Expressing half-open interval [a,b) as infinite intersection of open intervals. (Answer Verification) [closed]

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I need to express the half open interval $[a,b)$ where, $(a<b)$ as the infinite intersection of open intervals.

My answer/attempt is

$$[a,b)=bigcaplimits^infty_n=1(a-frac1n,b)$$ I'm not sure if this is correct however.

real-analysis general-topology measure-theory elementary-set-theory

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edited Sep 2 at 0:07

Shaun

7,43192972

asked Sep 1 at 23:10

kemb

634213

closed as off-topic by John Ma, Xander Henderson, kemb, Jyrki Lahtonen, Shailesh Sep 2 at 8:07

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Shailesh

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please ask one question at a time.
  – Shaun
  Sep 1 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun thank you. I'll only ask one question. I'm not sure why I received two downvotes. I even provided my solution and was just wondering if it was correct.
  – kemb
  Sep 2 at 0:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's probably because you asked two questions in one post.
  – Shaun
  Sep 2 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun. So should I just edit the question and remove the entire bullet point 2. with the second question to remove the downvotes?
  – kemb
  Sep 2 at 0:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Shaun I removed the second question.
  – kemb
  Sep 2 at 0:06
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

1

I need to express the half open interval $[a,b)$ where, $(a<b)$ as the infinite intersection of open intervals.

My answer/attempt is

$$[a,b)=bigcaplimits^infty_n=1(a-frac1n,b)$$ I'm not sure if this is correct however.

real-analysis general-topology measure-theory elementary-set-theory

share|cite|improve this question

edited Sep 2 at 0:07

Shaun

7,43192972

asked Sep 1 at 23:10

kemb

634213

closed as off-topic by John Ma, Xander Henderson, kemb, Jyrki Lahtonen, Shailesh Sep 2 at 8:07

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Shailesh

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please ask one question at a time.
  – Shaun
  Sep 1 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun thank you. I'll only ask one question. I'm not sure why I received two downvotes. I even provided my solution and was just wondering if it was correct.
  – kemb
  Sep 2 at 0:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's probably because you asked two questions in one post.
  – Shaun
  Sep 2 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun. So should I just edit the question and remove the entire bullet point 2. with the second question to remove the downvotes?
  – kemb
  Sep 2 at 0:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Shaun I removed the second question.
  – kemb
  Sep 2 at 0:06
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

I need to express the half open interval $[a,b)$ where, $(a<b)$ as the infinite intersection of open intervals.

My answer/attempt is

$$[a,b)=bigcaplimits^infty_n=1(a-frac1n,b)$$ I'm not sure if this is correct however.

real-analysis general-topology measure-theory elementary-set-theory

share|cite|improve this question

edited Sep 2 at 0:07

Shaun

7,43192972

asked Sep 1 at 23:10

kemb

634213

I need to express the half open interval $[a,b)$ where, $(a<b)$ as the infinite intersection of open intervals.

My answer/attempt is

$$[a,b)=bigcaplimits^infty_n=1(a-frac1n,b)$$ I'm not sure if this is correct however.

real-analysis general-topology measure-theory elementary-set-theory

share|cite|improve this question

edited Sep 2 at 0:07

Shaun

7,43192972

asked Sep 1 at 23:10

kemb

634213

share|cite|improve this question

share|cite|improve this question

edited Sep 2 at 0:07

Shaun

7,43192972

edited Sep 2 at 0:07

Shaun

7,43192972

edited Sep 2 at 0:07

Shaun

7,43192972

7,43192972

asked Sep 1 at 23:10

kemb

634213

asked Sep 1 at 23:10

kemb

634213

asked Sep 1 at 23:10

kemb

634213

634213

closed as off-topic by John Ma, Xander Henderson, kemb, Jyrki Lahtonen, Shailesh Sep 2 at 8:07

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Shailesh

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

closed as off-topic by John Ma, Xander Henderson, kemb, Jyrki Lahtonen, Shailesh Sep 2 at 8:07

