Mixing
Show that $[a,b] = bigcap_n=1^infty (a-1/n, b+1/n).$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Show that $[a,b] = bigcap_n=1^infty (a-1/n, b+1/n).$
Attempt: $a> a -1/n$ and $b < b+1/n$. This implies $[a,b] subset (a-1/n, b+1/n)$. This implies $[a,b] subset bigcap_n=1^infty (a-1/n, b+1/n) $.
Show that $(a,b) = bigcup_n=1^infty [a+1/n, b-1/n].$
Attempt: Similarly, $a< a+1/n$ and $b> b-1/n$. Therefore, $[a+1/n, b-1/n] subset (a,b)$. This implies $bigcup_n=1^infty [a+1/n, b-1/n] subset (a,b)$.
Is this correct? and could you give some hint for how to proceed the other side?
Thank you in advance.
elementary-set-theory
asked Aug 2 at 5:38
Sihyun Kim
701210
add a comment |Â
up vote
1
down vote
favorite
Show that $[a,b] = bigcap_n=1^infty (a-1/n, b+1/n).$
Attempt: $a> a -1/n$ and $b < b+1/n$. This implies $[a,b] subset (a-1/n, b+1/n)$. This implies $[a,b] subset bigcap_n=1^infty (a-1/n, b+1/n) $.
Show that $(a,b) = bigcup_n=1^infty [a+1/n, b-1/n].$
Attempt: Similarly, $a< a+1/n$ and $b> b-1/n$. Therefore, $[a+1/n, b-1/n] subset (a,b)$. This implies $bigcup_n=1^infty [a+1/n, b-1/n] subset (a,b)$.
Is this correct? and could you give some hint for how to proceed the other side?
Thank you in advance.
elementary-set-theory
asked Aug 2 at 5:38
Sihyun Kim
701210
3
You have only proved the containments in the easier directions: you need $[a,b]supseteqbigcap(a-1/n,b+1/n)$ etc. To prove this, take $x>b$ or $x<a$ and prove there is $n$ with $xnotin(a-1/n,b+1/n)$.
– Lord Shark the Unknown
Aug 2 at 5:42
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that $[a,b] = bigcap_n=1^infty (a-1/n, b+1/n).$
Attempt: $a> a -1/n$ and $b < b+1/n$. This implies $[a,b] subset (a-1/n, b+1/n)$. This implies $[a,b] subset bigcap_n=1^infty (a-1/n, b+1/n) $.
Show that $(a,b) = bigcup_n=1^infty [a+1/n, b-1/n].$
Attempt: Similarly, $a< a+1/n$ and $b> b-1/n$. Therefore, $[a+1/n, b-1/n] subset (a,b)$. This implies $bigcup_n=1^infty [a+1/n, b-1/n] subset (a,b)$.
Is this correct? and could you give some hint for how to proceed the other side?
Thank you in advance.
elementary-set-theory
asked Aug 2 at 5:38
Sihyun Kim
701210
Show that $[a,b] = bigcap_n=1^infty (a-1/n, b+1/n).$
Attempt: $a> a -1/n$ and $b < b+1/n$. This implies $[a,b] subset (a-1/n, b+1/n)$. This implies $[a,b] subset bigcap_n=1^infty (a-1/n, b+1/n) $.
Show that $(a,b) = bigcup_n=1^infty [a+1/n, b-1/n].$
Attempt: Similarly, $a< a+1/n$ and $b> b-1/n$. Therefore, $[a+1/n, b-1/n] subset (a,b)$. This implies $bigcup_n=1^infty [a+1/n, b-1/n] subset (a,b)$.
Is this correct? and could you give some hint for how to proceed the other side?
Thank you in advance.
elementary-set-theory
asked Aug 2 at 5:38
Sihyun Kim
701210
asked Aug 2 at 5:38
Sihyun Kim
701210
asked Aug 2 at 5:38
Sihyun Kim
701210
asked Aug 2 at 5:38
Sihyun Kim
701210
701210
3
You have only proved the containments in the easier directions: you need $[a,b]supseteqbigcap(a-1/n,b+1/n)$ etc. To prove this, take $x>b$ or $x<a$ and prove there is $n$ with $xnotin(a-1/n,b+1/n)$.
