Length of countably many intervals (2)
I already posted this question along with a related one but only the other one got answered, so I'm reposting this.
The following is a proof from my textbook. I have 2 questions, which are in bold.
Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.
Proof:
Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.
Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .
Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .
But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.
Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.
And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?
real-analysis
asked Nov 29 '18 at 14:15
ThomasThomas
730416
add a comment |
I already posted this question along with a related one but only the other one got answered, so I'm reposting this.
The following is a proof from my textbook. I have 2 questions, which are in bold.
Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.
Proof:
Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.
Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .
Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .
But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.
Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.
And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?
real-analysis
asked Nov 29 '18 at 14:15
ThomasThomas
730416
add a comment |
I already posted this question along with a related one but only the other one got answered, so I'm reposting this.
The following is a proof from my textbook. I have 2 questions, which are in bold.
Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.
Proof:
Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.
Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .
Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .
But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.
Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.
And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?
real-analysis
asked Nov 29 '18 at 14:15
ThomasThomas
730416
I already posted this question along with a related one but only the other one got answered, so I'm reposting this.
The following is a proof from my textbook. I have 2 questions, which are in bold.
Theorem: Let $(I_n)$ and $(J_k)$ be sequences of intervals such that $bigcup_{n=1}^infty I_n=bigcup_{k=1}^infty J_k$. If the $I_n$ are pairwise disjoint, then $sum_{n=1}^inftyell(I_n)lesum_{k=1}^inftyell(J_k)$. Thus, if the $J_k$ are also pairwise disjoint, then the two sums are equal.
Proof:
Suppose, to the contrary, that $sum_{n=1}^inftyell(I_n)>sum_{k=1}^inftyell(J_k)$.
Then, for some $N$, we must have $sum_{n=1}^Nell(I_n)>sum_{k=1}^inftyell(J_k)$ .
Of course, we also have $bigcup_{n=1}^N I_nsubsetbigcup_{k=1}^infty J_k$ .
But now, by expanding each $J_k$ slightly and shrinking each $I_n$ slightly, we may suppose that the $J_k$ are open and the $I_n$ are closed.
Thus, the $J_k$ form an open cover for the compact set $bigcup_{n=1}^N I_n$.
And here is the contradiction: Since we have $sum_{n=1}^Nell(I_n)>sum_{k=1}^Mell(J_k)$, for any $M$, the sets ($J_k$) form an open cover for that admits no finite subcover. Could you please explain this? Also, where do we use the fact that the $I_n$ are pairwise disjoint?
real-analysis
real-analysis
asked Nov 29 '18 at 14:15
ThomasThomas
730416
asked Nov 29 '18 at 14:15
ThomasThomas
730416
asked Nov 29 '18 at 14:15
ThomasThomas
730416
asked Nov 29 '18 at 14:15
ThomasThomas
730416
asked Nov 29 '18 at 14:15
ThomasThomas
730416
730416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.
This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.
You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
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%2f3018675%2flength-of-countably-many-intervals-2%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
Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.
This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.
You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
add a comment |
Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.
This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.
You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
add a comment |
Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.
This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.
You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
Assuming $J_k=[a_k, b_k]$(it could also be $J_k=[a_k, b_k)$ or $J_k=(a_k, b_k]$, it doesn't matter, you'll see) we get that $J_k subset (a_k-frac{epsilon}{2^k}, b_k+frac{epsilon}{2^k})$ and note that the sum of these extensions increases the sum of the lengths of $J_k$ by $epsilon$ which can be made arbitrarilty small.
This is how we show that $J_k$ is contained in an open set. We do a similar thing for the $I_n$ but by making them smaller and thus closed. The reason we do this is because then the union of finitely many $I_n$ will also be closed and (assuming they are also bounded, which you didn't say btw) their union will be compact and therefore they will have to be covered by finitely many of the open $J_k$.
You use the fact that the $I_n$ are disjoint when you write $l(I_1 cup ... cup I_n) = l(I_1)+...+l(I_n) $
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
answered Nov 29 '18 at 14:26
Sorin TircSorin Tirc
1,520113
1,520113
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3018675%2flength-of-countably-many-intervals-2%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
