general topology on real line [closed]

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Given open sets A and B on Real line. Construct example such that $Acap overline B$, $overline A cap B$, $overline {Acap B}$, $overline A cap overline B$ are distinct. Also give an example of two intervals such that $Acap overline B not subset overline {Acap B}$.

general-topology

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asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

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closed as off-topic by José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan Jan 31 at 21:31

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Given open sets A and B on Real line. Construct example such that $Acap overline B$, $overline A cap B$, $overline {Acap B}$, $overline A cap overline B$ are distinct. Also give an example of two intervals such that $Acap overline B not subset overline {Acap B}$.

general-topology

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asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

$endgroup$

closed as off-topic by José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan Jan 31 at 21:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome. You didn't actually ask a question—you issued a command. You'll get more useful information back from this community if you do your own work first, then ask a question at the point you get stuck.
  $endgroup$
  – Matthew Leingang
  Jan 31 at 19:09
  

  
  
  
  

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$begingroup$

Given open sets A and B on Real line. Construct example such that $Acap overline B$, $overline A cap B$, $overline {Acap B}$, $overline A cap overline B$ are distinct. Also give an example of two intervals such that $Acap overline B not subset overline {Acap B}$.

general-topology

share|cite|improve this question

asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

$endgroup$

Given open sets A and B on Real line. Construct example such that $Acap overline B$, $overline A cap B$, $overline {Acap B}$, $overline A cap overline B$ are distinct. Also give an example of two intervals such that $Acap overline B not subset overline {Acap B}$.

general-topology

general-topology

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asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

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asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

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share|cite|improve this question

asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

asked Jan 31 at 19:06

prabhjotsinghprabhjotsingh

43

43

closed as off-topic by José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan Jan 31 at 21:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan Jan 31 at 21:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Michael Hoppe, Lee Mosher, Jyrki Lahtonen, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome. You didn't actually ask a question—you issued a command. You'll get more useful information back from this community if you do your own work first, then ask a question at the point you get stuck.
  $endgroup$
  – Matthew Leingang
  Jan 31 at 19:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome. You didn't actually ask a question—you issued a command. You'll get more useful information back from this community if you do your own work first, then ask a question at the point you get stuck.
  $endgroup$
  – Matthew Leingang
  Jan 31 at 19:09
  

  
  
  
  

4

4

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Hi, welcome. You didn't actually ask a question—you issued a command. You'll get more useful information back from this community if you do your own work first, then ask a question at the point you get stuck.
$endgroup$
– Matthew Leingang
Jan 31 at 19:09

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Hi, welcome. You didn't actually ask a question—you issued a command. You'll get more useful information back from this community if you do your own work first, then ask a question at the point you get stuck.
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– Matthew Leingang
Jan 31 at 19:09

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2 Answers
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Let $A=(0,2)cup (3,4)$ and $B=(1,3).$

$Acap overline B=((0,2)cup (3,4))cap [1,3]=[1,2).$

$overline Acap B=([0,2]cup [3,4])cap (1,3) =(1,2].$

$overline Acap overline B=([0,2]cup [3,4])cap [1,3]=[1,2]cup {3}.$

$overline {Acap B}=overline {(1,2)}=[1,2].$

Let $A'=[0,1]$ and $B'=(1,2].$

$1in [0,1]cap [1,2]=A'capoverline {B'}.$

$1not in emptyset =A'cap B'=overline {A'cap B'}.$

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answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

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I think I found something, but I'm not quite sure whether it is a little bit too complicated:
begin{equation*}
A= bigcup_{ninmathbb{N}} bigg]frac{1}{n+1}, frac{1}{n}bigg[, quad B=bigcup_{ninmathbb{N}}bigg]frac{2n+3}{(n+1)(n+2)},frac{2n+1}{n(n+1)}bigg[.
end{equation*}
Try it with these sets.