– Lord Shark the Unknown
Aug 2 at 5:42
add a comment |Â
3
You have only proved the containments in the easier directions: you need $[a,b]supseteqbigcap(a-1/n,b+1/n)$ etc. To prove this, take $x>b$ or $x<a$ and prove there is $n$ with $xnotin(a-1/n,b+1/n)$.
– Lord Shark the Unknown
Aug 2 at 5:42
3
3
You have only proved the containments in the easier directions: you need $[a,b]supseteqbigcap(a-1/n,b+1/n)$ etc. To prove this, take $x>b$ or $x<a$ and prove there is $n$ with $xnotin(a-1/n,b+1/n)$.
– Lord Shark the Unknown
Aug 2 at 5:42
You have only proved the containments in the easier directions: you need $[a,b]supseteqbigcap(a-1/n,b+1/n)$ etc. To prove this, take $x>b$ or $x<a$ and prove there is $n$ with $xnotin(a-1/n,b+1/n)$.
– Lord Shark the Unknown
Aug 2 at 5:42
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
It is fine so far. For added correctness I would add "for any $n$" somewhere appropriate in both cases. For instance right before or right after the inequalities.
For the top one, I would continue by considering an $xnotin [a,b]$ and show that it is not in the intersection (i.e. that at least one of the sets you're intersecting doesn't contain $x$). For the bottom one I would continue by considering a $yin(a,b)$ and show that it is in the union (i.e. that at least one of the sets you're unioning contains $y$). The Archimedean property will be of use in both cases.
answered Aug 2 at 5:44
Arthur
98.2k793174
1
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
1
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
add a comment |Â
up vote
1
down vote
Here is another approach that might help you. I would try to avoid proving both directions separately, and treat this as a simplification problem, simplifying the right hand side to the left hand side.$%
requirebegingroup
begingroup
newcommandcalcbeginalign quad &
newcommandop[1]\ #1 quad & quad unicodex201c
newcommandhints[1]mbox#1 \ quad & quad phantomunicodex201c
newcommandhint[1]mbox#1 unicodex201d \ quad &
newcommandendcalcendalign
%$
For example, for the second problem, which $;x;$ are in the RHS set? We calculate, for any $;x;$,
$$calc
x in bigcup_n=1^infty [a+1/n, b-1/n].
opequivhintdefinition of $;bigcup;$
langle exists n : n ge 1 : a+1/n le x le b-1/n rangle
opequivhintarithmetic -- the only real moment of inspiration
langle exists n : n ge 1 : a+1/n le x le b-1/n ;land; a<x<b rangle
opequivhint...a simple calculation to isolate $;n;$...
langle exists n : n ge 1 : n ge ldots land n ge ldots rangle ;land; a<x<b
opequivhintchoose a large enough $;n;$
a<x<b
opequivhintdefinition of open interval
x in (a,b)
endcalc$$
That proves the second statement.
$%
endgroup
%$
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It is fine so far. For added correctness I would add "for any $n$" somewhere appropriate in both cases. For instance right before or right after the inequalities.
For the top one, I would continue by considering an $xnotin [a,b]$ and show that it is not in the intersection (i.e. that at least one of the sets you're intersecting doesn't contain $x$). For the bottom one I would continue by considering a $yin(a,b)$ and show that it is in the union (i.e. that at least one of the sets you're unioning contains $y$). The Archimedean property will be of use in both cases.
answered Aug 2 at 5:44
Arthur
98.2k793174
1
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
1
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
add a comment |Â
up vote
2
down vote
accepted
It is fine so far. For added correctness I would add "for any $n$" somewhere appropriate in both cases. For instance right before or right after the inequalities.
For the top one, I would continue by considering an $xnotin [a,b]$ and show that it is not in the intersection (i.e. that at least one of the sets you're intersecting doesn't contain $x$). For the bottom one I would continue by considering a $yin(a,b)$ and show that it is in the union (i.e. that at least one of the sets you're unioning contains $y$). The Archimedean property will be of use in both cases.
answered Aug 2 at 5:44
Arthur
98.2k793174
1
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
1
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It is fine so far. For added correctness I would add "for any $n$" somewhere appropriate in both cases. For instance right before or right after the inequalities.
