Some basic questions about two definition of limit point in Zorich‘s “Mathematical Analysis I”
$begingroup$
I'm new at the mathematic analysis and currently reading Zorich's book. When I read the definition about the limit point, I noticed that it's a bit different than Rudin's "Mathematical Principle"
Rudin's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an at least one point different from $p$ itself.
While the Zorich's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an infinite subset of $X$.
The book of Zorich then mentions Rudin's definition and says these two are equivalent and asks for verification. Here is what I tried.
From Zorich's definition, $forall N(p),exists Ssubset X(S is infinite)$, the Rudin's definition is quite obvious, just find another point in $S$
Here is what really have confused me.
From Rudin's definition, I begin with finding a subset $S$ of $N(p)$ that contains $p$, and then try to prove such $S$ can not be finite by contradiction.
Here is what I did, Let $S$ be a subset of $X$ such that $qin Ssubset X$, then assume $S$ is finite. Since $S$ is finite, there exists such $q^primein S$ such that $d(q^prime,p)=d(q,p)_{min}$, Then for the neiborhood of $p$ with radius $r<d(q^prime,p)$, there is no such $qin N(p)$ exists then contradict.
But then I realize that my proof is some sort of prove for a $S$ which is the set containing all $q$, not for some specific $q$. Since there might be infinite $q$, my proof is wrong.(from my understading)
Is it right or wrong, and why. I am really new at this. And it's my first question in this site, if I did something wrong pls tell me. THX.
real-analysis proof-verification
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
$endgroup$
add a comment |
$begingroup$
I'm new at the mathematic analysis and currently reading Zorich's book. When I read the definition about the limit point, I noticed that it's a bit different than Rudin's "Mathematical Principle"
Rudin's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an at least one point different from $p$ itself.
While the Zorich's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an infinite subset of $X$.
The book of Zorich then mentions Rudin's definition and says these two are equivalent and asks for verification. Here is what I tried.
From Zorich's definition, $forall N(p),exists Ssubset X(S is infinite)$, the Rudin's definition is quite obvious, just find another point in $S$
Here is what really have confused me.
From Rudin's definition, I begin with finding a subset $S$ of $N(p)$ that contains $p$, and then try to prove such $S$ can not be finite by contradiction.
Here is what I did, Let $S$ be a subset of $X$ such that $qin Ssubset X$, then assume $S$ is finite. Since $S$ is finite, there exists such $q^primein S$ such that $d(q^prime,p)=d(q,p)_{min}$, Then for the neiborhood of $p$ with radius $r<d(q^prime,p)$, there is no such $qin N(p)$ exists then contradict.
But then I realize that my proof is some sort of prove for a $S$ which is the set containing all $q$, not for some specific $q$. Since there might be infinite $q$, my proof is wrong.(from my understading)
Is it right or wrong, and why. I am really new at this. And it's my first question in this site, if I did something wrong pls tell me. THX.
real-analysis proof-verification
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
$endgroup$
$begingroup$
Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is.
$endgroup$
– xbh
Jan 23 at 12:34
$begingroup$
Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command prime by typing something like p^prime, as in $p^prime$.
$endgroup$
– xbh
Jan 23 at 12:37
$begingroup$
THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$.
$endgroup$
– 卢弘毅
Jan 23 at 12:40
$begingroup$
These are equivalent over any Hausdorff Space. Rudin's definition is more general though.
$endgroup$
– Brevan Ellefsen
Jan 23 at 21:40
add a comment |
$begingroup$
I'm new at the mathematic analysis and currently reading Zorich's book. When I read the definition about the limit point, I noticed that it's a bit different than Rudin's "Mathematical Principle"
Rudin's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an at least one point different from $p$ itself.
While the Zorich's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an infinite subset of $X$.
The book of Zorich then mentions Rudin's definition and says these two are equivalent and asks for verification. Here is what I tried.
From Zorich's definition, $forall N(p),exists Ssubset X(S is infinite)$, the Rudin's definition is quite obvious, just find another point in $S$
Here is what really have confused me.
