Mixing
An arbitrary intersection of open sets
Clash Royale CLAN TAG#URR8PPP
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I learned in class that the finite intersection of open sets is open. With that "big picture" in mind I want to prove the following:
Define $ A:=bigcaplimits_n=1^infty I_n $, where $I_n:=left(-frac1n, 1+frac1nright). textThen A = [0,1]$
It seems very obvious that A equals $ [0,1] $, but I am having trouble proving that the two sets are equal to each other. I think I have to show that $ A subset [0.1]$ and $ [0,1] subset A $. It is very irritating because I don't know where to begin.
Which direction should I take to prove the two sets are equal? Any help or hint in any form would be very much appreciated. I believe this question would be helpful for novices like me, because the answers I found from the "similar questions" tab do not discuss how to prove that the sets are identical.
elementary-set-theory real-numbers
asked 37 mins ago
Ko Byeongmin
153
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Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
I learned in class that the finite intersection of open sets is open. With that "big picture" in mind I want to prove the following:
Define $ A:=bigcaplimits_n=1^infty I_n $, where $I_n:=left(-frac1n, 1+frac1nright). textThen A = [0,1]$
It seems very obvious that A equals $ [0,1] $, but I am having trouble proving that the two sets are equal to each other. I think I have to show that $ A subset [0.1]$ and $ [0,1] subset A $. It is very irritating because I don't know where to begin.
Which direction should I take to prove the two sets are equal? Any help or hint in any form would be very much appreciated. I believe this question would be helpful for novices like me, because the answers I found from the "similar questions" tab do not discuss how to prove that the sets are identical.
elementary-set-theory real-numbers
asked 37 mins ago
Ko Byeongmin
153
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Did you mean -- arbitrary (finite or infinite) intersection of "closed" sets is closed? Looks like there is a typo.
– nature1729
21 mins ago
Oh that was a big error I made there!! I hope no one got confused because of my typo.... (Edited it!)
– Ko Byeongmin
18 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I learned in class that the finite intersection of open sets is open. With that "big picture" in mind I want to prove the following:
Define $ A:=bigcaplimits_n=1^infty I_n $, where $I_n:=left(-frac1n, 1+frac1nright). textThen A = [0,1]$
It seems very obvious that A equals $ [0,1] $, but I am having trouble proving that the two sets are equal to each other. I think I have to show that $ A subset [0.1]$ and $ [0,1] subset A $. It is very irritating because I don't know where to begin.
Which direction should I take to prove the two sets are equal? Any help or hint in any form would be very much appreciated. I believe this question would be helpful for novices like me, because the answers I found from the "similar questions" tab do not discuss how to prove that the sets are identical.
elementary-set-theory real-numbers
asked 37 mins ago
Ko Byeongmin
153
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I learned in class that the finite intersection of open sets is open. With that "big picture" in mind I want to prove the following:
Define $ A:=bigcaplimits_n=1^infty I_n $, where $I_n:=left(-frac1n, 1+frac1nright). textThen A = [0,1]$
It seems very obvious that A equals $ [0,1] $, but I am having trouble proving that the two sets are equal to each other. I think I have to show that $ A subset [0.1]$ and $ [0,1] subset A $. It is very irritating because I don't know where to begin.
Which direction should I take to prove the two sets are equal? Any help or hint in any form would be very much appreciated. I believe this question would be helpful for novices like me, because the answers I found from the "similar questions" tab do not discuss how to prove that the sets are identical.
elementary-set-theory real-numbers
elementary-set-theory real-numbers
asked 37 mins ago
Ko Byeongmin
153
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 37 mins ago
Ko Byeongmin
153
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 mins ago
asked 37 mins ago
Ko Byeongmin
153
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 37 mins ago
Ko Byeongmin
153
asked 37 mins ago
Ko Byeongmin
153
153
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ko Byeongmin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Did you mean -- arbitrary (finite or infinite) intersection of "closed" sets is closed? Looks like there is a typo.
– nature1729
21 mins ago
Oh that was a big error I made there!! I hope no one got confused because of my typo.... (Edited it!)
– Ko Byeongmin
18 mins ago
add a comment |Â
Did you mean -- arbitrary (finite or infinite) intersection of "closed" sets is closed? Looks like there is a typo.
– nature1729
21 mins ago
Oh that was a big error I made there!! I hope no one got confused because of my typo.... (Edited it!)
– Ko Byeongmin
18 mins ago
Did you mean -- arbitrary (finite or infinite) intersection of "closed" sets is closed? Looks like there is a typo.