For the top one, I would continue by considering an $xnotin [a,b]$ and show that it is not in the intersection (i.e. that at least one of the sets you're intersecting doesn't contain $x$). For the bottom one I would continue by considering a $yin(a,b)$ and show that it is in the union (i.e. that at least one of the sets you're unioning contains $y$). The Archimedean property will be of use in both cases.
answered Aug 2 at 5:44
Arthur
98.2k793174
It is fine so far. For added correctness I would add "for any $n$" somewhere appropriate in both cases. For instance right before or right after the inequalities.
For the top one, I would continue by considering an $xnotin [a,b]$ and show that it is not in the intersection (i.e. that at least one of the sets you're intersecting doesn't contain $x$). For the bottom one I would continue by considering a $yin(a,b)$ and show that it is in the union (i.e. that at least one of the sets you're unioning contains $y$). The Archimedean property will be of use in both cases.
answered Aug 2 at 5:44
Arthur
98.2k793174
answered Aug 2 at 5:44
Arthur
98.2k793174
answered Aug 2 at 5:44
Arthur
98.2k793174
answered Aug 2 at 5:44
Arthur
98.2k793174
98.2k793174
1
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
1
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
add a comment |Â
1
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
1
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
1
1
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
For the second one, let $y in (a,b)$. By Archiemedean property, $exists n$ such that $a+1/n<y$ and $b-1/n > y$. Then, $y in (a+1/n, b-1/n)$. Could give some more hint to proceed from here?
– Sihyun Kim
Aug 2 at 5:57
1
1
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
@SihyunKim Since $y in (a+1/n, b-1/n)$, we also have $y in [a+1/n, b-1/n]$, which is a set in the union. By definition of union, $y$ is contained in the whole union. This finishes the proof of $(a,b) subseteq bigcup_n=1^infty [a+1/n, b-1/n]$.
– Arthur
Aug 2 at 6:34
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
Thanks. I am wondering, for the first one, if we let $x notin [a,b]$, then how can we use the Arcimedean property as we did for the second proof?
– Sihyun Kim
Aug 2 at 6:45
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
@SihyunKim In that case, we need to split into cases: Either we have $x<a$, or we have $x>b$. If $x<a$, by the Archimedian property, there exists an $n$ such that ... On the other hand, if $x>b$, ... Either way there exists an $n$ such that $xnotin ldots$, and therefore by definition of intersection ...
– Arthur
Aug 2 at 6:49
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
I got it. Thanks again. :)
– Sihyun Kim
Aug 2 at 6:56
add a comment |Â
up vote
1
down vote
Here is another approach that might help you. I would try to avoid proving both directions separately, and treat this as a simplification problem, simplifying the right hand side to the left hand side.$%
requirebegingroup
begingroup
newcommandcalcbeginalign quad &
newcommandop[1]\ #1 quad & quad unicodex201c
newcommandhints[1]mbox#1 \ quad & quad phantomunicodex201c
newcommandhint[1]mbox#1 unicodex201d \ quad &
newcommandendcalcendalign
%$
For example, for the second problem, which $;x;$ are in the RHS set? We calculate, for any $;x;$,
$$calc
x in bigcup_n=1^infty [a+1/n, b-1/n].
opequivhintdefinition of $;bigcup;$
langle exists n : n ge 1 : a+1/n le x le b-1/n rangle
opequivhintarithmetic -- the only real moment of inspiration
langle exists n : n ge 1 : a+1/n le x le b-1/n ;land; a<x<b rangle
opequivhint...a simple calculation to isolate $;n;$...
langle exists n : n ge 1 : n ge ldots land n ge ldots rangle ;land; a<x<b
opequivhintchoose a large enough $;n;$
a<x<b
opequivhintdefinition of open interval
x in (a,b)
endcalc$$
That proves the second statement.
