If a set is open, does that imply that it has no boundary points?What concept does an open set axiomatise?in...
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If a set is open, does that imply that it has no boundary points?
What concept does an open set axiomatise?in topology a set is open IFF?Is the boundary of a clopen set the empty set?A question on boundary of open setFind open set according to a boundaryOpen sets intersecting on boundaryOpen covering of the empty setShowing a set is open using boundary.Every open set has a proper open subset. What spaces satisfy this property?Open set and boundary points
$begingroup$
I'm wondering if let's say E is an open set, does that imply that ∂E is the empty set?
general-topology
asked yesterday
2000mroliver2000mroliver
644
$endgroup$
add a comment |
$begingroup$
I'm wondering if let's say E is an open set, does that imply that ∂E is the empty set?
general-topology
asked yesterday
2000mroliver2000mroliver
644
$endgroup$
7
$begingroup$
No it means that $E cap partial E=emptyset$. This is equivalent to openness of $E$.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Note that some authors (E.T. Copson) do use the term "Boundary" To refer exclusivey to points not contained in the original set. Although typically boundary refers to any point which can be "approached" from E or its compliment.
$endgroup$
– Mathaddict
yesterday
2
$begingroup$
Since no one else has, let me mention that it works the other way - if $E$ has no boundary whatsoever, then it certainly doesn't contain any of it, so is open
$endgroup$
– Wojowu
yesterday
$begingroup$
Can this question actually be answered when a particular space and topology hasn't been specified? Would any of the current answers apply to discrete space?
$endgroup$
– Shufflepants
yesterday
add a comment |
$begingroup$
I'm wondering if let's say E is an open set, does that imply that ∂E is the empty set?
general-topology
asked yesterday
2000mroliver2000mroliver
644
$endgroup$
I'm wondering if let's say E is an open set, does that imply that ∂E is the empty set?
general-topology
general-topology
asked yesterday
2000mroliver2000mroliver
644
asked yesterday
2000mroliver2000mroliver
644
asked yesterday
2000mroliver2000mroliver
644
asked yesterday
2000mroliver2000mroliver
644
asked yesterday
2000mroliver2000mroliver
644
644
7
$begingroup$
No it means that $E cap partial E=emptyset$. This is equivalent to openness of $E$.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Note that some authors (E.T. Copson) do use the term "Boundary" To refer exclusivey to points not contained in the original set. Although typically boundary refers to any point which can be "approached" from E or its compliment.
$endgroup$
– Mathaddict
yesterday
2
$begingroup$
Since no one else has, let me mention that it works the other way - if $E$ has no boundary whatsoever, then it certainly doesn't contain any of it, so is open
$endgroup$
– Wojowu
yesterday
$begingroup$
Can this question actually be answered when a particular space and topology hasn't been specified? Would any of the current answers apply to discrete space?
$endgroup$
– Shufflepants
yesterday
add a comment |
7
$begingroup$
No it means that $E cap partial E=emptyset$. This is equivalent to openness of $E$.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Note that some authors (E.T. Copson) do use the term "Boundary" To refer exclusivey to points not contained in the original set. Although typically boundary refers to any point which can be "approached" from E or its compliment.
$endgroup$
– Mathaddict
yesterday
2
$begingroup$
Since no one else has, let me mention that it works the other way - if $E$ has no boundary whatsoever, then it certainly doesn't contain any of it, so is open
$endgroup$
– Wojowu
yesterday
$begingroup$
Can this question actually be answered when a particular space and topology hasn't been specified? Would any of the current answers apply to discrete space?
$endgroup$
– Shufflepants
yesterday
7
7
$begingroup$
No it means that $E cap partial E=emptyset$. This is equivalent to openness of $E$.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
No it means that $E cap partial E=emptyset$. This is equivalent to openness of $E$.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Note that some authors (E.T. Copson) do use the term "Boundary" To refer exclusivey to points not contained in the original set. Although typically boundary refers to any point which can be "approached" from E or its compliment.
$endgroup$
– Mathaddict
yesterday
$begingroup$
Note that some authors (E.T. Copson) do use the term "Boundary" To refer exclusivey to points not contained in the original set. Although typically boundary refers to any point which can be "approached" from E or its compliment.
$endgroup$
– Mathaddict
yesterday
2
2
$begingroup$
Since no one else has, let me mention that it works the other way - if $E$ has no boundary whatsoever, then it certainly doesn't contain any of it, so is open
$endgroup$
– Wojowu
yesterday
$begingroup$
Since no one else has, let me mention that it works the other way - if $E$ has no boundary whatsoever, then it certainly doesn't contain any of it, so is open
$endgroup$
– Wojowu
yesterday
$begingroup$
Can this question actually be answered when a particular space and topology hasn't been specified? Would any of the current answers apply to discrete space?
$endgroup$
– Shufflepants
yesterday
$begingroup$
Can this question actually be answered when a particular space and topology hasn't been specified? Would any of the current answers apply to discrete space?
$endgroup$
– Shufflepants
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it means the set does not contain its boundary points, which isn't the same as not having a boundary.