– nature1729
21 mins ago
Did you mean -- arbitrary (finite or infinite) intersection of "closed" sets is closed? Looks like there is a typo.
– nature1729
21 mins ago
Oh that was a big error I made there!! I hope no one got confused because of my typo.... (Edited it!)
– Ko Byeongmin
18 mins ago
Oh that was a big error I made there!! I hope no one got confused because of my typo.... (Edited it!)
– Ko Byeongmin
18 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Start by "proving" (actually it is evident) that $[0,1]subseteq I_n$ for every $n$ so that consequently: $$[0,1]subseteq Atag1$$
Then for $xnotin[0,1]$ prove that a positive integer $n$ exists with $xnotin I_n$, hence $xnotin A$
That proves: $[0,1]^complementsubseteq A^complement$ or equivalently $$Asubseteq [0,1]tag2$$
Taking $(1)$ and $(2)$ together we end up with:$$[0,1]=A$$qed
answered 31 mins ago
drhab
90.9k542124
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
add a comment |Â
up vote
2
down vote
Since $(forall ninmathbbN):Asubsetleft(-frac1n,1+frac1nright)$,$$Asubsetbigcap_ninmathbbNleft(-frac1n,1+frac1nright).tag1$$In order to prove that $(1)$ is actually an equality, take $xin A^complement$ and then prove that $xnotinleft(-frac1n,1+frac1nright)$, for some $ninmathbb N$.
answered 29 mins ago
José Carlos Santos
128k17103191
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
add a comment |Â
up vote
2
down vote
If $x in [0,1]$ then for any $n$: $-frac1n < 0 le x le 1 < 1+frac1n$, so $x in I_n$ for all $n$, hence $x in bigcap_n I_n = A$. So $[0,1] subseteq A$.
To see that $A subseteq [0,1]$, let $x$ be any point of $A$. Then suppose (for a contradiction) that $x notin [0,1]$ so either $x <0$ or $x > 1$. If $x < 0$, then as $-frac1n to 0$, there is some $n_0$ such that $x < -frac1n_0$ so $x notin I_n_0$ which contradicts $x in A$. If $x >1$ we find an $n_1$ such that $1+frac1n_1 < x$, as $1+frac1n to 1$, and then $x notin I_n_1$, another contradiction with $x in A$. This shows that either way $x notin [0,1]$ leads to a contradiction, so $x in [0,1]$ as required. So we have equality $[0,1] = A$.
The intersection of open sets can be any set at all (need not be closed): note that all sets of the form $mathbbRsetminus x$ are open and $$A = bigcap mathbbR setminus x: x notin A$$
Subsets that are countable intersections of open sets are called $G_delta$ sets and among those we have the irrationals, and yes, also all closed subsets (of metric spaces at least) can be written that way. But these sets can also be open, if we take $O_n = O$ for some fixed open set $O$, then also $bigcap_n O_n = O$ etc.
answered 23 mins ago
Henno Brandsma
96.7k343104
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
add a comment |Â
up vote
0
down vote
If $x>0, x<1$ then it is in all $I_n$. Hence, $(0,1)subseteq A$.
Assume, $a < 0$. Then there is a $n >0 $ such that, $a <-frac1n$. That $a not in I_n$. Hence it will not be in $A$.
Further, similarly, $a>1$ is not in $A$.
Now consider $0$. It is in every $I_n$. So it is in A. Similarly for $1$.
Thus, $A=0cup(0,1)cup1=[0,1]$.
