Separation Proof












0












$begingroup$


I have been banging my head against this proof for a few days now, as I can visualize why it is true in my head, but don't know how to prove it in words:



Let $A$ and $B$ be nonempty subsets of $mathbb{R}$. Show that if there exist disjoint, open sets $U$ and $V$ with $A subseteq U$ and $B subseteq V$, then $A$ and $B$ are separated.



I've seen two answers to this proof on here, but I don't fully understand either one of them, and the question was asked so long ago that neither of those users are active on here anymore, so I can't even ask them specific questions to help my understanding. I have tried proving it directly, but immediately get bogged down in multiple cases of what $A$ looks like in $U$ while $B$ looks a certain way in $V$, and vice versa. I have also tried assuming that $A$ and $B$ aren't separated, and I can't find a way to reach a contradiction (or contrapositive) from that. I would greatly appreciate any assistance as it is the only proof from my homework that I haven't been able to figure out on my own.






general-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    How do you define “separated”?
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 23:47












  • $begingroup$
    The definition I was given is as follows: Two nonempty sets $A,B subseteq mathbb{R}$ are separated if $overline{A} cap B$ and $A cap overline{B}$ are both empty, where $overline{A}$ is the closure of $A$.
    $endgroup$
    – automattik
    Dec 9 '18 at 23:51
















0












$begingroup$


I have been banging my head against this proof for a few days now, as I can visualize why it is true in my head, but don't know how to prove it in words:



Let $A$ and $B$ be nonempty subsets of $mathbb{R}$. Show that if there exist disjoint, open sets $U$ and $V$ with $A subseteq U$ and $B subseteq V$, then $A$ and $B$ are separated.



I've seen two answers to this proof on here, but I don't fully understand either one of them, and the question was asked so long ago that neither of those users are active on here anymore, so I can't even ask them specific questions to help my understanding. I have tried proving it directly, but immediately get bogged down in multiple cases of what $A$ looks like in $U$ while $B$ looks a certain way in $V$, and vice versa. I have also tried assuming that $A$ and $B$ aren't separated, and I can't find a way to reach a contradiction (or contrapositive) from that. I would greatly appreciate any assistance as it is the only proof from my homework that I haven't been able to figure out on my own.






general-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    How do you define “separated”?
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 23:47












  • $begingroup$
    The definition I was given is as follows: Two nonempty sets $A,B subseteq mathbb{R}$ are separated if $overline{A} cap B$ and $A cap overline{B}$ are both empty, where $overline{A}$ is the closure of $A$.
    $endgroup$
    – automattik
    Dec 9 '18 at 23:51














0












0








0





$begingroup$


I have been banging my head against this proof for a few days now, as I can visualize why it is true in my head, but don't know how to prove it in words:



Let $A$ and $B$ be nonempty subsets of $mathbb{R}$. Show that if there exist disjoint, open sets $U$ and $V$ with $A subseteq U$ and $B subseteq V$, then $A$ and $B$ are separated.



I've seen two answers to this proof on here, but I don't fully understand either one of them, and the question was asked so long ago that neither of those users are active on here anymore, so I can't even ask them specific questions to help my understanding. I have tried proving it directly, but immediately get bogged down in multiple cases of what $A$ looks like in $U$ while $B$ looks a certain way in $V$, and vice versa. I have also tried assuming that $A$ and $B$ aren't separated, and I can't find a way to reach a contradiction (or contrapositive) from that. I would greatly appreciate any assistance as it is the only proof from my homework that I haven't been able to figure out on my own.






general-topology







share|cite|improve this question









$endgroup$




I have been banging my head against this proof for a few days now, as I can visualize why it is true in my head, but don't know how to prove it in words:



Let $A$ and $B$ be nonempty subsets of $mathbb{R}$. Show that if there exist disjoint, open sets $U$ and $V$ with $A subseteq U$ and $B subseteq V$, then $A$ and $B$ are separated.



I've seen two answers to this proof on here, but I don't fully understand either one of them, and the question was asked so long ago that neither of those users are active on here anymore, so I can't even ask them specific questions to help my understanding. I have tried proving it directly, but immediately get bogged down in multiple cases of what $A$ looks like in $U$ while $B$ looks a certain way in $V$, and vice versa. I have also tried assuming that $A$ and $B$ aren't separated, and I can't find a way to reach a contradiction (or contrapositive) from that. I would greatly appreciate any assistance as it is the only proof from my homework that I haven't been able to figure out on my own.






general-topology




general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 23:45









automattikautomattik

517




517












  • $begingroup$
    How do you define “separated”?
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 23:47












