Is $(0,1]$ a closed or open set?Open/Closed Setsclosed and open set - set $S$ is open if and only if its complement is closed?How do I determine if a set is open or closed??Is if a set is not open and its complement not closed, is the set closed and its complement open?Is this set neither open nor closed or closed?Relatively open / Relatively closed setsShowing set is closed and other sets not closed.A closed subset of $[0,1]$ with no interiors and has measure exactly $1$(please check my work) Topology: interior,boundary,limit points, isolated points.Set $E = mathbb R setminus n in mathbb N $ is Open in $mathbb R$ ? Is it Closed in $mathbb R$?
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Is $(0,1]$ a closed or open set?
Open/Closed Setsclosed and open set - set $S$ is open if and only if its complement is closed?How do I determine if a set is open or closed??Is if a set is not open and its complement not closed, is the set closed and its complement open?Is this set neither open nor closed or closed?Relatively open / Relatively closed setsShowing set is closed and other sets not closed.A closed subset of $[0,1]$ with no interiors and has measure exactly $1$(please check my work) Topology: interior,boundary,limit points, isolated points.Set $E = mathbb R setminus n in mathbb N $ is Open in $mathbb R$ ? Is it Closed in $mathbb R$?
$begingroup$
Is $A=(0,1]$ a closed or open set?
I think it's not an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.
If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$
real-analysis
asked yesterday
QwertfordQwertford
323212
$endgroup$
|
show 3 more comments
$begingroup$
Is $A=(0,1]$ a closed or open set?
I think it's not an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.
If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$
real-analysis
asked yesterday
QwertfordQwertford
323212
$endgroup$
$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
yesterday
5
$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
yesterday
9
$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
yesterday
1
$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
yesterday
2
$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
yesterday
|
show 3 more comments
$begingroup$
Is $A=(0,1]$ a closed or open set?
I think it's not an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.
If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$
real-analysis
asked yesterday
QwertfordQwertford
323212
$endgroup$
Is $A=(0,1]$ a closed or open set?
I think it's not an open set because it is not a subset of its interior points. Mainly, $1in A$ but $1notin A^circ$.
If A is closed, then the complement is open. However, the complement $A^c$ is not open because it is not a subset of its interior points. Mainly, $0 in A^c$ but $0notin (A^c)^circ$
real-analysis
real-analysis
asked yesterday
QwertfordQwertford
323212
asked yesterday
QwertfordQwertford
323212
edited 18 hours ago
Qwertford
asked yesterday
QwertfordQwertford
323212
asked yesterday
QwertfordQwertford
323212
asked yesterday
QwertfordQwertford
323212
323212
$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
yesterday
5
$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
yesterday
9
$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
yesterday
1
$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
yesterday
2
$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
yesterday
|
show 3 more comments
$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
yesterday
5
$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
yesterday
9
$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
yesterday
1
$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
yesterday
2
$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
yesterday
$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
yesterday
$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
yesterday
5
5
$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
yesterday
$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
yesterday
9
9
$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
yesterday
$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
yesterday
1
1
$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
yesterday
$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
yesterday
2
2
$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
yesterday
$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
yesterday
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.
answered yesterday
J.G.J.G.
31.8k23250
$endgroup$
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
2
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
add a comment |
$begingroup$
It's important that you specify where you are considering the subset $A$.
If $A subset X$ with $ X = mathbbR$, J.G. is absolutely right in the usual topology of $mathbbR$.
If $A subset X$ with $ X = [0,1]$, $A^c = 0 $, which is closed in the usual topology, then $A = (0,1]$ is open.
In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.
edited yesterday
Norrius
1055
answered yesterday
521124521124
78110
$endgroup$
add a comment |
$begingroup$
A set is not a door.
It is not the case that a set is either open or closed. It can also be neither or both.
Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbbR$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.
answered yesterday
ArnoArno
1,2381615
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.
answered yesterday
J.G.J.G.
31.8k23250
$endgroup$
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
2
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
add a comment |
$begingroup$
Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.
answered yesterday
J.G.J.G.
31.8k23250
$endgroup$
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
2
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
add a comment |
$begingroup$
Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.
answered yesterday
J.G.J.G.
31.8k23250
$endgroup$
Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-infty,,0]cup (1,,infty)$, which doesn't contain a neighbourhood of $0$.
answered yesterday
J.G.J.G.
31.8k23250
answered yesterday
J.G.J.G.
31.8k23250
answered yesterday
J.G.J.G.