answered 13 mins ago
nature1729
39328
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Start by "proving" (actually it is evident) that $[0,1]subseteq I_n$ for every $n$ so that consequently: $$[0,1]subseteq Atag1$$
Then for $xnotin[0,1]$ prove that a positive integer $n$ exists with $xnotin I_n$, hence $xnotin A$
That proves: $[0,1]^complementsubseteq A^complement$ or equivalently $$Asubseteq [0,1]tag2$$
Taking $(1)$ and $(2)$ together we end up with:$$[0,1]=A$$qed
answered 31 mins ago
drhab
90.9k542124
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
add a comment |Â
up vote
3
down vote
accepted
Start by "proving" (actually it is evident) that $[0,1]subseteq I_n$ for every $n$ so that consequently: $$[0,1]subseteq Atag1$$
Then for $xnotin[0,1]$ prove that a positive integer $n$ exists with $xnotin I_n$, hence $xnotin A$
That proves: $[0,1]^complementsubseteq A^complement$ or equivalently $$Asubseteq [0,1]tag2$$
Taking $(1)$ and $(2)$ together we end up with:$$[0,1]=A$$qed
answered 31 mins ago
drhab
90.9k542124
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Start by "proving" (actually it is evident) that $[0,1]subseteq I_n$ for every $n$ so that consequently: $$[0,1]subseteq Atag1$$
Then for $xnotin[0,1]$ prove that a positive integer $n$ exists with $xnotin I_n$, hence $xnotin A$
That proves: $[0,1]^complementsubseteq A^complement$ or equivalently $$Asubseteq [0,1]tag2$$
Taking $(1)$ and $(2)$ together we end up with:$$[0,1]=A$$qed
answered 31 mins ago
drhab
90.9k542124
Start by "proving" (actually it is evident) that $[0,1]subseteq I_n$ for every $n$ so that consequently: $$[0,1]subseteq Atag1$$
Then for $xnotin[0,1]$ prove that a positive integer $n$ exists with $xnotin I_n$, hence $xnotin A$
That proves: $[0,1]^complementsubseteq A^complement$ or equivalently $$Asubseteq [0,1]tag2$$
Taking $(1)$ and $(2)$ together we end up with:$$[0,1]=A$$qed
answered 31 mins ago
drhab
90.9k542124
answered 31 mins ago
drhab
90.9k542124
answered 31 mins ago
drhab
90.9k542124
answered 31 mins ago
drhab
90.9k542124
90.9k542124
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
add a comment |Â
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
Thanks a lot! Here is my unremarkable proof for (2): Set $ x notin [0,1] $. Let $ varepsilon = minx $. Then there exists a positive integer $ N_1 $ s.t. $ |-frac1n-0|<varepsilon $ if $ n geq N_1 $. Similarly, there exists a positive integer $ N_2 $ s.t. $ |1+frac1n-1|<varepsilon $ if $ n geq N_2 $. Take $ N = maxN_1, N_2. If $ n $ is larger than $ N $, then $ x $ do not belong to $ I_n $.
– Ko Byeongmin
11 mins ago
add a comment |Â
up vote
2
down vote
Since $(forall ninmathbbN):Asubsetleft(-frac1n,1+frac1nright)$,$$Asubsetbigcap_ninmathbbNleft(-frac1n,1+frac1nright).tag1$$In order to prove that $(1)$ is actually an equality, take $xin A^complement$ and then prove that $xnotinleft(-frac1n,1+frac1nright)$, for some $ninmathbb N$.
answered 29 mins ago
José Carlos Santos
128k17103191
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
add a comment |Â
up vote
2
down vote
Since $(forall ninmathbbN):Asubsetleft(-frac1n,1+frac1nright)$,$$Asubsetbigcap_ninmathbbNleft(-frac1n,1+frac1nright).tag1$$In order to prove that $(1)$ is actually an equality, take $xin A^complement$ and then prove that $xnotinleft(-frac1n,1+frac1nright)$, for some $ninmathbb N$.
answered 29 mins ago
José Carlos Santos
128k17103191
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $(forall ninmathbbN):Asubsetleft(-frac1n,1+frac1nright)$,$$Asubsetbigcap_ninmathbbNleft(-frac1n,1+frac1nright).tag1$$In order to prove that $(1)$ is actually an equality, take $xin A^complement$ and then prove that $xnotinleft(-frac1n,1+frac1nright)$, for some $ninmathbb N$.
answered 29 mins ago
José Carlos Santos
128k17103191
Since $(forall ninmathbbN):Asubsetleft(-frac1n,1+frac1nright)$,$$Asubsetbigcap_ninmathbbNleft(-frac1n,1+frac1nright).tag1$$In order to prove that $(1)$ is actually an equality, take $xin A^complement$ and then prove that $xnotinleft(-frac1n,1+frac1nright)$, for some $ninmathbb N$.
answered 29 mins ago
José Carlos Santos
128k17103191
answered 29 mins ago
José Carlos Santos
128k17103191
answered 29 mins ago
José Carlos Santos
128k17103191
answered 29 mins ago
José Carlos Santos
128k17103191
128k17103191
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
add a comment |Â
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
Thanks a lot! I hope that one day I could be good at math like you and help other people in this forum..
– Ko Byeongmin
21 secs ago
add a comment |Â
up vote
2
down vote
If $x in [0,1]$ then for any $n$: $-frac1n < 0 le x le 1 < 1+frac1n$, so $x in I_n$ for all $n$, hence $x in bigcap_n I_n = A$. So $[0,1] subseteq A$.
