$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be it's bases?

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$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?

My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist]

Can anyone please correct me if I go wrong anywhere?

general-topology

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edited Jul 28 at 6:48

celtschk

28.1k65495

asked Jul 28 at 1:54

cmi

6029

add a comment | 

up vote
-1
down vote

favorite

$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?

My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist]

Can anyone please correct me if I go wrong anywhere?

general-topology

share|cite|improve this question

edited Jul 28 at 6:48

celtschk

28.1k65495

asked Jul 28 at 1:54

cmi

6029

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?

My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist]

Can anyone please correct me if I go wrong anywhere?

general-topology

share|cite|improve this question

edited Jul 28 at 6:48

celtschk

28.1k65495

asked Jul 28 at 1:54

cmi

6029

$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?

My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist]

Can anyone please correct me if I go wrong anywhere?

general-topology

share|cite|improve this question

edited Jul 28 at 6:48

celtschk

28.1k65495

asked Jul 28 at 1:54

cmi

6029

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 6:48

celtschk

28.1k65495

edited Jul 28 at 6:48

celtschk

28.1k65495

edited Jul 28 at 6:48

celtschk

28.1k65495

28.1k65495

asked Jul 28 at 1:54

cmi

6029

asked Jul 28 at 1:54

cmi

6029

asked Jul 28 at 1:54

cmi

6029

6029

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.

I would put it this way:

One basis for the order topology on $X:=left(-sqrt2,sqrt2right)capBbb Q$ is $$bigl(a,b)mid a,bin Xbigrcupbigl(-infty,b)cap Xmid bin Xbigrcupbigl(a,infty)cap Xmid ain Xbigr.$$

At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.

Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.

---

Added: As an exercise, you should try to prove that $bigl(a,b)mid a,bin Xbigr$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).

share|cite|improve this answer

edited Jul 28 at 2:22

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
  – Henning Makholm
  Jul 28 at 2:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
  – Cameron Buie
  Jul 28 at 2:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
  – Henning Makholm
  Jul 28 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
  – Cameron Buie
  Jul 28 at 2:23
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let r = $sqrt 2.$ A simple base is

(a,b) $cap$ Q : a,b in (-r,r) .

Exercise. Show a subbase is

(-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) .

In both definitions a,b can be futher restricted to rationals.

share|cite|improve this answer

answered Jul 28 at 5:24

William Elliot

5,0722414

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.

I would put it this way:

One basis for the order topology on $X:=left(-sqrt2,sqrt2right)capBbb Q$ is $$bigl(a,b)mid a,bin Xbigrcupbigl(-infty,b)cap Xmid bin Xbigrcupbigl(a,infty)cap Xmid ain Xbigr.$$

At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.

Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.

---

Added: As an exercise, you should try to prove that $bigl(a,b)mid a,bin Xbigr$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).

share|cite|improve this answer

edited Jul 28 at 2:22

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
  – Henning Makholm
  Jul 28 at 2:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
  – Cameron Buie
  Jul 28 at 2:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
  – Henning Makholm
  Jul 28 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
  – Cameron Buie
  Jul 28 at 2:23
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.

I would put it this way:

One basis for the order topology on $X:=left(-sqrt2,sqrt2right)capBbb Q$ is $$bigl(a,b)mid a,bin Xbigrcupbigl(-infty,b)cap Xmid bin Xbigrcupbigl(a,infty)cap Xmid ain Xbigr.$$

At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.

Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.

---

Added: As an exercise, you should try to prove that $bigl(a,b)mid a,bin Xbigr$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).

share|cite|improve this answer

edited Jul 28 at 2:22

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
  – Henning Makholm
  Jul 28 at 2:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
  – Cameron Buie
  Jul 28 at 2:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
  – Henning Makholm
  Jul 28 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
  – Cameron Buie
  Jul 28 at 2:23
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.

I would put it this way:

One basis for the order topology on $X:=left(-sqrt2,sqrt2right)capBbb Q$ is $$bigl(a,b)mid a,bin Xbigrcupbigl(-infty,b)cap Xmid bin Xbigrcupbigl(a,infty)cap Xmid ain Xbigr.$$

At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.

Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.

---

Added: As an exercise, you should try to prove that $bigl(a,b)mid a,bin Xbigr$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).

share|cite|improve this answer

edited Jul 28 at 2:22

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.

I would put it this way:

One basis for the order topology on $X:=left(-sqrt2,sqrt2right)capBbb Q$ is $$bigl(a,b)mid a,bin Xbigrcupbigl(-infty,b)cap Xmid bin Xbigrcupbigl(a,infty)cap Xmid ain Xbigr.$$

At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.

Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.

---

Added: As an exercise, you should try to prove that $bigl(a,b)mid a,bin Xbigr$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).

share|cite|improve this answer

edited Jul 28 at 2:22

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 2:22

edited Jul 28 at 2:22

edited Jul 28 at 2:22

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

answered Jul 28 at 2:16

Cameron Buie

83.5k771152

83.5k771152

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
  – Henning Makholm
  Jul 28 at 2:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
  – Cameron Buie
  Jul 28 at 2:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
  – Henning Makholm
  Jul 28 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
  – Cameron Buie
  Jul 28 at 2:23
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
  – Henning Makholm
  Jul 28 at 2:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
  – Cameron Buie
  Jul 28 at 2:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
  – Henning Makholm
  Jul 28 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
  – Cameron Buie
  Jul 28 at 2:23
  

  
  
  
  

1

1

It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
– Henning Makholm
Jul 28 at 2:17

It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
– Henning Makholm
Jul 28 at 2:17

@Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
– Cameron Buie
Jul 28 at 2:18

@Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
– Cameron Buie
Jul 28 at 2:18

1

1

Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
– Henning Makholm
Jul 28 at 2:21

Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
– Henning Makholm
Jul 28 at 2:21

@Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
– Cameron Buie
Jul 28 at 2:23

@Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
– Cameron Buie
Jul 28 at 2:23

add a comment | 

up vote
0
down vote

Let r = $sqrt 2.$ A simple base is

(a,b) $cap$ Q : a,b in (-r,r) .

Exercise. Show a subbase is

(-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) .

In both definitions a,b can be futher restricted to rationals.

share|cite|improve this answer

answered Jul 28 at 5:24

William Elliot

5,0722414

add a comment | 

up vote
0
down vote

Let r = $sqrt 2.$ A simple base is

(a,b) $cap$ Q : a,b in (-r,r) .

Exercise. Show a subbase is

(-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) .

In both definitions a,b can be futher restricted to rationals.

share|cite|improve this answer

answered Jul 28 at 5:24

William Elliot

5,0722414

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let r = $sqrt 2.$ A simple base is

(a,b) $cap$ Q : a,b in (-r,r) .

Exercise. Show a subbase is

(-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) .

In both definitions a,b can be futher restricted to rationals.

share|cite|improve this answer

answered Jul 28 at 5:24

William Elliot

5,0722414

Let r = $sqrt 2.$ A simple base is

(a,b) $cap$ Q : a,b in (-r,r) .

Exercise. Show a subbase is

(-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) .

In both definitions a,b can be futher restricted to rationals.

share|cite|improve this answer

answered Jul 28 at 5:24

William Elliot

5,0722414

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 5:24

William Elliot

5,0722414

answered Jul 28 at 5:24

William Elliot

5,0722414

answered Jul 28 at 5:24

William Elliot

5,0722414

5,0722414

add a comment | 

add a comment | 

 

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