  • $begingroup$
    The definition I was given is as follows: Two nonempty sets $A,B subseteq mathbb{R}$ are separated if $overline{A} cap B$ and $A cap overline{B}$ are both empty, where $overline{A}$ is the closure of $A$.
    $endgroup$
    – automattik
    Dec 9 '18 at 23:51


















  • $begingroup$
    How do you define “separated”?
    $endgroup$
    – José Carlos Santos
    Dec 9 '18 at 23:47












  • $begingroup$
    The definition I was given is as follows: Two nonempty sets $A,B subseteq mathbb{R}$ are separated if $overline{A} cap B$ and $A cap overline{B}$ are both empty, where $overline{A}$ is the closure of $A$.
    $endgroup$
    – automattik
    Dec 9 '18 at 23:51
















$begingroup$
How do you define “separated”?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:47






$begingroup$
How do you define “separated”?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:47














$begingroup$
The definition I was given is as follows: Two nonempty sets $A,B subseteq mathbb{R}$ are separated if $overline{A} cap B$ and $A cap overline{B}$ are both empty, where $overline{A}$ is the closure of $A$.
$endgroup$
– automattik
Dec 9 '18 at 23:51




$begingroup$
The definition I was given is as follows: Two nonempty sets $A,B subseteq mathbb{R}$ are separated if $overline{A} cap B$ and $A cap overline{B}$ are both empty, where $overline{A}$ is the closure of $A$.
$endgroup$
– automattik
Dec 9 '18 at 23:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

I will prove that $overline Acap B=emptyset$. Since $Asubset U$ and $U$ and $V$ are disjoint, then $Asubset V^complement$. But $V$ is open and so $V^complement$ is closed. So, and since $Asubset V^complement$, $overline Asubset V^complement$. But then, since $Bsubset V$, $overline Acap B=emptyset$. For the same reason, $Acapoverline B=emptyset$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
    $endgroup$
    – automattik
    Dec 10 '18 at 0:07










  • $begingroup$
    Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:56






  • 1




    $begingroup$
    That sentenc with the $x$ use useless and I've deleted it. What do you think now?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 1:01










  • $begingroup$
    I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
    $endgroup$
    – automattik
    Dec 10 '18 at 1:04



















1












$begingroup$

Hint: what can you say about the complements of $U$ and $V$ and their relationships with the closures of $A$ and $B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:02












  • $begingroup$
    Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
    $endgroup$
    – Rob Arthan
    Dec 10 '18 at 0:10













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2 Answers
2






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2 Answers
2






active

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active

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oldest

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1












$begingroup$

I will prove that $overline Acap B=emptyset$. Since $Asubset U$ and $U$ and $V$ are disjoint, then $Asubset V^complement$. But $V$ is open and so $V^complement$ is closed. So, and since $Asubset V^complement$, $overline Asubset V^complement$. But then, since $Bsubset V$, $overline Acap B=emptyset$. For the same reason, $Acapoverline B=emptyset$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
    $endgroup$
    – automattik
    Dec 10 '18 at 0:07










  • $begingroup$
    Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:56






  • 1




    $begingroup$
    That sentenc with the $x$ use useless and I've deleted it. What do you think now?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 1:01










  • $begingroup$
    I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
    $endgroup$
    – automattik
    Dec 10 '18 at 1:04
















1












$begingroup$

I will prove that $overline Acap B=emptyset$. Since $Asubset U$ and $U$ and $V$ are disjoint, then $Asubset V^complement$. But $V$ is open and so $V^complement$ is closed. So, and since $Asubset V^complement$, $overline Asubset V^complement$. But then, since $Bsubset V$, $overline Acap B=emptyset$. For the same reason, $Acapoverline B=emptyset$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
    $endgroup$
    – automattik
    Dec 10 '18 at 0:07










  • $begingroup$
    Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:56






  • 1




    $begingroup$
    That sentenc with the $x$ use useless and I've deleted it. What do you think now?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 1:01










  • $begingroup$
    I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
    $endgroup$
    – automattik
    Dec 10 '18 at 1:04














1












1








1





$begingroup$

I will prove that $overline Acap B=emptyset$. Since $Asubset U$ and $U$ and $V$ are disjoint, then $Asubset V^complement$. But $V$ is open and so $V^complement$ is closed. So, and since $Asubset V^complement$, $overline Asubset V^complement$. But then, since $Bsubset V$, $overline Acap B=emptyset$. For the same reason, $Acapoverline B=emptyset$.






share|cite|improve this answer











$endgroup$



I will prove that $overline Acap B=emptyset$. Since $Asubset U$ and $U$ and $V$ are disjoint, then $Asubset V^complement$. But $V$ is open and so $V^complement$ is closed. So, and since $Asubset V^complement$, $overline Asubset V^complement$. But then, since $Bsubset V$, $overline Acap B=emptyset$. For the same reason, $Acapoverline B=emptyset$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 1:00

























answered Dec 9 '18 at 23:57









José Carlos SantosJosé Carlos Santos

169k23132237




169k23132237












  • $begingroup$
    So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
    $endgroup$
    – automattik
    Dec 10 '18 at 0:07










