$
No. Consider the integers with the subspace topology from $mathbb{R}$. For $a in mathbb{Z}$, ${a}$ is an open neighbourhood of $a$ that has empty intersection with $mathbb{Z} setminus {a}$.
$endgroup$
– greelious
Jan 10 at 23:06
$begingroup$
No. Consider the integers with the subspace topology from $mathbb{R}$. For $a in mathbb{Z}$, ${a}$ is an open neighbourhood of $a$ that has empty intersection with $mathbb{Z} setminus {a}$.
$endgroup$
– greelious
Jan 10 at 23:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. Or, more accurately, yes, but the thing you wrote after "my question is" is not the same as the definition: for a point $p$ to be a limit point of a set $S$, every neighbourhood of $p$ must contain some element of $S setminus {p}$.
It is then easy to see that not all points have this property: for example, for any $p in mathbb{Z}$, there is no point of $mathbb{Z}$ in the neighbourhood $(p - 1, p + 1)$ of $p$.
answered Jan 10 at 23:04
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
For limit points the requirement is that every neighborhood of that point has a point of set different from that point.
As the result every neighborhood will have infinitely many points of the set different from that point. Another way of saying it is the limit point is actually the limit of a sequence of distinct points of the set converging to that point.
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069284%2fquestion-regarding-isolation-and-limit-points%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Or, more accurately, yes, but the thing you wrote after "my question is" is not the same as the definition: for a point $p$ to be a limit point of a set $S$, every neighbourhood of $p$ must contain some element of $S setminus {p}$.
It is then easy to see that not all points have this property: for example, for any $p in mathbb{Z}$, there is no point of $mathbb{Z}$ in the neighbourhood $(p - 1, p + 1)$ of $p$.
answered Jan 10 at 23:04
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
No. Or, more accurately, yes, but the thing you wrote after "my question is" is not the same as the definition: for a point $p$ to be a limit point of a set $S$, every neighbourhood of $p$ must contain some element of $S setminus {p}$.
It is then easy to see that not all points have this property: for example, for any $p in mathbb{Z}$, there is no point of $mathbb{Z}$ in the neighbourhood $(p - 1, p + 1)$ of $p$.
answered Jan 10 at 23:04
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
No. Or, more accurately, yes, but the thing you wrote after "my question is" is not the same as the definition: for a point $p$ to be a limit point of a set $S$, every neighbourhood of $p$ must contain some element of $S setminus {p}$.
It is then easy to see that not all points have this property: for example, for any $p in mathbb{Z}$, there is no point of $mathbb{Z}$ in the neighbourhood $(p - 1, p + 1)$ of $p$.
answered Jan 10 at 23:04
user3482749user3482749
4,266919
$endgroup$
No. Or, more accurately, yes, but the thing you wrote after "my question is" is not the same as the definition: for a point $p$ to be a limit point of a set $S$, every neighbourhood of $p$ must contain some element of $S setminus {p}$.
It is then easy to see that not all points have this property: for example, for any $p in mathbb{Z}$, there is no point of $mathbb{Z}$ in the neighbourhood $(p - 1, p + 1)$ of $p$.
answered Jan 10 at 23:04
user3482749user3482749
4,266919
answered Jan 10 at 23:04
user3482749user3482749
4,266919
answered Jan 10 at 23:04
user3482749user3482749
4,266919
answered Jan 10 at 23:04
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
$begingroup$
For limit points the requirement is that every neighborhood of that point has a point of set different from that point.
As the result every neighborhood will have infinitely many points of the set different from that point. Another way of saying it is the limit point is actually the limit of a sequence of distinct points of the set converging to that point.
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
For limit points the requirement is that every neighborhood of that point has a point of set different from that point.
As the result every neighborhood will have infinitely many points of the set different from that point. Another way of saying it is the limit point is actually the limit of a sequence of distinct points of the set converging to that point.
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
For limit points the requirement is that every neighborhood of that point has a point of set different from that point.
As the result every neighborhood will have infinitely many points of the set different from that point. Another way of saying it is the limit point is actually the limit of a sequence of distinct points of the set converging to that point.
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
For limit points the requirement is that every neighborhood of that point has a point of set different from that point.
As the result every neighborhood will have infinitely many points of the set different from that point. Another way of saying it is the limit point is actually the limit of a sequence of distinct points of the set converging to that point.
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 10 at 23:06
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069284%2fquestion-regarding-isolation-and-limit-points%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
No. Consider the integers with the subspace topology from $mathbb{R}$. For $a in mathbb{Z}$, ${a}$ is an open neighbourhood of $a$ that has empty intersection with $mathbb{Z} setminus {a}$.
$endgroup$
– greelious
Jan 10 at 23:06