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Shailesh

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please ask one question at a time.
  – Shaun
  Sep 1 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun thank you. I'll only ask one question. I'm not sure why I received two downvotes. I even provided my solution and was just wondering if it was correct.
  – kemb
  Sep 2 at 0:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's probably because you asked two questions in one post.
  – Shaun
  Sep 2 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun. So should I just edit the question and remove the entire bullet point 2. with the second question to remove the downvotes?
  – kemb
  Sep 2 at 0:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Shaun I removed the second question.
  – kemb
  Sep 2 at 0:06
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please ask one question at a time.
  – Shaun
  Sep 1 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun thank you. I'll only ask one question. I'm not sure why I received two downvotes. I even provided my solution and was just wondering if it was correct.
  – kemb
  Sep 2 at 0:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's probably because you asked two questions in one post.
  – Shaun
  Sep 2 at 0:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun. So should I just edit the question and remove the entire bullet point 2. with the second question to remove the downvotes?
  – kemb
  Sep 2 at 0:04
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Shaun I removed the second question.
  – kemb
  Sep 2 at 0:06
  

  
  
  
  

Please ask one question at a time.
– Shaun
Sep 1 at 23:53

Please ask one question at a time.
– Shaun
Sep 1 at 23:53

@Shaun thank you. I'll only ask one question. I'm not sure why I received two downvotes. I even provided my solution and was just wondering if it was correct.
– kemb
Sep 2 at 0:02

@Shaun thank you. I'll only ask one question. I'm not sure why I received two downvotes. I even provided my solution and was just wondering if it was correct.
– kemb
Sep 2 at 0:02

It's probably because you asked two questions in one post.
– Shaun
Sep 2 at 0:03

It's probably because you asked two questions in one post.
– Shaun
Sep 2 at 0:03

@Shaun. So should I just edit the question and remove the entire bullet point 2. with the second question to remove the downvotes?
– kemb
Sep 2 at 0:04

@Shaun. So should I just edit the question and remove the entire bullet point 2. with the second question to remove the downvotes?
– kemb
Sep 2 at 0:04

1

1

@Shaun I removed the second question.
– kemb
Sep 2 at 0:06

@Shaun I removed the second question.
– kemb
Sep 2 at 0:06

 | 
show 5 more comments

2 Answers
2

active

oldest

votes

up vote
6
down vote



accepted

To prove that $L=[a,b) =displaystyle bigcap_n=1^inftyleft(a-frac1n,bright)=R$, you need to show that one is the subset of the other. To show $L subseteq R$, we start with some $x in [a,b)$. Then $a-frac1n<aleq x$ for all $n in mathbbN$, so $x in R$.

Now we start with $y in R$, this means for all $n geq 1$, we have $a-frac1n<y<b$.

If $y notin L$, that would mean $y<a$. So let $y=a-epsilon$, with $epsilon >0$, by the Archimedean property, there exists $k in mathbbN$ such that $kepsilon >1$. This would mean,
$$y=a-epsilon <a-frac1k.$$
But then $y notin left(a-frac1k,bright)$. This contradicts the assumption we made at the very beginning. So $y in L$.

Similarly you can check the other part.

share|cite|improve this answer

answered Sep 1 at 23:36

Anurag A

22.6k12244

add a comment | 

up vote
2
down vote

The solution is correct if your intervals make sense.
For instance $(a+1/n,b) $ makes sense if $a+1/n<b$

I would change the $1/n$ to $(b-a)/n$ to take care of that.

share|cite|improve this answer

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote



accepted

To prove that $L=[a,b) =displaystyle bigcap_n=1^inftyleft(a-frac1n,bright)=R$, you need to show that one is the subset of the other. To show $L subseteq R$, we start with some $x in [a,b)$. Then $a-frac1n<aleq x$ for all $n in mathbbN$, so $x in R$.

Now we start with $y in R$, this means for all $n geq 1$, we have $a-frac1n<y<b$.

If $y notin L$, that would mean $y<a$. So let $y=a-epsilon$, with $epsilon >0$, by the Archimedean property, there exists $k in mathbbN$ such that $kepsilon >1$. This would mean,
$$y=a-epsilon <a-frac1k.$$
But then $y notin left(a-frac1k,bright)$. This contradicts the assumption we made at the very beginning. So $y in L$.

Similarly you can check the other part.

share|cite|improve this answer

answered Sep 1 at 23:36

Anurag A

22.6k12244

add a comment | 

up vote
6
down vote



accepted

To prove that $L=[a,b) =displaystyle bigcap_n=1^inftyleft(a-frac1n,bright)=R$, you need to show that one is the subset of the other. To show $L subseteq R$, we start with some $x in [a,b)$. Then $a-frac1n<aleq x$ for all $n in mathbbN$, so $x in R$.