From Rudin's definition, I begin with finding a subset $S$ of $N(p)$ that contains $p$, and then try to prove such $S$ can not be finite by contradiction.
Here is what I did, Let $S$ be a subset of $X$ such that $qin Ssubset X$, then assume $S$ is finite. Since $S$ is finite, there exists such $q^primein S$ such that $d(q^prime,p)=d(q,p)_{min}$, Then for the neiborhood of $p$ with radius $r<d(q^prime,p)$, there is no such $qin N(p)$ exists then contradict.
But then I realize that my proof is some sort of prove for a $S$ which is the set containing all $q$, not for some specific $q$. Since there might be infinite $q$, my proof is wrong.(from my understading)
Is it right or wrong, and why. I am really new at this. And it's my first question in this site, if I did something wrong pls tell me. THX.
real-analysis proof-verification
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
$endgroup$
I'm new at the mathematic analysis and currently reading Zorich's book. When I read the definition about the limit point, I noticed that it's a bit different than Rudin's "Mathematical Principle"
Rudin's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an at least one point different from $p$ itself.
While the Zorich's definition is
A point $p$ $in$ $R$ is a limit point of $Xsubset R$ if every neiborhood of the point $p$ contains an infinite subset of $X$.
The book of Zorich then mentions Rudin's definition and says these two are equivalent and asks for verification. Here is what I tried.
From Zorich's definition, $forall N(p),exists Ssubset X(S is infinite)$, the Rudin's definition is quite obvious, just find another point in $S$
Here is what really have confused me.
From Rudin's definition, I begin with finding a subset $S$ of $N(p)$ that contains $p$, and then try to prove such $S$ can not be finite by contradiction.
Here is what I did, Let $S$ be a subset of $X$ such that $qin Ssubset X$, then assume $S$ is finite. Since $S$ is finite, there exists such $q^primein S$ such that $d(q^prime,p)=d(q,p)_{min}$, Then for the neiborhood of $p$ with radius $r<d(q^prime,p)$, there is no such $qin N(p)$ exists then contradict.
But then I realize that my proof is some sort of prove for a $S$ which is the set containing all $q$, not for some specific $q$. Since there might be infinite $q$, my proof is wrong.(from my understading)
Is it right or wrong, and why. I am really new at this. And it's my first question in this site, if I did something wrong pls tell me. THX.
real-analysis proof-verification
real-analysis proof-verification
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
edited Jan 23 at 12:38
卢弘毅
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
asked Jan 23 at 12:24
卢弘毅卢弘毅
11
11
$begingroup$
Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is.
$endgroup$
– xbh
Jan 23 at 12:34
$begingroup$
Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command prime by typing something like p^prime, as in $p^prime$.
$endgroup$
– xbh
Jan 23 at 12:37
$begingroup$
THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$.
$endgroup$
– 卢弘毅
Jan 23 at 12:40
$begingroup$
These are equivalent over any Hausdorff Space. Rudin's definition is more general though.
$endgroup$
– Brevan Ellefsen
Jan 23 at 21:40
add a comment |
$begingroup$
Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is.
$endgroup$
– xbh
Jan 23 at 12:34
$begingroup$
Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command prime by typing something like p^prime, as in $p^prime$.
$endgroup$
– xbh
Jan 23 at 12:37
$begingroup$
THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$.
$endgroup$
– 卢弘毅
Jan 23 at 12:40
$begingroup$
These are equivalent over any Hausdorff Space. Rudin's definition is more general though.
$endgroup$
– Brevan Ellefsen
Jan 23 at 21:40
$begingroup$
Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is.
$endgroup$
– xbh
Jan 23 at 12:34
$begingroup$
Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is.
$endgroup$
– xbh
Jan 23 at 12:34
$begingroup$
Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command prime by typing something like p^prime, as in $p^prime$.
$endgroup$
– xbh
Jan 23 at 12:37
$begingroup$
Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command prime by typing something like p^prime, as in $p^prime$.
$endgroup$
– xbh
Jan 23 at 12:37
$begingroup$
THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$.
$endgroup$
– 卢弘毅
Jan 23 at 12:40
$begingroup$
THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$.
