Length of countably many intervals
$begingroup$
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. This does not seem like a valid mathematical step to me. In a proof, are we allowed to change the question? Please help.
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 as well?
real-analysis
asked Nov 29 '18 at 1:50
ThomasThomas
730416
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add a comment |
$begingroup$
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. This does not seem like a valid mathematical step to me. In a proof, are we allowed to change the question? Please help.
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 as well?
real-analysis
asked Nov 29 '18 at 1:50
ThomasThomas
730416
$endgroup$
add a comment |
$begingroup$
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. This does not seem like a valid mathematical step to me. In a proof, are we allowed to change the question? Please help.
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 as well?
real-analysis
asked Nov 29 '18 at 1:50
ThomasThomas
730416
$endgroup$
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. This does not seem like a valid mathematical step to me. In a proof, are we allowed to change the question? Please help.
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 as well?
real-analysis
real-analysis
asked Nov 29 '18 at 1:50
ThomasThomas
730416
asked Nov 29 '18 at 1:50
ThomasThomas
730416
asked Nov 29 '18 at 1:50
ThomasThomas
730416
asked Nov 29 '18 at 1:50
ThomasThomas
730416
asked Nov 29 '18 at 1:50
ThomasThomas
730416
730416
add a comment |
add a comment |
1 Answer
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This is valid because you have a little bit of room to play with in the inequality. To be explicit, let's suppose that there is a $gamma > 0$ such that
$$sum_{n = 1}^{N} |I_n| > sum_{k = 1}^{infty} |J_k| + 2gamma.$$
Replace each $J_k$ with a slightly bigger open set $tilde{J_k}$, where $|tilde{J_k}| le |J_k| + frac{gamma}{2^k}$; this can be done explicitly by replacing a (open/closed/semi-open) interval with endpoints $a, b$ by the open interval $(a-gamma/2^{k + 1}, b + gamma/2^{k + 1})$.
Carry out a similar (but completely explicit shrinking) process with the $I_n$, getting $tilde{I_n}$ with $|tilde{I_n}| ge |I_n| - frac{gamma}{2^n}$. Summing over all $n$ and $k$, you ought to find that
$$sum_{n = 1}^{N} |tilde{I_n}| > sum_{k = 1}^{infty} |tilde{J_k}|$$
where the new collections have the open/closed properties you were after.
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
$endgroup$
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
add a comment |
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$begingroup$
This is valid because you have a little bit of room to play with in the inequality. To be explicit, let's suppose that there is a $gamma > 0$ such that
$$sum_{n = 1}^{N} |I_n| > sum_{k = 1}^{infty} |J_k| + 2gamma.$$
Replace each $J_k$ with a slightly bigger open set $tilde{J_k}$, where $|tilde{J_k}| le |J_k| + frac{gamma}{2^k}$; this can be done explicitly by replacing a (open/closed/semi-open) interval with endpoints $a, b$ by the open interval $(a-gamma/2^{k + 1}, b + gamma/2^{k + 1})$.
Carry out a similar (but completely explicit shrinking) process with the $I_n$, getting $tilde{I_n}$ with $|tilde{I_n}| ge |I_n| - frac{gamma}{2^n}$. Summing over all $n$ and $k$, you ought to find that
$$sum_{n = 1}^{N} |tilde{I_n}| > sum_{k = 1}^{infty} |tilde{J_k}|$$
where the new collections have the open/closed properties you were after.
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
$endgroup$
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
add a comment |
$begingroup$
This is valid because you have a little bit of room to play with in the inequality. To be explicit, let's suppose that there is a $gamma > 0$ such that
$$sum_{n = 1}^{N} |I_n| > sum_{k = 1}^{infty} |J_k| + 2gamma.$$
Replace each $J_k$ with a slightly bigger open set $tilde{J_k}$, where $|tilde{J_k}| le |J_k| + frac{gamma}{2^k}$; this can be done explicitly by replacing a (open/closed/semi-open) interval with endpoints $a, b$ by the open interval $(a-gamma/2^{k + 1}, b + gamma/2^{k + 1})$.
Carry out a similar (but completely explicit shrinking) process with the $I_n$, getting $tilde{I_n}$ with $|tilde{I_n}| ge |I_n| - frac{gamma}{2^n}$. Summing over all $n$ and $k$, you ought to find that
$$sum_{n = 1}^{N} |tilde{I_n}| > sum_{k = 1}^{infty} |tilde{J_k}|$$
where the new collections have the open/closed properties you were after.
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
$endgroup$
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
add a comment |
$begingroup$
This is valid because you have a little bit of room to play with in the inequality. To be explicit, let's suppose that there is a $gamma > 0$ such that
$$sum_{n = 1}^{N} |I_n| > sum_{k = 1}^{infty} |J_k| + 2gamma.$$
Replace each $J_k$ with a slightly bigger open set $tilde{J_k}$, where $|tilde{J_k}| le |J_k| + frac{gamma}{2^k}$; this can be done explicitly by replacing a (open/closed/semi-open) interval with endpoints $a, b$ by the open interval $(a-gamma/2^{k + 1}, b + gamma/2^{k + 1})$.
Carry out a similar (but completely explicit shrinking) process with the $I_n$, getting $tilde{I_n}$ with $|tilde{I_n}| ge |I_n| - frac{gamma}{2^n}$. Summing over all $n$ and $k$, you ought to find that
$$sum_{n = 1}^{N} |tilde{I_n}| > sum_{k = 1}^{infty} |tilde{J_k}|$$
where the new collections have the open/closed properties you were after.
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
$endgroup$
This is valid because you have a little bit of room to play with in the inequality. To be explicit, let's suppose that there is a $gamma > 0$ such that
$$sum_{n = 1}^{N} |I_n| > sum_{k = 1}^{infty} |J_k| + 2gamma.$$
Replace each $J_k$ with a slightly bigger open set $tilde{J_k}$, where $|tilde{J_k}| le |J_k| + frac{gamma}{2^k}$; this can be done explicitly by replacing a (open/closed/semi-open) interval with endpoints $a, b$ by the open interval $(a-gamma/2^{k + 1}, b + gamma/2^{k + 1})$.
Carry out a similar (but completely explicit shrinking) process with the $I_n$, getting $tilde{I_n}$ with $|tilde{I_n}| ge |I_n| - frac{gamma}{2^n}$. Summing over all $n$ and $k$, you ought to find that
$$sum_{n = 1}^{N} |tilde{I_n}| > sum_{k = 1}^{infty} |tilde{J_k}|$$
where the new collections have the open/closed properties you were after.
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
answered Nov 29 '18 at 2:11
T. BongersT. Bongers
22.9k54661
22.9k54661
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
add a comment |
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Thank you so much! Could you answer the 2nd question as well?
$endgroup$
– Thomas
Nov 29 '18 at 3:26
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
$begingroup$
Also, where do we use the fact that the $I_n$ are pairwise disjoint?
$endgroup$
– Thomas
Nov 29 '18 at 3:41
add a comment |
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