Now we start with $y in R$, this means for all $n geq 1$, we have $a-frac1n<y<b$.

If $y notin L$, that would mean $y<a$. So let $y=a-epsilon$, with $epsilon >0$, by the Archimedean property, there exists $k in mathbbN$ such that $kepsilon >1$. This would mean,
$$y=a-epsilon <a-frac1k.$$
But then $y notin left(a-frac1k,bright)$. This contradicts the assumption we made at the very beginning. So $y in L$.

Similarly you can check the other part.

share|cite|improve this answer

answered Sep 1 at 23:36

Anurag A

22.6k12244

add a comment | 

up vote
6
down vote



accepted

up vote
6
down vote



accepted

To prove that $L=[a,b) =displaystyle bigcap_n=1^inftyleft(a-frac1n,bright)=R$, you need to show that one is the subset of the other. To show $L subseteq R$, we start with some $x in [a,b)$. Then $a-frac1n<aleq x$ for all $n in mathbbN$, so $x in R$.

Now we start with $y in R$, this means for all $n geq 1$, we have $a-frac1n<y<b$.

If $y notin L$, that would mean $y<a$. So let $y=a-epsilon$, with $epsilon >0$, by the Archimedean property, there exists $k in mathbbN$ such that $kepsilon >1$. This would mean,
$$y=a-epsilon <a-frac1k.$$
But then $y notin left(a-frac1k,bright)$. This contradicts the assumption we made at the very beginning. So $y in L$.

Similarly you can check the other part.

share|cite|improve this answer

answered Sep 1 at 23:36

Anurag A

22.6k12244

To prove that $L=[a,b) =displaystyle bigcap_n=1^inftyleft(a-frac1n,bright)=R$, you need to show that one is the subset of the other. To show $L subseteq R$, we start with some $x in [a,b)$. Then $a-frac1n<aleq x$ for all $n in mathbbN$, so $x in R$.

Now we start with $y in R$, this means for all $n geq 1$, we have $a-frac1n<y<b$.

If $y notin L$, that would mean $y<a$. So let $y=a-epsilon$, with $epsilon >0$, by the Archimedean property, there exists $k in mathbbN$ such that $kepsilon >1$. This would mean,
$$y=a-epsilon <a-frac1k.$$
But then $y notin left(a-frac1k,bright)$. This contradicts the assumption we made at the very beginning. So $y in L$.

Similarly you can check the other part.

share|cite|improve this answer

answered Sep 1 at 23:36

Anurag A

22.6k12244

share|cite|improve this answer

share|cite|improve this answer

answered Sep 1 at 23:36

Anurag A

22.6k12244

answered Sep 1 at 23:36

Anurag A

22.6k12244

answered Sep 1 at 23:36

Anurag A

22.6k12244

22.6k12244

add a comment | 

add a comment | 

up vote
2
down vote

The solution is correct if your intervals make sense.
For instance $(a+1/n,b) $ makes sense if $a+1/n<b$

I would change the $1/n$ to $(b-a)/n$ to take care of that.

share|cite|improve this answer

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

add a comment | 

up vote
2
down vote

The solution is correct if your intervals make sense.
For instance $(a+1/n,b) $ makes sense if $a+1/n<b$

I would change the $1/n$ to $(b-a)/n$ to take care of that.

share|cite|improve this answer

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

add a comment | 

up vote
2
down vote

up vote
2
down vote

The solution is correct if your intervals make sense.
For instance $(a+1/n,b) $ makes sense if $a+1/n<b$

I would change the $1/n$ to $(b-a)/n$ to take care of that.

share|cite|improve this answer

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

The solution is correct if your intervals make sense.
For instance $(a+1/n,b) $ makes sense if $a+1/n<b$

I would change the $1/n$ to $(b-a)/n$ to take care of that.

share|cite|improve this answer

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

share|cite|improve this answer

share|cite|improve this answer

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

answered Sep 1 at 23:21

Mohammad Riazi-Kermani

30.6k41852

30.6k41852

add a comment | 

add a comment |