To see that $A subseteq [0,1]$, let $x$ be any point of $A$. Then suppose (for a contradiction) that $x notin [0,1]$ so either $x <0$ or $x > 1$. If $x < 0$, then as $-frac1n to 0$, there is some $n_0$ such that $x < -frac1n_0$ so $x notin I_n_0$ which contradicts $x in A$. If $x >1$ we find an $n_1$ such that $1+frac1n_1 < x$, as $1+frac1n to 1$, and then $x notin I_n_1$, another contradiction with $x in A$. This shows that either way $x notin [0,1]$ leads to a contradiction, so $x in [0,1]$ as required. So we have equality $[0,1] = A$.
The intersection of open sets can be any set at all (need not be closed): note that all sets of the form $mathbbRsetminus x$ are open and $$A = bigcap mathbbR setminus x: x notin A$$
Subsets that are countable intersections of open sets are called $G_delta$ sets and among those we have the irrationals, and yes, also all closed subsets (of metric spaces at least) can be written that way. But these sets can also be open, if we take $O_n = O$ for some fixed open set $O$, then also $bigcap_n O_n = O$ etc.
answered 23 mins ago
Henno Brandsma
96.7k343104
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
add a comment |Â
up vote
2
down vote
If $x in [0,1]$ then for any $n$: $-frac1n < 0 le x le 1 < 1+frac1n$, so $x in I_n$ for all $n$, hence $x in bigcap_n I_n = A$. So $[0,1] subseteq A$.
To see that $A subseteq [0,1]$, let $x$ be any point of $A$. Then suppose (for a contradiction) that $x notin [0,1]$ so either $x <0$ or $x > 1$. If $x < 0$, then as $-frac1n to 0$, there is some $n_0$ such that $x < -frac1n_0$ so $x notin I_n_0$ which contradicts $x in A$. If $x >1$ we find an $n_1$ such that $1+frac1n_1 < x$, as $1+frac1n to 1$, and then $x notin I_n_1$, another contradiction with $x in A$. This shows that either way $x notin [0,1]$ leads to a contradiction, so $x in [0,1]$ as required. So we have equality $[0,1] = A$.
The intersection of open sets can be any set at all (need not be closed): note that all sets of the form $mathbbRsetminus x$ are open and $$A = bigcap mathbbR setminus x: x notin A$$
Subsets that are countable intersections of open sets are called $G_delta$ sets and among those we have the irrationals, and yes, also all closed subsets (of metric spaces at least) can be written that way. But these sets can also be open, if we take $O_n = O$ for some fixed open set $O$, then also $bigcap_n O_n = O$ etc.
answered 23 mins ago
Henno Brandsma
96.7k343104
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $x in [0,1]$ then for any $n$: $-frac1n < 0 le x le 1 < 1+frac1n$, so $x in I_n$ for all $n$, hence $x in bigcap_n I_n = A$. So $[0,1] subseteq A$.
To see that $A subseteq [0,1]$, let $x$ be any point of $A$. Then suppose (for a contradiction) that $x notin [0,1]$ so either $x <0$ or $x > 1$. If $x < 0$, then as $-frac1n to 0$, there is some $n_0$ such that $x < -frac1n_0$ so $x notin I_n_0$ which contradicts $x in A$. If $x >1$ we find an $n_1$ such that $1+frac1n_1 < x$, as $1+frac1n to 1$, and then $x notin I_n_1$, another contradiction with $x in A$. This shows that either way $x notin [0,1]$ leads to a contradiction, so $x in [0,1]$ as required. So we have equality $[0,1] = A$.
The intersection of open sets can be any set at all (need not be closed): note that all sets of the form $mathbbRsetminus x$ are open and $$A = bigcap mathbbR setminus x: x notin A$$
Subsets that are countable intersections of open sets are called $G_delta$ sets and among those we have the irrationals, and yes, also all closed subsets (of metric spaces at least) can be written that way. But these sets can also be open, if we take $O_n = O$ for some fixed open set $O$, then also $bigcap_n O_n = O$ etc.
answered 23 mins ago
Henno Brandsma
96.7k343104
If $x in [0,1]$ then for any $n$: $-frac1n < 0 le x le 1 < 1+frac1n$, so $x in I_n$ for all $n$, hence $x in bigcap_n I_n = A$. So $[0,1] subseteq A$.
