Mixing
Proving set of interiors is open
Clash Royale CLAN TAG#URR8PPP
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Prove: The set of interior points of any set $A$, written int($A$), is an open set.
Let $pin$ int($A$), then by definition $p$ must belong to some open interval $S_psubset A$. Now since we know that the real line itself is open then $S_psubset mathbbR$. Now suppose we pick some $qin A$. I want $qin$ int($A$).
I am stuck here I am not sure if my logic is correct, any suggestions would be greatly appreciated.
Definition of interior: Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A: $pin S_psubset A$
Definition of open: A set $A$ is open iff each of its points is an interior point
general-topology
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
 |Â
show 1 more comment
up vote
1
down vote
favorite
Prove: The set of interior points of any set $A$, written int($A$), is an open set.
Let $pin$ int($A$), then by definition $p$ must belong to some open interval $S_psubset A$. Now since we know that the real line itself is open then $S_psubset mathbbR$. Now suppose we pick some $qin A$. I want $qin$ int($A$).
I am stuck here I am not sure if my logic is correct, any suggestions would be greatly appreciated.
Definition of interior: Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A: $pin S_psubset A$
Definition of open: A set $A$ is open iff each of its points is an interior point
general-topology
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
What's your definition of the interior of a set ?
– Tryss
Feb 20 '15 at 16:02
Are you working in a general topological space, or in the real line?
– Stefan Hamcke
Feb 20 '15 at 16:06
First, it does not follow that if $q in A$ then $q in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation.
– Sloan
Feb 20 '15 at 16:06
1
Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A:
– Wolfy
Feb 20 '15 at 16:08
1
If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess.
– Samrat Mukhopadhyay
Feb 20 '15 at 16:12
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove: The set of interior points of any set $A$, written int($A$), is an open set.
Let $pin$ int($A$), then by definition $p$ must belong to some open interval $S_psubset A$. Now since we know that the real line itself is open then $S_psubset mathbbR$. Now suppose we pick some $qin A$. I want $qin$ int($A$).
I am stuck here I am not sure if my logic is correct, any suggestions would be greatly appreciated.
Definition of interior: Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A: $pin S_psubset A$
Definition of open: A set $A$ is open iff each of its points is an interior point
general-topology
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
Prove: The set of interior points of any set $A$, written int($A$), is an open set.
Let $pin$ int($A$), then by definition $p$ must belong to some open interval $S_psubset A$. Now since we know that the real line itself is open then $S_psubset mathbbR$. Now suppose we pick some $qin A$. I want $qin$ int($A$).
I am stuck here I am not sure if my logic is correct, any suggestions would be greatly appreciated.
Definition of interior: Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A: $pin S_psubset A$
Definition of open: A set $A$ is open iff each of its points is an interior point
general-topology
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
edited Feb 20 '15 at 16:10
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
asked Feb 20 '15 at 15:58
Wolfy
2,06011034
2,06011034
What's your definition of the interior of a set ?
– Tryss
Feb 20 '15 at 16:02
Are you working in a general topological space, or in the real line?
– Stefan Hamcke
Feb 20 '15 at 16:06
First, it does not follow that if $q in A$ then $q in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation.
– Sloan
Feb 20 '15 at 16:06
1
Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A:
– Wolfy
Feb 20 '15 at 16:08
1
If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess.
– Samrat Mukhopadhyay
Feb 20 '15 at 16:12
 |Â
show 1 more comment
What's your definition of the interior of a set ?
– Tryss
Feb 20 '15 at 16:02
Are you working in a general topological space, or in the real line?
– Stefan Hamcke
Feb 20 '15 at 16:06
First, it does not follow that if $q in A$ then $q in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation.
– Sloan
Feb 20 '15 at 16:06
1
Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A:
– Wolfy
Feb 20 '15 at 16:08
1
If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess.
– Samrat Mukhopadhyay
Feb 20 '15 at 16:12
What's your definition of the interior of a set ?
– Tryss
Feb 20 '15 at 16:02
What's your definition of the interior of a set ?
– Tryss
Feb 20 '15 at 16:02
Are you working in a general topological space, or in the real line?
– Stefan Hamcke
Feb 20 '15 at 16:06
Are you working in a general topological space, or in the real line?
– Stefan Hamcke
Feb 20 '15 at 16:06
First, it does not follow that if $q in A$ then $q in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation.
– Sloan
Feb 20 '15 at 16:06
First, it does not follow that if $q in A$ then $q in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation.
– Sloan
Feb 20 '15 at 16:06
1
1
Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A:
– Wolfy
Feb 20 '15 at 16:08
Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A:
– Wolfy
Feb 20 '15 at 16:08
1
1
If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess.
– Samrat Mukhopadhyay
Feb 20 '15 at 16:12
If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess.
– Samrat Mukhopadhyay
Feb 20 '15 at 16:12
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Hint 1:
Any union of open sets is open. (This is axiomatic.)
Hint 2:
We would like to show that the interior is a union of open intervals. Can you see which intervals in particular? Use the definition of interior points here.
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
add a comment |Â
up vote
6
down vote
Given $A$, we want to show that the set of interior points $operatornameint(A)$ is open, meaning that each point $pinoperatornameint(A)$ is an interior point of $operatornameint(A)$.
By definition, $pin S_psubseteq A$ for some open interval $S_p$. Now consider some $qin S_p$. Since $q$ belongs to the open interval $S_psubseteq A$, so $q$ is also an interior point of $A$. Therefore $S_psubseteqoperatornameint(A)$, so $pin S_psubseteq operatornameint(A)$ and $p$ is also an interior point of $operatornameint(A)$.
