Alternative solution to a problem involving an enumeration of rationals
$begingroup$
I'm working on the same problem as in this post. I understand the solutions provided in the answers. The question basically asks us to find an enumeration of the rationals ${r_n}_{n≥1}$ such that the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$ in $mathbb{R}$ is non-empty.
I found the solutions rather complicated. What's wrong with my approach? For all $n in mathbb{N}$, let $r_n = 2^{n^2}$. This way, I know $[-infty,1)$ is in the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$. Doesn't that complete the problem? What did I miss?
real-analysis general-topology rational-numbers
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
asked Jan 21 at 22:05
user1691278user1691278
49439
$endgroup$
add a comment |
$begingroup$
I'm working on the same problem as in this post. I understand the solutions provided in the answers. The question basically asks us to find an enumeration of the rationals ${r_n}_{n≥1}$ such that the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$ in $mathbb{R}$ is non-empty.
I found the solutions rather complicated. What's wrong with my approach? For all $n in mathbb{N}$, let $r_n = 2^{n^2}$. This way, I know $[-infty,1)$ is in the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$. Doesn't that complete the problem? What did I miss?
real-analysis general-topology rational-numbers
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
asked Jan 21 at 22:05
user1691278user1691278
49439
$endgroup$
3
$begingroup$
Well, you did not enumerate all rationals!
$endgroup$
– Mindlack
Jan 21 at 22:07
$begingroup$
@Mindlack Oops. I didn't know that we needed to have all the rationals.
$endgroup$
– user1691278
Jan 21 at 22:07
1
$begingroup$
The question is confusing. Any numeration of the rationals usually lists positive numbers, so the negative numbers would not be covered in any union. I suspect the original question would include negative rationals, or else getting a complement in $R_+$..
$endgroup$
– herb steinberg
Jan 21 at 22:17
$begingroup$
@Mindlack Why not an official answer?
$endgroup$
– Paul Frost
Jan 22 at 12:59
add a comment |
$begingroup$
I'm working on the same problem as in this post. I understand the solutions provided in the answers. The question basically asks us to find an enumeration of the rationals ${r_n}_{n≥1}$ such that the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$ in $mathbb{R}$ is non-empty.
I found the solutions rather complicated. What's wrong with my approach? For all $n in mathbb{N}$, let $r_n = 2^{n^2}$. This way, I know $[-infty,1)$ is in the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$. Doesn't that complete the problem? What did I miss?
real-analysis general-topology rational-numbers
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
asked Jan 21 at 22:05
user1691278user1691278
49439
$endgroup$
I'm working on the same problem as in this post. I understand the solutions provided in the answers. The question basically asks us to find an enumeration of the rationals ${r_n}_{n≥1}$ such that the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$ in $mathbb{R}$ is non-empty.
I found the solutions rather complicated. What's wrong with my approach? For all $n in mathbb{N}$, let $r_n = 2^{n^2}$. This way, I know $[-infty,1)$ is in the complement of $bigcup_{n=1}^{infty}{left(r_{n}-frac{1}{n},r_{n}+frac{1}{n}right)}$. Doesn't that complete the problem? What did I miss?
real-analysis general-topology rational-numbers
real-analysis general-topology rational-numbers
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
asked Jan 21 at 22:05
user1691278user1691278
49439
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
asked Jan 21 at 22:05
user1691278user1691278
49439
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
edited Jan 21 at 23:26
Andrés E. Caicedo
65.5k8159250
65.5k8159250
asked Jan 21 at 22:05
user1691278user1691278
49439
asked Jan 21 at 22:05
user1691278user1691278
49439
asked Jan 21 at 22:05
user1691278user1691278
49439
49439
3
$begingroup$
Well, you did not enumerate all rationals!
$endgroup$
– Mindlack
Jan 21 at 22:07
$begingroup$
@Mindlack Oops. I didn't know that we needed to have all the rationals.
$endgroup$
– user1691278
Jan 21 at 22:07
1
$begingroup$
The question is confusing. Any numeration of the rationals usually lists positive numbers, so the negative numbers would not be covered in any union. I suspect the original question would include negative rationals, or else getting a complement in $R_+$..
$endgroup$
– herb steinberg
Jan 21 at 22:17
$begingroup$
@Mindlack Why not an official answer?
$endgroup$
– Paul Frost
Jan 22 at 12:59
add a comment |
3
$begingroup$
Well, you did not enumerate all rationals!
$endgroup$
– Mindlack
Jan 21 at 22:07
$begingroup$
@Mindlack Oops. I didn't know that we needed to have all the rationals.
$endgroup$
– user1691278
Jan 21 at 22:07
1
$begingroup$
The question is confusing. Any numeration of the rationals usually lists positive numbers, so the negative numbers would not be covered in any union. I suspect the original question would include negative rationals, or else getting a complement in $R_+$..
$endgroup$
– herb steinberg
Jan 21 at 22:17
$begingroup$
@Mindlack Why not an official answer?
$endgroup$
– Paul Frost
Jan 22 at 12:59
3
3
$begingroup$
Well, you did not enumerate all rationals!
$endgroup$
– Mindlack
Jan 21 at 22:07
$begingroup$
Well, you did not enumerate all rationals!
$endgroup$
– Mindlack
Jan 21 at 22:07
$begingroup$
@Mindlack Oops. I didn't know that we needed to have all the rationals.
$endgroup$
– user1691278
Jan 21 at 22:07
$begingroup$
@Mindlack Oops. I didn't know that we needed to have all the rationals.
$endgroup$
– user1691278
Jan 21 at 22:07
1
1
$begingroup$
The question is confusing. Any numeration of the rationals usually lists positive numbers, so the negative numbers would not be covered in any union. I suspect the original question would include negative rationals, or else getting a complement in $R_+$..
$endgroup$
– herb steinberg
Jan 21 at 22:17
$begingroup$
The question is confusing. Any numeration of the rationals usually lists positive numbers, so the negative numbers would not be covered in any union. I suspect the original question would include negative rationals, or else getting a complement in $R_+$..
$endgroup$
– herb steinberg
Jan 21 at 22:17
$begingroup$
@Mindlack Why not an official answer?
$endgroup$
– Paul Frost
Jan 22 at 12:59
$begingroup$
@Mindlack Why not an official answer?
$endgroup$
– Paul Frost
Jan 22 at 12:59
add a comment |
0
active
oldest
votes
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%2f3082484%2falternative-solution-to-a-problem-involving-an-enumeration-of-rationals%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3082484%2falternative-solution-to-a-problem-involving-an-enumeration-of-rationals%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
3
$begingroup$
Well, you did not enumerate all rationals!
$endgroup$
– Mindlack
Jan 21 at 22:07
$begingroup$
@Mindlack Oops. I didn't know that we needed to have all the rationals.
$endgroup$
– user1691278
Jan 21 at 22:07
1
$begingroup$
The question is confusing. Any numeration of the rationals usually lists positive numbers, so the negative numbers would not be covered in any union. I suspect the original question would include negative rationals, or else getting a complement in $R_+$..
$endgroup$
– herb steinberg
Jan 21 at 22:17
$begingroup$
@Mindlack Why not an official answer?
$endgroup$
– Paul Frost
Jan 22 at 12:59