To see that $A subseteq [0,1]$, let $x$ be any point of $A$. Then suppose (for a contradiction) that $x notin [0,1]$ so either $x <0$ or $x > 1$. If $x < 0$, then as $-frac1n to 0$, there is some $n_0$ such that $x < -frac1n_0$ so $x notin I_n_0$ which contradicts $x in A$. If $x >1$ we find an $n_1$ such that $1+frac1n_1 < x$, as $1+frac1n to 1$, and then $x notin I_n_1$, another contradiction with $x in A$. This shows that either way $x notin [0,1]$ leads to a contradiction, so $x in [0,1]$ as required. So we have equality $[0,1] = A$.
The intersection of open sets can be any set at all (need not be closed): note that all sets of the form $mathbbRsetminus x$ are open and $$A = bigcap mathbbR setminus x: x notin A$$
Subsets that are countable intersections of open sets are called $G_delta$ sets and among those we have the irrationals, and yes, also all closed subsets (of metric spaces at least) can be written that way. But these sets can also be open, if we take $O_n = O$ for some fixed open set $O$, then also $bigcap_n O_n = O$ etc.
answered 23 mins ago
Henno Brandsma
96.7k343104
edited 8 mins ago
answered 23 mins ago
Henno Brandsma
96.7k343104
answered 23 mins ago
Henno Brandsma
96.7k343104
answered 23 mins ago
Henno Brandsma
96.7k343104
96.7k343104
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
add a comment |Â
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
Thank you for your great proof! I think I get what you mean. And thanks again for correcting my typo I made!! I can't state enough how much I appreciate your answer.
– Ko Byeongmin
1 min ago
add a comment |Â
up vote
0
down vote
If $x>0, x<1$ then it is in all $I_n$. Hence, $(0,1)subseteq A$.
Assume, $a < 0$. Then there is a $n >0 $ such that, $a <-frac1n$. That $a not in I_n$. Hence it will not be in $A$.
Further, similarly, $a>1$ is not in $A$.
Now consider $0$. It is in every $I_n$. So it is in A. Similarly for $1$.
Thus, $A=0cup(0,1)cup1=[0,1]$.
answered 13 mins ago
nature1729
39328
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0
down vote
If $x>0, x<1$ then it is in all $I_n$. Hence, $(0,1)subseteq A$.
Assume, $a < 0$. Then there is a $n >0 $ such that, $a <-frac1n$. That $a not in I_n$. Hence it will not be in $A$.
Further, similarly, $a>1$ is not in $A$.
Now consider $0$. It is in every $I_n$. So it is in A. Similarly for $1$.
Thus, $A=0cup(0,1)cup1=[0,1]$.
answered 13 mins ago
nature1729
39328
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $x>0, x<1$ then it is in all $I_n$. Hence, $(0,1)subseteq A$.
Assume, $a < 0$. Then there is a $n >0 $ such that, $a <-frac1n$. That $a not in I_n$. Hence it will not be in $A$.
Further, similarly, $a>1$ is not in $A$.
Now consider $0$. It is in every $I_n$. So it is in A. Similarly for $1$.
Thus, $A=0cup(0,1)cup1=[0,1]$.
answered 13 mins ago
nature1729
39328
If $x>0, x<1$ then it is in all $I_n$. Hence, $(0,1)subseteq A$.
Assume, $a < 0$. Then there is a $n >0 $ such that, $a <-frac1n$. That $a not in I_n$. Hence it will not be in $A$.
Further, similarly, $a>1$ is not in $A$.
Now consider $0$. It is in every $I_n$. So it is in A. Similarly for $1$.
Thus, $A=0cup(0,1)cup1=[0,1]$.
answered 13 mins ago
nature1729
39328
answered 13 mins ago
nature1729
39328
answered 13 mins ago
nature1729
39328
answered 13 mins ago
nature1729
39328
39328
add a comment |Â
add a comment |Â
Ko Byeongmin is a new contributor. Be nice, and check out our Code of Conduct.
Ko Byeongmin is a new contributor. Be nice, and check out our Code of Conduct.
Ko Byeongmin is a new contributor. Be nice, and check out our Code of Conduct.
Ko Byeongmin is a new contributor. Be nice, and check out our Code of Conduct.
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Did you mean -- arbitrary (finite or infinite) intersection of "closed" sets is closed? Looks like there is a typo.
– nature1729
21 mins ago
Oh that was a big error I made there!! I hope no one got confused because of my typo.... (Edited it!)
– Ko Byeongmin
18 mins ago