Supremum Proof Question
Let $a<b$ be real numbers and consider the set $T=mathbb{Q}cap[a,b]$. Show $sup T=b$.
I can show that $bgeq x$ for all $xin T$ and thus an upper bound, but am not sure how to go about showing it is the least upper bound.
supremum-and-infimum
edited Jan 10 '17 at 13:00
Cettt
1,749621
asked Aug 25 '15 at 18:11
Laura
5114
add a comment |
Let $a<b$ be real numbers and consider the set $T=mathbb{Q}cap[a,b]$. Show $sup T=b$.
I can show that $bgeq x$ for all $xin T$ and thus an upper bound, but am not sure how to go about showing it is the least upper bound.
supremum-and-infimum
edited Jan 10 '17 at 13:00
Cettt
1,749621
asked Aug 25 '15 at 18:11
Laura
5114
Take any other upper bound and show it must be greater than $b$
– Gregory Grant
Aug 25 '15 at 18:13
2
Use the fact that between any two distinct reals is a rational number.
– Akiva Weinberger
Aug 25 '15 at 18:14
add a comment |
Let $a<b$ be real numbers and consider the set $T=mathbb{Q}cap[a,b]$. Show $sup T=b$.
I can show that $bgeq x$ for all $xin T$ and thus an upper bound, but am not sure how to go about showing it is the least upper bound.
supremum-and-infimum
edited Jan 10 '17 at 13:00
Cettt
1,749621
asked Aug 25 '15 at 18:11
Laura
5114
Let $a<b$ be real numbers and consider the set $T=mathbb{Q}cap[a,b]$. Show $sup T=b$.
I can show that $bgeq x$ for all $xin T$ and thus an upper bound, but am not sure how to go about showing it is the least upper bound.
supremum-and-infimum
supremum-and-infimum
edited Jan 10 '17 at 13:00
Cettt
1,749621
asked Aug 25 '15 at 18:11
Laura
5114
edited Jan 10 '17 at 13:00
Cettt
1,749621
asked Aug 25 '15 at 18:11
Laura
5114
edited Jan 10 '17 at 13:00
Cettt
1,749621
edited Jan 10 '17 at 13:00
Cettt
1,749621
edited Jan 10 '17 at 13:00
Cettt
1,749621
1,749621
asked Aug 25 '15 at 18:11
Laura
5114
asked Aug 25 '15 at 18:11
Laura
5114
asked Aug 25 '15 at 18:11
Laura
5114
5114
Take any other upper bound and show it must be greater than $b$
– Gregory Grant
Aug 25 '15 at 18:13
2
Use the fact that between any two distinct reals is a rational number.
– Akiva Weinberger
Aug 25 '15 at 18:14
add a comment |
Take any other upper bound and show it must be greater than $b$
– Gregory Grant
Aug 25 '15 at 18:13
2
Use the fact that between any two distinct reals is a rational number.
– Akiva Weinberger
Aug 25 '15 at 18:14
Take any other upper bound and show it must be greater than $b$
– Gregory Grant
Aug 25 '15 at 18:13
Take any other upper bound and show it must be greater than $b$
– Gregory Grant
Aug 25 '15 at 18:13
2
2
Use the fact that between any two distinct reals is a rational number.
– Akiva Weinberger
Aug 25 '15 at 18:14
Use the fact that between any two distinct reals is a rational number.
– Akiva Weinberger
Aug 25 '15 at 18:14
add a comment |
2 Answers
2
active
oldest
votes
Let $b=d_1.d_2d_3dots d_kdots$
Then the members of the sequence $b-0.dots d_kdots, b-0.0dots d_kdots, b-0.00dots d_kdots dots$ are all rational, with $b-0.000dots d_kdotsto b$ as $ktoinfty$, hence the supremum of this sequence is $b$.
For example, let $b=pi$ and consider $b_0=3, b_1=3.1, b_2=3.14, b_3=3.141,dots$. We have $b_ninmathbb{Q}, lim_limits{ntoinfty}b_ntopiinmathbb{I}$.
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
add a comment |
Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $mathbb{Q}$ there is some rational number $r$ such that
$$M<r<b. (1)$$
Again by denseness of the rationals, $T$ is not empty. Take some $x in T$. Then $a leq x leq M$. Combining this with the above yields $r in T$, which gives $r leq M$, a contradiction with $(1)$.
answered Sep 12 '17 at 20:39
M. Van
2,555311
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
Let $b=d_1.d_2d_3dots d_kdots$
Then the members of the sequence $b-0.dots d_kdots, b-0.0dots d_kdots, b-0.00dots d_kdots dots$ are all rational, with $b-0.000dots d_kdotsto b$ as $ktoinfty$, hence the supremum of this sequence is $b$.