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answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Let $A=(0,2)cup (3,4)$ and $B=(1,3).$

$Acap overline B=((0,2)cup (3,4))cap [1,3]=[1,2).$

$overline Acap B=([0,2]cup [3,4])cap (1,3) =(1,2].$

$overline Acap overline B=([0,2]cup [3,4])cap [1,3]=[1,2]cup {3}.$

$overline {Acap B}=overline {(1,2)}=[1,2].$

Let $A'=[0,1]$ and $B'=(1,2].$

$1in [0,1]cap [1,2]=A'capoverline {B'}.$

$1not in emptyset =A'cap B'=overline {A'cap B'}.$

share|cite|improve this answer

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

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add a comment | 

1

$begingroup$

Let $A=(0,2)cup (3,4)$ and $B=(1,3).$

$Acap overline B=((0,2)cup (3,4))cap [1,3]=[1,2).$

$overline Acap B=([0,2]cup [3,4])cap (1,3) =(1,2].$

$overline Acap overline B=([0,2]cup [3,4])cap [1,3]=[1,2]cup {3}.$

$overline {Acap B}=overline {(1,2)}=[1,2].$

Let $A'=[0,1]$ and $B'=(1,2].$

$1in [0,1]cap [1,2]=A'capoverline {B'}.$

$1not in emptyset =A'cap B'=overline {A'cap B'}.$

share|cite|improve this answer

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $A=(0,2)cup (3,4)$ and $B=(1,3).$

$Acap overline B=((0,2)cup (3,4))cap [1,3]=[1,2).$

$overline Acap B=([0,2]cup [3,4])cap (1,3) =(1,2].$

$overline Acap overline B=([0,2]cup [3,4])cap [1,3]=[1,2]cup {3}.$

$overline {Acap B}=overline {(1,2)}=[1,2].$

Let $A'=[0,1]$ and $B'=(1,2].$

$1in [0,1]cap [1,2]=A'capoverline {B'}.$

$1not in emptyset =A'cap B'=overline {A'cap B'}.$

share|cite|improve this answer

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

Let $A=(0,2)cup (3,4)$ and $B=(1,3).$

$Acap overline B=((0,2)cup (3,4))cap [1,3]=[1,2).$

$overline Acap B=([0,2]cup [3,4])cap (1,3) =(1,2].$

$overline Acap overline B=([0,2]cup [3,4])cap [1,3]=[1,2]cup {3}.$

$overline {Acap B}=overline {(1,2)}=[1,2].$

Let $A'=[0,1]$ and $B'=(1,2].$

$1in [0,1]cap [1,2]=A'capoverline {B'}.$

$1not in emptyset =A'cap B'=overline {A'cap B'}.$

share|cite|improve this answer

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

share|cite|improve this answer

share|cite|improve this answer

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

answered Jan 31 at 21:31

DanielWainfleetDanielWainfleet

35.8k31648

35.8k31648

add a comment | 

add a comment | 

0

$begingroup$

I think I found something, but I'm not quite sure whether it is a little bit too complicated:
begin{equation*}
A= bigcup_{ninmathbb{N}} bigg]frac{1}{n+1}, frac{1}{n}bigg[, quad B=bigcup_{ninmathbb{N}}bigg]frac{2n+3}{(n+1)(n+2)},frac{2n+1}{n(n+1)}bigg[.
end{equation*}
Try it with these sets.

share|cite|improve this answer

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

$endgroup$

add a comment | 

0

$begingroup$

I think I found something, but I'm not quite sure whether it is a little bit too complicated:
begin{equation*}
A= bigcup_{ninmathbb{N}} bigg]frac{1}{n+1}, frac{1}{n}bigg[, quad B=bigcup_{ninmathbb{N}}bigg]frac{2n+3}{(n+1)(n+2)},frac{2n+1}{n(n+1)}bigg[.
end{equation*}
Try it with these sets.

share|cite|improve this answer

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

$endgroup$

add a comment | 

0

0

0

$begingroup$

I think I found something, but I'm not quite sure whether it is a little bit too complicated:
begin{equation*}
A= bigcup_{ninmathbb{N}} bigg]frac{1}{n+1}, frac{1}{n}bigg[, quad B=bigcup_{ninmathbb{N}}bigg]frac{2n+3}{(n+1)(n+2)},frac{2n+1}{n(n+1)}bigg[.
end{equation*}
Try it with these sets.

share|cite|improve this answer

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

$endgroup$

I think I found something, but I'm not quite sure whether it is a little bit too complicated:
begin{equation*}
A= bigcup_{ninmathbb{N}} bigg]frac{1}{n+1}, frac{1}{n}bigg[, quad B=bigcup_{ninmathbb{N}}bigg]frac{2n+3}{(n+1)(n+2)},frac{2n+1}{n(n+1)}bigg[.
end{equation*}
Try it with these sets.

share|cite|improve this answer

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

share|cite|improve this answer

share|cite|improve this answer

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

answered Jan 31 at 20:20

Joseph ExpoJoseph Expo

465

465

add a comment | 

add a comment | 

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