Usbrth

Are there continuous functions who are the same in an interval but differ in at least one other point?

For what reasons would an animal species NOT cross a *horizontal* land bridge?

how can a perfect fourth interval be considered either consonant or dissonant?

Homework question about an engine pulling a train

"is" operation returns false even though two objects have same id

Why doesn't a hydraulic lever violate conservation of energy?

Circular reasoning in L'Hopital's rule

Working through the single responsibility principle (SRP) in Python when calls are expensive

Is every episode of "Where are my Pants?" identical?

Presidential Pardon

Is there a writing software that you can sort scenes like slides in PowerPoint?

Student Loan from years ago pops up and is taking my salary

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

Can the DM override racial traits?

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

Did the new image of black hole confirm the general theory of relativity?

60's-70's movie: home appliances revolting against the owners

US Healthcare consultation for visitors

Can withdrawing asylum be illegal?

"... to apply for a visa" or "... and applied for a visa"?

Is an up-to-date browser secure on an out-of-date OS?

How to support a colleague who finds meetings extremely tiring?

Can each chord in a progression create its own key?

Huge performance difference of the command find with and without using %M option to show permissions

If $E$ is a subset of the metric space $X$. Find a converging sequence in $E$ that converges to a point $g$ that is not the limit point of $E$.

0

2

$begingroup$

I am interested in this problem but there is a lapse in my understanding.

I am tasked with finding a converging sequence in the set $E$ that has a convergence to a point that is not a limit point of the set $E$. My issue comes with not understanding what it means for a sequence to be in the set $E$.

I have come to the conclusion that sequence does not have an infinite range but I can't come up with any example that would make it true. If it is in $E$, how do I notate it is in $E$ and converging to a point that is not the limit point of E.

I'm guessing something like $0$ would converge to $0$ but what if $0$ was a limit point. Then that breaks.

real-analysis

share|cite|improve this question

edited Apr 8 at 7:04

ZeroXLR

1,528519

asked Apr 8 at 6:52

John FoeJohn Foe

1

New contributor

John Foe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

2

$begingroup$

I am interested in this problem but there is a lapse in my understanding.

I am tasked with finding a converging sequence in the set $E$ that has a convergence to a point that is not a limit point of the set $E$. My issue comes with not understanding what it means for a sequence to be in the set $E$.

I have come to the conclusion that sequence does not have an infinite range but I can't come up with any example that would make it true. If it is in $E$, how do I notate it is in $E$ and converging to a point that is not the limit point of E.

I'm guessing something like $0$ would converge to $0$ but what if $0$ was a limit point. Then that breaks.

real-analysis

share|cite|improve this question

edited Apr 8 at 7:04

ZeroXLR

1,528519

asked Apr 8 at 6:52

John FoeJohn Foe

1

New contributor

John Foe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I am interested in this problem but there is a lapse in my understanding.

I am tasked with finding a converging sequence in the set $E$ that has a convergence to a point that is not a limit point of the set $E$. My issue comes with not understanding what it means for a sequence to be in the set $E$.

I have come to the conclusion that sequence does not have an infinite range but I can't come up with any example that would make it true. If it is in $E$, how do I notate it is in $E$ and converging to a point that is not the limit point of E.

I'm guessing something like $0$ would converge to $0$ but what if $0$ was a limit point. Then that breaks.

real-analysis

share|cite|improve this question

edited Apr 8 at 7:04

ZeroXLR

1,528519

asked Apr 8 at 6:52

John FoeJohn Foe

1

New contributor

John Foe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 8 at 7:04

ZeroXLR

1,528519

asked Apr 8 at 6:52

John FoeJohn Foe

1

New contributor

John Foe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 8 at 7:04

ZeroXLR

1,528519

asked Apr 8 at 6:52

John FoeJohn Foe

1

New contributor

John Foe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 8 at 7:04

ZeroXLR

1,528519

edited Apr 8 at 7:04

ZeroXLR

1,528519

asked Apr 8 at 6:52

John FoeJohn Foe

1

New contributor

John Foe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 8 at 6:52

John FoeJohn Foe

1

1 Answer
1

active

oldest

votes

0

$begingroup$

Your guess of using a singleton like $0$ has the right general idea because singletons have no limit points in any topological space. So for instance, you can take $E = 0$ in $mathbbR$ and consider the sequence $x_n = 0$. Obviously this converges to $0$ but $0$ is not a limit point of $E$. This is because of the way limit points are defined:

$p$ in a space $X$ is a limit point of $E subseteq X$ if all open sets $U$ containing $p$ also contains a point of $E$ different from $p$.

But in our example, all open sets around $0$ fail to contain an element of $E$ different from $0$ because it is the only element of $E = 0$!

share|cite|improve this answer

edited Apr 8 at 7:49

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519

$endgroup$

add a comment | 

Your Answer

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
);



);

John Foe is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179272%2fif-e-is-a-subset-of-the-metric-space-x-find-a-converging-sequence-in-e-th%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Your guess of using a singleton like $0$ has the right general idea because singletons have no limit points in any topological space. So for instance, you can take $E = 0$ in $mathbbR$ and consider the sequence $x_n = 0$. Obviously this converges to $0$ but $0$ is not a limit point of $E$. This is because of the way limit points are defined:

$p$ in a space $X$ is a limit point of $E subseteq X$ if all open sets $U$ containing $p$ also contains a point of $E$ different from $p$.

But in our example, all open sets around $0$ fail to contain an element of $E$ different from $0$ because it is the only element of $E = 0$!

share|cite|improve this answer

edited Apr 8 at 7:49

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519

$endgroup$

add a comment | 

0

$begingroup$

Your guess of using a singleton like $0$ has the right general idea because singletons have no limit points in any topological space. So for instance, you can take $E = 0$ in $mathbbR$ and consider the sequence $x_n = 0$. Obviously this converges to $0$ but $0$ is not a limit point of $E$. This is because of the way limit points are defined:

$p$ in a space $X$ is a limit point of $E subseteq X$ if all open sets $U$ containing $p$ also contains a point of $E$ different from $p$.

But in our example, all open sets around $0$ fail to contain an element of $E$ different from $0$ because it is the only element of $E = 0$!

share|cite|improve this answer

edited Apr 8 at 7:49

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519

$endgroup$

add a comment | 

$begingroup$

Your guess of using a singleton like $0$ has the right general idea because singletons have no limit points in any topological space. So for instance, you can take $E = 0$ in $mathbbR$ and consider the sequence $x_n = 0$. Obviously this converges to $0$ but $0$ is not a limit point of $E$. This is because of the way limit points are defined:

$p$ in a space $X$ is a limit point of $E subseteq X$ if all open sets $U$ containing $p$ also contains a point of $E$ different from $p$.

But in our example, all open sets around $0$ fail to contain an element of $E$ different from $0$ because it is the only element of $E = 0$!

share|cite|improve this answer

edited Apr 8 at 7:49

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519

$endgroup$

share|cite|improve this answer

edited Apr 8 at 7:49

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519

answered Apr 8 at 7:09

ZeroXLRZeroXLR

1,528519