$endgroup$
– 卢弘毅
Jan 23 at 12:40
$begingroup$
These are equivalent over any Hausdorff Space. Rudin's definition is more general though.
$endgroup$
– Brevan Ellefsen
Jan 23 at 21:40
$begingroup$
These are equivalent over any Hausdorff Space. Rudin's definition is more general though.
$endgroup$
– Brevan Ellefsen
Jan 23 at 21:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I don't understand your proof, and therefore I cannot comment on it. However, if $pinmathbb R$ is such that some neighborhood $U$ of $p$ only has finitely many elements of $X$, then $p$ is not a limit point of $X$. In fact, let $F=Ucap X$. I am assuming that it is finite. Take $varepsilon>0$ such that $(x-varepsilon,x+varepsilon)cap F=emptyset$. Then $Ucap(x-varepsilon,x+varepsilon)$ is a neighborhood of $x$ which has no element of $X$.
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
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%2f3084407%2fsome-basic-questions-about-two-definition-of-limit-point-in-zorich-s-mathematic%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't understand your proof, and therefore I cannot comment on it. However, if $pinmathbb R$ is such that some neighborhood $U$ of $p$ only has finitely many elements of $X$, then $p$ is not a limit point of $X$. In fact, let $F=Ucap X$. I am assuming that it is finite. Take $varepsilon>0$ such that $(x-varepsilon,x+varepsilon)cap F=emptyset$. Then $Ucap(x-varepsilon,x+varepsilon)$ is a neighborhood of $x$ which has no element of $X$.
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
add a comment |
$begingroup$
I don't understand your proof, and therefore I cannot comment on it. However, if $pinmathbb R$ is such that some neighborhood $U$ of $p$ only has finitely many elements of $X$, then $p$ is not a limit point of $X$. In fact, let $F=Ucap X$. I am assuming that it is finite. Take $varepsilon>0$ such that $(x-varepsilon,x+varepsilon)cap F=emptyset$. Then $Ucap(x-varepsilon,x+varepsilon)$ is a neighborhood of $x$ which has no element of $X$.
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
add a comment |
$begingroup$
I don't understand your proof, and therefore I cannot comment on it. However, if $pinmathbb R$ is such that some neighborhood $U$ of $p$ only has finitely many elements of $X$, then $p$ is not a limit point of $X$. In fact, let $F=Ucap X$. I am assuming that it is finite. Take $varepsilon>0$ such that $(x-varepsilon,x+varepsilon)cap F=emptyset$. Then $Ucap(x-varepsilon,x+varepsilon)$ is a neighborhood of $x$ which has no element of $X$.
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
I don't understand your proof, and therefore I cannot comment on it. However, if $pinmathbb R$ is such that some neighborhood $U$ of $p$ only has finitely many elements of $X$, then $p$ is not a limit point of $X$. In fact, let $F=Ucap X$. I am assuming that it is finite. Take $varepsilon>0$ such that $(x-varepsilon,x+varepsilon)cap F=emptyset$. Then $Ucap(x-varepsilon,x+varepsilon)$ is a neighborhood of $x$ which has no element of $X$.
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 23 at 12:31
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
add a comment |
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
Thanks a lot, I just edited my $S$ as a set that contains all possible $q$
$endgroup$
– 卢弘毅
Jan 23 at 12:42
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
$begingroup$
I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $qin Ssubset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$?
$endgroup$
– José Carlos Santos
Jan 23 at 13:10
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%2f3084407%2fsome-basic-questions-about-two-definition-of-limit-point-in-zorich-s-mathematic%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
$begingroup$
Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is.
$endgroup$
– xbh
Jan 23 at 12:34
$begingroup$
Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command prime by typing something like p^prime, as in $p^prime$.
$endgroup$
– xbh
Jan 23 at 12:37
$begingroup$
THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$.
$endgroup$
– 卢弘毅
Jan 23 at 12:40
$begingroup$
These are equivalent over any Hausdorff Space. Rudin's definition is more general though.
$endgroup$
– Brevan Ellefsen
Jan 23 at 21:40