Is $bigcuplimits_i=1^infty left(-frac1i,frac1iright)$ finite?What are the cases of not using (countable) induction?How to prove this set is denumerable?Infinite set A has the same cardinality as $cup_a in A B_a$What does the notation $bigcup_ninmathbb N A_n$ mean in sets?Cardinality of the set $sum_i=1^infty a_iover 5^i subset mathbb R$Supremum, max, min, inf of $S=bigcup_n in mathbb Nleft[frac12^n,8-frac1nright)$Confused over a statement regarding power setWhich Notation Makes The Most Sense?Proving that for $E_1subset E_2subset… limlimits_nrightarrowinftyE_n=bigcuplimits_n=0^inftyE_n$ and working with measure/limit/sets$I$ is infinite, $A_k$ is countably infinite, and $A_i$ is countable for all $i neq k$. Is $prodlimits_iin IA_i$ countable?
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Is $bigcuplimits_i=1^infty left(-frac1i,frac1iright)$ finite?
What are the cases of not using (countable) induction?How to prove this set is denumerable?Infinite set A has the same cardinality as $cup_a in A B_a$What does the notation $bigcup_ninmathbb N A_n$ mean in sets?Cardinality of the set $sum_i=1^infty a_iover 5^i subset mathbb R$Supremum, max, min, inf of $S=bigcup_n in mathbb Nleft[frac12^n,8-frac1nright)$Confused over a statement regarding power setWhich Notation Makes The Most Sense?Proving that for $E_1subset E_2subset… limlimits_nrightarrowinftyE_n=bigcuplimits_n=0^inftyE_n$ and working with measure/limit/sets$I$ is infinite, $A_k$ is countably infinite, and $A_i$ is countable for all $i neq k$. Is $prodlimits_iin IA_i$ countable?
$begingroup$
In a practice exam, there is a question asking if $$
bigcuplimits_i=1^infty left(-frac1i,frac1iright)
$$ is finite, countably infinite, or uncountable. The solution to the practice exam says this is finite. Is this true?
I think it should be uncountable. $left(-frac11,frac11right)$ is uncountable, and is the first interval. The union of this interval with any other interval will also be uncountable.
The only way I can see this being true is if I have misunderstood the notation, and it is saying the number of intervals is finite. However, I don't believe this is the case.
elementary-set-theory
edited Mar 21 at 18:11
Brian
1,268216
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
$endgroup$
add a comment |
$begingroup$
In a practice exam, there is a question asking if $$
bigcuplimits_i=1^infty left(-frac1i,frac1iright)
$$ is finite, countably infinite, or uncountable. The solution to the practice exam says this is finite. Is this true?
I think it should be uncountable. $left(-frac11,frac11right)$ is uncountable, and is the first interval. The union of this interval with any other interval will also be uncountable.
The only way I can see this being true is if I have misunderstood the notation, and it is saying the number of intervals is finite. However, I don't believe this is the case.
elementary-set-theory
edited Mar 21 at 18:11
Brian
1,268216
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
$endgroup$
1
$begingroup$
Your analysis makes sense to me. Perhaps it is a misprint and was meant to be an intersection instead of a union.
$endgroup$
– David K
Mar 21 at 17:42
$begingroup$
@David That makes sense. And just as a sanity check, if it were an intersection rather than a union, the final answer would be 0, which is finite?
$endgroup$
– Jacob Jones
Mar 21 at 17:44
$begingroup$
I agree with that too.
$endgroup$
– David K
Mar 21 at 17:46
add a comment |
$begingroup$
In a practice exam, there is a question asking if $$
bigcuplimits_i=1^infty left(-frac1i,frac1iright)
$$ is finite, countably infinite, or uncountable. The solution to the practice exam says this is finite. Is this true?
I think it should be uncountable. $left(-frac11,frac11right)$ is uncountable, and is the first interval. The union of this interval with any other interval will also be uncountable.
The only way I can see this being true is if I have misunderstood the notation, and it is saying the number of intervals is finite. However, I don't believe this is the case.
elementary-set-theory
edited Mar 21 at 18:11
Brian
1,268216
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
$endgroup$
In a practice exam, there is a question asking if $$
bigcuplimits_i=1^infty left(-frac1i,frac1iright)
$$ is finite, countably infinite, or uncountable. The solution to the practice exam says this is finite. Is this true?
I think it should be uncountable. $left(-frac11,frac11right)$ is uncountable, and is the first interval. The union of this interval with any other interval will also be uncountable.
The only way I can see this being true is if I have misunderstood the notation, and it is saying the number of intervals is finite. However, I don't believe this is the case.
elementary-set-theory
elementary-set-theory
edited Mar 21 at 18:11
Brian
1,268216
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
edited Mar 21 at 18:11
Brian
1,268216
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
edited Mar 21 at 18:11
Brian
1,268216
edited Mar 21 at 18:11
Brian
1,268216
edited Mar 21 at 18:11
Brian
1,268216
1,268216
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
asked Mar 21 at 17:39
Jacob JonesJacob Jones
14111
14111
1
$begingroup$
Your analysis makes sense to me. Perhaps it is a misprint and was meant to be an intersection instead of a union.
$endgroup$
– David K
Mar 21 at 17:42
$begingroup$
@David That makes sense. And just as a sanity check, if it were an intersection rather than a union, the final answer would be 0, which is finite?
$endgroup$
– Jacob Jones
Mar 21 at 17:44
$begingroup$
I agree with that too.
$endgroup$
– David K
Mar 21 at 17:46
add a comment |
1
$begingroup$
Your analysis makes sense to me. Perhaps it is a misprint and was meant to be an intersection instead of a union.