For example, let $b=pi$ and consider $b_0=3, b_1=3.1, b_2=3.14, b_3=3.141,dots$. We have $b_ninmathbb{Q}, lim_limits{ntoinfty}b_ntopiinmathbb{I}$.
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
add a comment |
Let $b=d_1.d_2d_3dots d_kdots$
Then the members of the sequence $b-0.dots d_kdots, b-0.0dots d_kdots, b-0.00dots d_kdots dots$ are all rational, with $b-0.000dots d_kdotsto b$ as $ktoinfty$, hence the supremum of this sequence is $b$.
For example, let $b=pi$ and consider $b_0=3, b_1=3.1, b_2=3.14, b_3=3.141,dots$. We have $b_ninmathbb{Q}, lim_limits{ntoinfty}b_ntopiinmathbb{I}$.
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
add a comment |
Let $b=d_1.d_2d_3dots d_kdots$
Then the members of the sequence $b-0.dots d_kdots, b-0.0dots d_kdots, b-0.00dots d_kdots dots$ are all rational, with $b-0.000dots d_kdotsto b$ as $ktoinfty$, hence the supremum of this sequence is $b$.
For example, let $b=pi$ and consider $b_0=3, b_1=3.1, b_2=3.14, b_3=3.141,dots$. We have $b_ninmathbb{Q}, lim_limits{ntoinfty}b_ntopiinmathbb{I}$.
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
Let $b=d_1.d_2d_3dots d_kdots$
Then the members of the sequence $b-0.dots d_kdots, b-0.0dots d_kdots, b-0.00dots d_kdots dots$ are all rational, with $b-0.000dots d_kdotsto b$ as $ktoinfty$, hence the supremum of this sequence is $b$.
For example, let $b=pi$ and consider $b_0=3, b_1=3.1, b_2=3.14, b_3=3.141,dots$. We have $b_ninmathbb{Q}, lim_limits{ntoinfty}b_ntopiinmathbb{I}$.
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
edited Aug 26 '15 at 4:38
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
answered Aug 25 '15 at 18:55
JonMark Perry
11.2k92237
11.2k92237
add a comment |
add a comment |
Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $mathbb{Q}$ there is some rational number $r$ such that
$$M<r<b. (1)$$
Again by denseness of the rationals, $T$ is not empty. Take some $x in T$. Then $a leq x leq M$. Combining this with the above yields $r in T$, which gives $r leq M$, a contradiction with $(1)$.
answered Sep 12 '17 at 20:39
M. Van
2,555311
add a comment |
Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $mathbb{Q}$ there is some rational number $r$ such that
$$M<r<b. (1)$$
Again by denseness of the rationals, $T$ is not empty. Take some $x in T$. Then $a leq x leq M$. Combining this with the above yields $r in T$, which gives $r leq M$, a contradiction with $(1)$.
answered Sep 12 '17 at 20:39
M. Van
2,555311
add a comment |
Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $mathbb{Q}$ there is some rational number $r$ such that
$$M<r<b. (1)$$
Again by denseness of the rationals, $T$ is not empty. Take some $x in T$. Then $a leq x leq M$. Combining this with the above yields $r in T$, which gives $r leq M$, a contradiction with $(1)$.
answered Sep 12 '17 at 20:39
M. Van
2,555311
Suppose there is a smaller upper bound for $T$, say this is some $M<b$. Then by denseness of $mathbb{Q}$ there is some rational number $r$ such that
$$M<r<b. (1)$$
Again by denseness of the rationals, $T$ is not empty. Take some $x in T$. Then $a leq x leq M$. Combining this with the above yields $r in T$, which gives $r leq M$, a contradiction with $(1)$.
answered Sep 12 '17 at 20:39
M. Van
2,555311
answered Sep 12 '17 at 20:39
M. Van
2,555311
answered Sep 12 '17 at 20:39
M. Van
2,555311
answered Sep 12 '17 at 20:39
M. Van
2,555311
2,555311
add a comment |
add a comment |
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Take any other upper bound and show it must be greater than $b$
– Gregory Grant
Aug 25 '15 at 18:13
2
Use the fact that between any two distinct reals is a rational number.
– Akiva Weinberger
Aug 25 '15 at 18:14