Consider the following set $E subset mathbb{R^2}$, an open disc centered in $(0,0)$ with radius $1$:
$$E=left{(x,y) in mathbb{R^2} ;vert; x^2+y^2 < 1right}$$
Then this set $E$ has boundary points (namely all the points on the circle $x^2+y^2=1$), but it doesn't contain them - $E$ is an open set.
answered yesterday
StackTDStackTD
23.2k2153
$endgroup$
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
2
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
add a comment |
$begingroup$
No.
The boundary of the open interval $(0,1)$ in $Bbb R$, for example, is ${0,1}$. The set has boundary points, but does not contain them.
edited yesterday
Henno Brandsma
112k348120
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it means the set does not contain its boundary points, which isn't the same as not having a boundary.
Consider the following set $E subset mathbb{R^2}$, an open disc centered in $(0,0)$ with radius $1$:
$$E=left{(x,y) in mathbb{R^2} ;vert; x^2+y^2 < 1right}$$
Then this set $E$ has boundary points (namely all the points on the circle $x^2+y^2=1$), but it doesn't contain them - $E$ is an open set.
answered yesterday
StackTDStackTD
23.2k2153
$endgroup$
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
2
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
add a comment |
$begingroup$
No, it means the set does not contain its boundary points, which isn't the same as not having a boundary.
Consider the following set $E subset mathbb{R^2}$, an open disc centered in $(0,0)$ with radius $1$:
$$E=left{(x,y) in mathbb{R^2} ;vert; x^2+y^2 < 1right}$$
Then this set $E$ has boundary points (namely all the points on the circle $x^2+y^2=1$), but it doesn't contain them - $E$ is an open set.
answered yesterday
StackTDStackTD
23.2k2153
$endgroup$
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
2
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
add a comment |
$begingroup$
No, it means the set does not contain its boundary points, which isn't the same as not having a boundary.
Consider the following set $E subset mathbb{R^2}$, an open disc centered in $(0,0)$ with radius $1$:
$$E=left{(x,y) in mathbb{R^2} ;vert; x^2+y^2 < 1right}$$
Then this set $E$ has boundary points (namely all the points on the circle $x^2+y^2=1$), but it doesn't contain them - $E$ is an open set.
answered yesterday
StackTDStackTD
23.2k2153
$endgroup$
No, it means the set does not contain its boundary points, which isn't the same as not having a boundary.
Consider the following set $E subset mathbb{R^2}$, an open disc centered in $(0,0)$ with radius $1$:
$$E=left{(x,y) in mathbb{R^2} ;vert; x^2+y^2 < 1right}$$
Then this set $E$ has boundary points (namely all the points on the circle $x^2+y^2=1$), but it doesn't contain them - $E$ is an open set.
answered yesterday
StackTDStackTD
23.2k2153
answered yesterday
StackTDStackTD
23.2k2153
answered yesterday
StackTDStackTD
23.2k2153
answered yesterday
StackTDStackTD
23.2k2153
23.2k2153
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
2
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
add a comment |
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
2
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
This is only true under the normal familiar topology, right? Under the discrete topology, since every subset is open, doesn't every subset contain its boundary?
$endgroup$
– Shufflepants
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
$begingroup$
@Shufflepants All topological examples are depending on the topology. If nothing else is said, the default topology on $mathbb R^n$ is the one induced by the euclidean norm. (But the equivalence is also true for discrete spaces.)
$endgroup$
– Paŭlo Ebermann
yesterday
2
2
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
$begingroup$
@Shuffle Boundary is closure minus interior. In a discrete space, every subset has empty boundary.
$endgroup$
– Matt Samuel
yesterday
add a comment |
$begingroup$
No.
The boundary of the open interval $(0,1)$ in $Bbb R$, for example, is ${0,1}$. The set has boundary points, but does not contain them.
edited yesterday
Henno Brandsma
112k348120
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
add a comment |
$begingroup$
No.
The boundary of the open interval $(0,1)$ in $Bbb R$, for example, is ${0,1}$. The set has boundary points, but does not contain them.
edited yesterday
Henno Brandsma
112k348120
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
add a comment |
$begingroup$
No.
The boundary of the open interval $(0,1)$ in $Bbb R$, for example, is ${0,1}$. The set has boundary points, but does not contain them.
edited yesterday
Henno Brandsma
112k348120
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
No.
The boundary of the open interval $(0,1)$ in $Bbb R$, for example, is ${0,1}$. The set has boundary points, but does not contain them.
edited yesterday
Henno Brandsma
112k348120
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
edited yesterday
Henno Brandsma
112k348120
edited yesterday
Henno Brandsma
112k348120
edited yesterday
Henno Brandsma
112k348120
112k348120
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
answered yesterday
Hagen von EitzenHagen von Eitzen
282k23272506
282k23272506
add a comment |
add a comment |
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7
$begingroup$
No it means that $E cap partial E=emptyset$. This is equivalent to openness of $E$.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
Note that some authors (E.T. Copson) do use the term "Boundary" To refer exclusivey to points not contained in the original set. Although typically boundary refers to any point which can be "approached" from E or its compliment.
$endgroup$
– Mathaddict
yesterday
2
$begingroup$
Since no one else has, let me mention that it works the other way - if $E$ has no boundary whatsoever, then it certainly doesn't contain any of it, so is open
$endgroup$
– Wojowu
yesterday
$begingroup$
Can this question actually be answered when a particular space and topology hasn't been specified? Would any of the current answers apply to discrete space?
$endgroup$
– Shufflepants
yesterday