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show $[0,1]$ is uncountable using outer measure
$begingroup$
This is a question from Real Analysis by Royden, 4th edition. (#5, pg. 34)
Using properties of outer measure, prove that $[0,1]$ is uncountable.
I believe that I am going to have to assume otherwise and use countable subadditivity in some way to produce a contradiction. I am not sure how to do so though. What other properties of outer measure might I need?
Thank you for any help!
real-analysis measure-theory lebesgue-measure
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
$endgroup$
add a comment |
$begingroup$
This is a question from Real Analysis by Royden, 4th edition. (#5, pg. 34)
Using properties of outer measure, prove that $[0,1]$ is uncountable.
I believe that I am going to have to assume otherwise and use countable subadditivity in some way to produce a contradiction. I am not sure how to do so though. What other properties of outer measure might I need?
Thank you for any help!
real-analysis measure-theory lebesgue-measure
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
$endgroup$
add a comment |
$begingroup$
This is a question from Real Analysis by Royden, 4th edition. (#5, pg. 34)
Using properties of outer measure, prove that $[0,1]$ is uncountable.
I believe that I am going to have to assume otherwise and use countable subadditivity in some way to produce a contradiction. I am not sure how to do so though. What other properties of outer measure might I need?
Thank you for any help!
real-analysis measure-theory lebesgue-measure
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
$endgroup$
This is a question from Real Analysis by Royden, 4th edition. (#5, pg. 34)
Using properties of outer measure, prove that $[0,1]$ is uncountable.
I believe that I am going to have to assume otherwise and use countable subadditivity in some way to produce a contradiction. I am not sure how to do so though. What other properties of outer measure might I need?
Thank you for any help!
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
asked Sep 10 '14 at 19:16
RebekahRebekah
443614
443614
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Do you already know points have measure zero?
If so, then enumerate all points in $[0,1]$. The "countable" union of all these points is the entire interval, with measure $1$. The "countable" sum of the measures of all these points, on the other hand, is...
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
$endgroup$
2
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
add a comment |
$begingroup$
Let [0 1] is countable then outer measure of [0 1] is "zero" because outer measure of countable st is zero . and we also know that outer measure of [a b] is "b-a" so m*[0 1]=1-0=1.
This is contradiction our assumption so [0 1] is uncountable.
answered Jan 22 '17 at 13:55
user409137user409137
111
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Do you already know points have measure zero?
If so, then enumerate all points in $[0,1]$. The "countable" union of all these points is the entire interval, with measure $1$. The "countable" sum of the measures of all these points, on the other hand, is...
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
$endgroup$
2
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
add a comment |
$begingroup$
Do you already know points have measure zero?
If so, then enumerate all points in $[0,1]$. The "countable" union of all these points is the entire interval, with measure $1$. The "countable" sum of the measures of all these points, on the other hand, is...
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
$endgroup$
2
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
add a comment |
$begingroup$
Do you already know points have measure zero?
If so, then enumerate all points in $[0,1]$. The "countable" union of all these points is the entire interval, with measure $1$. The "countable" sum of the measures of all these points, on the other hand, is...
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
$endgroup$
Do you already know points have measure zero?
If so, then enumerate all points in $[0,1]$. The "countable" union of all these points is the entire interval, with measure $1$. The "countable" sum of the measures of all these points, on the other hand, is...
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
answered Sep 10 '14 at 19:21
BaronVTBaronVT
11.6k11337
11.6k11337
2
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
add a comment |
2
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
2
2
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
$begingroup$
In any proof of the fact that [0,1] is uncountable, you are bound to use the completeness axiom of the real numbers (after all, it is this completeness axiom that gives birth to the irrationals from the rationals, which previously formed a countable set!). For example, in Cantor's proof, it was used in the fact that every real number has a decimal representation. Hence, in Baron VT's proof also, the completeness axiom is exploited somewhere! Can you figure out, where? These kind of exercises will help you develop your mathematical feelings more and more!
$endgroup$
– Somabha Mukherjee
Sep 10 '14 at 20:24
add a comment |
$begingroup$
Let [0 1] is countable then outer measure of [0 1] is "zero" because outer measure of countable st is zero . and we also know that outer measure of [a b] is "b-a" so m*[0 1]=1-0=1.
This is contradiction our assumption so [0 1] is uncountable.
answered Jan 22 '17 at 13:55
user409137user409137
111
$endgroup$
add a comment |
$begingroup$
Let [0 1] is countable then outer measure of [0 1] is "zero" because outer measure of countable st is zero . and we also know that outer measure of [a b] is "b-a" so m*[0 1]=1-0=1.
This is contradiction our assumption so [0 1] is uncountable.
answered Jan 22 '17 at 13:55
user409137user409137
111
$endgroup$
add a comment |
$begingroup$
Let [0 1] is countable then outer measure of [0 1] is "zero" because outer measure of countable st is zero . and we also know that outer measure of [a b] is "b-a" so m*[0 1]=1-0=1.
This is contradiction our assumption so [0 1] is uncountable.
answered Jan 22 '17 at 13:55
user409137user409137
111
$endgroup$
Let [0 1] is countable then outer measure of [0 1] is "zero" because outer measure of countable st is zero . and we also know that outer measure of [a b] is "b-a" so m*[0 1]=1-0=1.
This is contradiction our assumption so [0 1] is uncountable.
answered Jan 22 '17 at 13:55
user409137user409137
111
answered Jan 22 '17 at 13:55
user409137user409137
111
answered Jan 22 '17 at 13:55
user409137user409137
111
answered Jan 22 '17 at 13:55
user409137user409137
111
111
add a comment |
add a comment |
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