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint 1:
Any union of open sets is open. (This is axiomatic.)
Hint 2:
We would like to show that the interior is a union of open intervals. Can you see which intervals in particular? Use the definition of interior points here.
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
add a comment |Â
up vote
1
down vote
accepted
Hint 1:
Any union of open sets is open. (This is axiomatic.)
Hint 2:
We would like to show that the interior is a union of open intervals. Can you see which intervals in particular? Use the definition of interior points here.
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint 1:
Any union of open sets is open. (This is axiomatic.)
Hint 2:
We would like to show that the interior is a union of open intervals. Can you see which intervals in particular? Use the definition of interior points here.
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
Hint 1:
Any union of open sets is open. (This is axiomatic.)
Hint 2:
We would like to show that the interior is a union of open intervals. Can you see which intervals in particular? Use the definition of interior points here.
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
answered Feb 20 '15 at 16:12
Gyu Eun Lee
12.8k2252
12.8k2252
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
add a comment |Â
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am not sure if I understand your hints, we want me to state that since any union of open sets is open then let the point $p$ which be containted in the union of open intervals. Using the definition stated above the union of open intervals is contained in $A$. I am not sure if I am following...
– Wolfy
Feb 20 '15 at 16:18
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
I am saying that you can write $int(A)$ as a union of open intervals. If you can do that, then thanks to Hint 1 it is automatic that $int(A)$ is open.
– Gyu Eun Lee
Feb 20 '15 at 16:33
add a comment |Â
up vote
6
down vote
Given $A$, we want to show that the set of interior points $operatornameint(A)$ is open, meaning that each point $pinoperatornameint(A)$ is an interior point of $operatornameint(A)$.
By definition, $pin S_psubseteq A$ for some open interval $S_p$. Now consider some $qin S_p$. Since $q$ belongs to the open interval $S_psubseteq A$, so $q$ is also an interior point of $A$. Therefore $S_psubseteqoperatornameint(A)$, so $pin S_psubseteq operatornameint(A)$ and $p$ is also an interior point of $operatornameint(A)$.
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
add a comment |Â
up vote
6
down vote
Given $A$, we want to show that the set of interior points $operatornameint(A)$ is open, meaning that each point $pinoperatornameint(A)$ is an interior point of $operatornameint(A)$.
By definition, $pin S_psubseteq A$ for some open interval $S_p$. Now consider some $qin S_p$. Since $q$ belongs to the open interval $S_psubseteq A$, so $q$ is also an interior point of $A$. Therefore $S_psubseteqoperatornameint(A)$, so $pin S_psubseteq operatornameint(A)$ and $p$ is also an interior point of $operatornameint(A)$.
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Given $A$, we want to show that the set of interior points $operatornameint(A)$ is open, meaning that each point $pinoperatornameint(A)$ is an interior point of $operatornameint(A)$.
By definition, $pin S_psubseteq A$ for some open interval $S_p$. Now consider some $qin S_p$. Since $q$ belongs to the open interval $S_psubseteq A$, so $q$ is also an interior point of $A$. Therefore $S_psubseteqoperatornameint(A)$, so $pin S_psubseteq operatornameint(A)$ and $p$ is also an interior point of $operatornameint(A)$.
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
Given $A$, we want to show that the set of interior points $operatornameint(A)$ is open, meaning that each point $pinoperatornameint(A)$ is an interior point of $operatornameint(A)$.
By definition, $pin S_psubseteq A$ for some open interval $S_p$. Now consider some $qin S_p$. Since $q$ belongs to the open interval $S_psubseteq A$, so $q$ is also an interior point of $A$. Therefore $S_psubseteqoperatornameint(A)$, so $pin S_psubseteq operatornameint(A)$ and $p$ is also an interior point of $operatornameint(A)$.
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
answered Feb 20 '15 at 16:19
Mario Carneiro
17.6k33687
17.6k33687
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
add a comment |Â
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
Clean, Thanks man
– Wolfy
Feb 20 '15 at 16:22
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
No, recall that the goal was to show that $p$ is interior to $operatornameint(A)$. I wrote "also" because we started out assuming only that $p$ is an interior point of $A$.
– Mario Carneiro
Feb 20 '15 at 16:25
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
I see, well Thank you that is exactly what I was looking for
– Wolfy
Feb 20 '15 at 16:43
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
@MorganWeiss Don't forget to upvote and/or accept any answers you find useful (use the arrow buttons and checkmark next to the answers).
– Mario Carneiro
Feb 21 '15 at 1:42
add a comment |Â
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What's your definition of the interior of a set ?
– Tryss
Feb 20 '15 at 16:02
Are you working in a general topological space, or in the real line?
– Stefan Hamcke
Feb 20 '15 at 16:06
First, it does not follow that if $q in A$ then $q in int(A)$. Surely you meant something else. Also, not providing what you know about interiors and open sets will hinder progress. For instance, one definition of the interior of a set is the union of all open sets contained in $A$. Clearly this is open and needs no further investigation.
– Sloan
Feb 20 '15 at 16:06
1
Let $A$ be a set of real numbers. A point $pin A$ is an interior point iff $p$ belongs to some open interval $S_p$ which is contained in A:
– Wolfy
Feb 20 '15 at 16:08
1
If you are trying to do this proof for a metric space then you can use the definition of metric function. For a general topological space you can use the definition of base, I guess.
– Samrat Mukhopadhyay
Feb 20 '15 at 16:12