$%
endgroup
%$
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
add a comment |Â
up vote
1
down vote
Here is another approach that might help you. I would try to avoid proving both directions separately, and treat this as a simplification problem, simplifying the right hand side to the left hand side.$%
requirebegingroup
begingroup
newcommandcalcbeginalign quad &
newcommandop[1]\ #1 quad & quad unicodex201c
newcommandhints[1]mbox#1 \ quad & quad phantomunicodex201c
newcommandhint[1]mbox#1 unicodex201d \ quad &
newcommandendcalcendalign
%$
For example, for the second problem, which $;x;$ are in the RHS set? We calculate, for any $;x;$,
$$calc
x in bigcup_n=1^infty [a+1/n, b-1/n].
opequivhintdefinition of $;bigcup;$
langle exists n : n ge 1 : a+1/n le x le b-1/n rangle
opequivhintarithmetic -- the only real moment of inspiration
langle exists n : n ge 1 : a+1/n le x le b-1/n ;land; a<x<b rangle
opequivhint...a simple calculation to isolate $;n;$...
langle exists n : n ge 1 : n ge ldots land n ge ldots rangle ;land; a<x<b
opequivhintchoose a large enough $;n;$
a<x<b
opequivhintdefinition of open interval
x in (a,b)
endcalc$$
That proves the second statement.
$%
endgroup
%$
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is another approach that might help you. I would try to avoid proving both directions separately, and treat this as a simplification problem, simplifying the right hand side to the left hand side.$%
requirebegingroup
begingroup
newcommandcalcbeginalign quad &
newcommandop[1]\ #1 quad & quad unicodex201c
newcommandhints[1]mbox#1 \ quad & quad phantomunicodex201c
newcommandhint[1]mbox#1 unicodex201d \ quad &
newcommandendcalcendalign
%$
For example, for the second problem, which $;x;$ are in the RHS set? We calculate, for any $;x;$,
$$calc
x in bigcup_n=1^infty [a+1/n, b-1/n].
opequivhintdefinition of $;bigcup;$
langle exists n : n ge 1 : a+1/n le x le b-1/n rangle
opequivhintarithmetic -- the only real moment of inspiration
langle exists n : n ge 1 : a+1/n le x le b-1/n ;land; a<x<b rangle
opequivhint...a simple calculation to isolate $;n;$...
langle exists n : n ge 1 : n ge ldots land n ge ldots rangle ;land; a<x<b
opequivhintchoose a large enough $;n;$
a<x<b
opequivhintdefinition of open interval
x in (a,b)
endcalc$$
That proves the second statement.
$%
endgroup
%$
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
Here is another approach that might help you. I would try to avoid proving both directions separately, and treat this as a simplification problem, simplifying the right hand side to the left hand side.$%
requirebegingroup
begingroup
newcommandcalcbeginalign quad &
newcommandop[1]\ #1 quad & quad unicodex201c
newcommandhints[1]mbox#1 \ quad & quad phantomunicodex201c
newcommandhint[1]mbox#1 unicodex201d \ quad &
newcommandendcalcendalign
%$
For example, for the second problem, which $;x;$ are in the RHS set? We calculate, for any $;x;$,
$$calc
x in bigcup_n=1^infty [a+1/n, b-1/n].
opequivhintdefinition of $;bigcup;$
langle exists n : n ge 1 : a+1/n le x le b-1/n rangle
opequivhintarithmetic -- the only real moment of inspiration
langle exists n : n ge 1 : a+1/n le x le b-1/n ;land; a<x<b rangle
opequivhint...a simple calculation to isolate $;n;$...
langle exists n : n ge 1 : n ge ldots land n ge ldots rangle ;land; a<x<b
opequivhintchoose a large enough $;n;$
a<x<b
opequivhintdefinition of open interval
x in (a,b)
endcalc$$
That proves the second statement.
$%
endgroup
%$
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
answered Aug 2 at 20:33
Marnix Klooster
4,12212144
4,12212144
add a comment |Â
add a comment |Â
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3
You have only proved the containments in the easier directions: you need $[a,b]supseteqbigcap(a-1/n,b+1/n)$ etc. To prove this, take $x>b$ or $x<a$ and prove there is $n$ with $xnotin(a-1/n,b+1/n)$.
– Lord Shark the Unknown
Aug 2 at 5:42