A sequence of rationals converging to an irrational point (proving that $mathbb Q$ is not locally compact)
$begingroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
$endgroup$
add a comment |
$begingroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
$endgroup$
add a comment |
$begingroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
$endgroup$
Here is my attempt to prove that $mathbb Q$ is not locally compact. (My questions are below the proof.)
Suppose $mathbb Q$ is locally compact. Then it is locally compact at every point. Let $xin mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $mathbb Q$ and a neighborhood $U$ of $x$ in $mathbb Q$ such that $xin Usubseteq Ksubseteq Q$. (The neighborhood $U$ is of the form $(a,b)cap mathbb Q$ for some $a,binmathbb Q$; here $xin (a,b)subset mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $overline K$. But since $K$ is a compact subspace of the Hausdorff space $mathbb Q$, it is closed in $mathbb Q$, so $overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $Ksubseteq Q$.
The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $mathbb Q$.
And otherwise does the proof look okay?
sequences-and-series general-topology compactness
sequences-and-series general-topology compactness
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
asked Dec 9 '18 at 4:00
user531587user531587
25413
25413
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
$endgroup$
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%2f3031986%2fa-sequence-of-rationals-converging-to-an-irrational-point-proving-that-mathbb%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
$begingroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
$endgroup$
add a comment |
$begingroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
$endgroup$
add a comment |
$begingroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
$endgroup$
Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $mathbb{Q}$.
As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $rin (a,b)$. Then for any $nin mathbb{N}$, there exists a rational number $x_n$ such that
$$
r < x_n < r+frac{1}{n}.
$$
This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_nin (a,b)$ for all $ninmathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)capmathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
answered Dec 9 '18 at 4:28
QuokaQuoka
1,247312
1,247312
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%2f3031986%2fa-sequence-of-rationals-converging-to-an-irrational-point-proving-that-mathbb%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