$endgroup$
– David K
Mar 21 at 17:42
$begingroup$
@David That makes sense. And just as a sanity check, if it were an intersection rather than a union, the final answer would be 0, which is finite?
$endgroup$
– Jacob Jones
Mar 21 at 17:44
$begingroup$
I agree with that too.
$endgroup$
– David K
Mar 21 at 17:46
1
1
$begingroup$
Your analysis makes sense to me. Perhaps it is a misprint and was meant to be an intersection instead of a union.
$endgroup$
– David K
Mar 21 at 17:42
$begingroup$
Your analysis makes sense to me. Perhaps it is a misprint and was meant to be an intersection instead of a union.
$endgroup$
– David K
Mar 21 at 17:42
$begingroup$
@David That makes sense. And just as a sanity check, if it were an intersection rather than a union, the final answer would be 0, which is finite?
$endgroup$
– Jacob Jones
Mar 21 at 17:44
$begingroup$
@David That makes sense. And just as a sanity check, if it were an intersection rather than a union, the final answer would be 0, which is finite?
$endgroup$
– Jacob Jones
Mar 21 at 17:44
$begingroup$
I agree with that too.
$endgroup$
– David K
Mar 21 at 17:46
$begingroup$
I agree with that too.
$endgroup$
– David K
Mar 21 at 17:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$bigcup_i=1^inftyleft(-frac1i,frac1iright)=(-1,1)cupleft(-frac12,frac12right)cupleft(-frac13,frac13right)cupdots=(-1,1)$$
If your intervals are subsets of $mathbbR$, then this interval is uncountable.
answered Mar 21 at 17:48
csch2csch2
6251314
$endgroup$
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
add a comment |
$begingroup$
I think the question is asking you if the union is a finite, countable, or uncountable union. In which case since one index $i=1$ is enough to describe the resulting interval, this is a finite union. But you are correct these intervals are uncountable (assuming we are working in $mathbbR$), So you may also want to check the space you are working with.
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
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oldest
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votes
$begingroup$
$$bigcup_i=1^inftyleft(-frac1i,frac1iright)=(-1,1)cupleft(-frac12,frac12right)cupleft(-frac13,frac13right)cupdots=(-1,1)$$
If your intervals are subsets of $mathbbR$, then this interval is uncountable.
answered Mar 21 at 17:48
csch2csch2
6251314
$endgroup$
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
add a comment |
$begingroup$
$$bigcup_i=1^inftyleft(-frac1i,frac1iright)=(-1,1)cupleft(-frac12,frac12right)cupleft(-frac13,frac13right)cupdots=(-1,1)$$
If your intervals are subsets of $mathbbR$, then this interval is uncountable.
answered Mar 21 at 17:48
csch2csch2
6251314
$endgroup$
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
add a comment |
$begingroup$
$$bigcup_i=1^inftyleft(-frac1i,frac1iright)=(-1,1)cupleft(-frac12,frac12right)cupleft(-frac13,frac13right)cupdots=(-1,1)$$
If your intervals are subsets of $mathbbR$, then this interval is uncountable.
answered Mar 21 at 17:48
csch2csch2
6251314
$endgroup$
$$bigcup_i=1^inftyleft(-frac1i,frac1iright)=(-1,1)cupleft(-frac12,frac12right)cupleft(-frac13,frac13right)cupdots=(-1,1)$$
If your intervals are subsets of $mathbbR$, then this interval is uncountable.
answered Mar 21 at 17:48
csch2csch2
6251314
answered Mar 21 at 17:48
csch2csch2
6251314
answered Mar 21 at 17:48
csch2csch2
6251314
answered Mar 21 at 17:48
csch2csch2
6251314
6251314
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
add a comment |
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Is assuming the intervals are subsets of R a reasonable assumption when it is not specified?
$endgroup$
– Jacob Jones
Mar 21 at 17:50
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
$begingroup$
Maybe. It depends on the context of the question. If you are instead working with $mathbbQ$, then the interval would be countably infinite.
$endgroup$
– csch2
Mar 21 at 17:52
add a comment |
$begingroup$
I think the question is asking you if the union is a finite, countable, or uncountable union. In which case since one index $i=1$ is enough to describe the resulting interval, this is a finite union. But you are correct these intervals are uncountable (assuming we are working in $mathbbR$), So you may also want to check the space you are working with.
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
add a comment |
$begingroup$
I think the question is asking you if the union is a finite, countable, or uncountable union. In which case since one index $i=1$ is enough to describe the resulting interval, this is a finite union. But you are correct these intervals are uncountable (assuming we are working in $mathbbR$), So you may also want to check the space you are working with.
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
add a comment |
$begingroup$
I think the question is asking you if the union is a finite, countable, or uncountable union. In which case since one index $i=1$ is enough to describe the resulting interval, this is a finite union. But you are correct these intervals are uncountable (assuming we are working in $mathbbR$), So you may also want to check the space you are working with.
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
$endgroup$
I think the question is asking you if the union is a finite, countable, or uncountable union. In which case since one index $i=1$ is enough to describe the resulting interval, this is a finite union. But you are correct these intervals are uncountable (assuming we are working in $mathbbR$), So you may also want to check the space you are working with.
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
answered Mar 21 at 17:47
Al-Fahad Al-QadhiAl-Fahad Al-Qadhi
213
213
add a comment |
add a comment |
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1
$begingroup$
Your analysis makes sense to me. Perhaps it is a misprint and was meant to be an intersection instead of a union.
$endgroup$
– David K
Mar 21 at 17:42
$begingroup$
@David That makes sense. And just as a sanity check, if it were an intersection rather than a union, the final answer would be 0, which is finite?
$endgroup$
– Jacob Jones
Mar 21 at 17:44
$begingroup$
I agree with that too.
$endgroup$
– David K
Mar 21 at 17:46