  • $begingroup$
    Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:56






  • 1




    $begingroup$
    That sentenc with the $x$ use useless and I've deleted it. What do you think now?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 1:01










  • $begingroup$
    I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
    $endgroup$
    – automattik
    Dec 10 '18 at 1:04


















  • $begingroup$
    So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
    $endgroup$
    – automattik
    Dec 10 '18 at 0:07










  • $begingroup$
    Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:56






  • 1




    $begingroup$
    That sentenc with the $x$ use useless and I've deleted it. What do you think now?
    $endgroup$
    – José Carlos Santos
    Dec 10 '18 at 1:01










  • $begingroup$
    I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
    $endgroup$
    – automattik
    Dec 10 '18 at 1:04
















$begingroup$
So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
$endgroup$
– automattik
Dec 10 '18 at 0:07




$begingroup$
So by assuming $x in overline{A} cap B$, we are assuming that $A$ and $B$ are not separated and then arriving at a contradiction?
$endgroup$
– automattik
Dec 10 '18 at 0:07












$begingroup$
Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
$endgroup$
– automattik
Dec 10 '18 at 0:56




$begingroup$
Also, what happened to $x$ in your proof? You start out with it, and then it never shows up again.
$endgroup$
– automattik
Dec 10 '18 at 0:56




1




1




$begingroup$
That sentenc with the $x$ use useless and I've deleted it. What do you think now?
$endgroup$
– José Carlos Santos
Dec 10 '18 at 1:01




$begingroup$
That sentenc with the $x$ use useless and I've deleted it. What do you think now?
$endgroup$
– José Carlos Santos
Dec 10 '18 at 1:01












$begingroup$
I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
$endgroup$
– automattik
Dec 10 '18 at 1:04




$begingroup$
I actually really like it. And I suppose you could've left the $x$ sentence in there and just made another sentence at the end saying, "This contradicts that $x in overline{A} cap B$," but either way your proof helped me to make sense of another proof I saw elsewhere on this site, so I greatly appreciate your efforts!
$endgroup$
– automattik
Dec 10 '18 at 1:04











1












$begingroup$

Hint: what can you say about the complements of $U$ and $V$ and their relationships with the closures of $A$ and $B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:02












  • $begingroup$
    Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
    $endgroup$
    – Rob Arthan
    Dec 10 '18 at 0:10


















1












$begingroup$

Hint: what can you say about the complements of $U$ and $V$ and their relationships with the closures of $A$ and $B$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:02












  • $begingroup$
    Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
    $endgroup$
    – Rob Arthan
    Dec 10 '18 at 0:10
















1












1








1





$begingroup$

Hint: what can you say about the complements of $U$ and $V$ and their relationships with the closures of $A$ and $B$.






share|cite|improve this answer









$endgroup$



Hint: what can you say about the complements of $U$ and $V$ and their relationships with the closures of $A$ and $B$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 23:56









Rob ArthanRob Arthan

29.5k42967




29.5k42967












  • $begingroup$
    I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:02












  • $begingroup$
    Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
    $endgroup$
    – Rob Arthan
    Dec 10 '18 at 0:10




















  • $begingroup$
    I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
    $endgroup$
    – automattik
    Dec 10 '18 at 0:02












  • $begingroup$
    Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
    $endgroup$
    – Rob Arthan
    Dec 10 '18 at 0:10


















$begingroup$
I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
$endgroup$
– automattik
Dec 10 '18 at 0:02






$begingroup$
I believe this is what one of the other proofs I mentioned was getting at, which I did not quite follow. Since $U$ and $V$ are both open, the complement of $U$ must be closed. We know that $B$ must be a proper subset of this since it is not a subset of $U$.
$endgroup$
– automattik
Dec 10 '18 at 0:02














$begingroup$
Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
$endgroup$
– Rob Arthan
Dec 10 '18 at 0:10






$begingroup$
Try drawing a picture showing $A$, $B$, $U$, $V$, the complements of $U$ and $V$ and the closures of $A$ and $B$. Remember that if $X$ is closed and $Y subseteq X$ then $mbox{cl}(Y) subseteq X$ and think about what this means if $X$ is given as $overline{Z}$ for some open set $Z$.
$endgroup$
– Rob Arthan
Dec 10 '18 at 0:10




















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How to send String Array data to Server using php in android

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0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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Is anime1.com a legal site for watching anime?

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up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
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