31.8k23250
answered yesterday
J.G.J.G.
31.8k23250
31.8k23250
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
2
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
add a comment |
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
2
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
$begingroup$
note that "its complement" is only what it is if we assume reals as the topology. $(0, 1]$ is open in itself when using the $D = |x-y|$ metric because the complement is the empty set.
$endgroup$
– John Dvorak
yesterday
2
2
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
$begingroup$
@JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
$endgroup$
– J.G.
yesterday
add a comment |
$begingroup$
It's important that you specify where you are considering the subset $A$.
If $A subset X$ with $ X = mathbbR$, J.G. is absolutely right in the usual topology of $mathbbR$.
If $A subset X$ with $ X = [0,1]$, $A^c = 0 $, which is closed in the usual topology, then $A = (0,1]$ is open.
In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.
edited yesterday
Norrius
1055
answered yesterday
521124521124
78110
$endgroup$
add a comment |
$begingroup$
It's important that you specify where you are considering the subset $A$.
If $A subset X$ with $ X = mathbbR$, J.G. is absolutely right in the usual topology of $mathbbR$.
If $A subset X$ with $ X = [0,1]$, $A^c = 0 $, which is closed in the usual topology, then $A = (0,1]$ is open.
In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.
edited yesterday
Norrius
1055
answered yesterday
521124521124
78110
$endgroup$
add a comment |
$begingroup$
It's important that you specify where you are considering the subset $A$.
If $A subset X$ with $ X = mathbbR$, J.G. is absolutely right in the usual topology of $mathbbR$.
If $A subset X$ with $ X = [0,1]$, $A^c = 0 $, which is closed in the usual topology, then $A = (0,1]$ is open.
In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.
edited yesterday
Norrius
1055
answered yesterday
521124521124
78110
$endgroup$
It's important that you specify where you are considering the subset $A$.
If $A subset X$ with $ X = mathbbR$, J.G. is absolutely right in the usual topology of $mathbbR$.
If $A subset X$ with $ X = [0,1]$, $A^c = 0 $, which is closed in the usual topology, then $A = (0,1]$ is open.
In other words, it's important to specify in what topologic space $X$ you are considering $A$ as a subset. There are some stranger metrics which may define some different open sets where things can be different.
edited yesterday
Norrius
1055
answered yesterday
521124521124
78110
edited yesterday
Norrius
1055
edited yesterday
Norrius
1055
edited yesterday
Norrius
1055
1055
answered yesterday
521124521124
78110
answered yesterday
521124521124
78110
answered yesterday
521124521124
78110
78110
add a comment |
add a comment |
$begingroup$
A set is not a door.
It is not the case that a set is either open or closed. It can also be neither or both.
Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbbR$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.
answered yesterday
ArnoArno
1,2381615
$endgroup$
add a comment |
$begingroup$
A set is not a door.
It is not the case that a set is either open or closed. It can also be neither or both.
Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbbR$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.
answered yesterday
ArnoArno
1,2381615
$endgroup$
add a comment |
$begingroup$
A set is not a door.
It is not the case that a set is either open or closed. It can also be neither or both.
Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbbR$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.
answered yesterday
ArnoArno
1,2381615
$endgroup$
A set is not a door.
It is not the case that a set is either open or closed. It can also be neither or both.
Indeed, your arguments correctly establish that $(0,1]$ is neither open nor closed as a subset of $mathbbR$ with the usual topology. The empty set $emptyset$ is always both open and closed, no matter what the ambient space is.
answered yesterday
ArnoArno
1,2381615
answered yesterday
ArnoArno
1,2381615
answered yesterday
ArnoArno
1,2381615
answered yesterday
ArnoArno
1,2381615
1,2381615
add a comment |
add a comment |
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$begingroup$
$(0, 1]$ is a semi-open or semi-closed set.
$endgroup$
– Paras Khosla
yesterday
5
$begingroup$
Depends on the topology!
$endgroup$
– Jakobian
yesterday
9
$begingroup$
Unlike doors, subsets of topological spaces may be both open and closed, and they may be neither open nor closed. This is an example of how using every day words to name precise mathematical definitions can be misleading.
$endgroup$
– Simon
yesterday
1
$begingroup$
The answer to your question is no.
$endgroup$
– Robert Shore
yesterday
2
$begingroup$
@Simon doors can also be open and closed at the same time! You just need them to be adjoint to two different entrances at once.
$endgroup$
– John Dvorak
yesterday