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Is $(0,1]$ a closed or open set?

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4

1

$begingroup$

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

share|cite|improve this question

edited Mar 24 at 10:06

Qwertford

asked Mar 23 at 16:17

QwertfordQwertford

340213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(0, 1]$ is a semi-open or semi-closed set.
  $endgroup$
  – Paras Khosla
  Mar 23 at 16:18
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Depends on the topology!
  $endgroup$
  – Jakobian
  Mar 23 at 16:21
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 23 at 16:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer to your question is no.
  $endgroup$
  – Robert Shore
  Mar 23 at 16:42
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 23 at 18:38
  

  
  
  
  

 | 
show 2 more comments

4

1

$begingroup$

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

share|cite|improve this question

edited Mar 24 at 10:06

Qwertford

asked Mar 23 at 16:17

QwertfordQwertford

340213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(0, 1]$ is a semi-open or semi-closed set.
  $endgroup$
  – Paras Khosla
  Mar 23 at 16:18
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Depends on the topology!
  $endgroup$
  – Jakobian
  Mar 23 at 16:21
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 23 at 16:23
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer to your question is no.
  $endgroup$
  – Robert Shore
  Mar 23 at 16:42
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 23 at 18:38
  

  
  
  
  

 | 
show 2 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

share|cite|improve this question

edited Mar 24 at 10:06

Qwertford

asked Mar 23 at 16:17

QwertfordQwertford

340213

$endgroup$

share|cite|improve this question

edited Mar 24 at 10:06

Qwertford

asked Mar 23 at 16:17

QwertfordQwertford

340213

share|cite|improve this question

edited Mar 24 at 10:06

Qwertford

asked Mar 23 at 16:17

QwertfordQwertford

340213

asked Mar 23 at 16:17

QwertfordQwertford

340213

asked Mar 23 at 16:17

QwertfordQwertford

340213

3 Answers
3

active

oldest

votes

12

$begingroup$

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 $0$.

share|cite|improve this answer

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 23 at 18:24
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
  $endgroup$
  – J.G.
  Mar 23 at 18:36
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

edited Mar 23 at 18:01

Norrius

1055

answered Mar 23 at 17:15

521124521124

78110

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 23 at 18:45

ArnoArno

1,2381615

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add a comment | 

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12

$begingroup$

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 $0$.

share|cite|improve this answer

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 23 at 18:24
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
  $endgroup$
  – J.G.
  Mar 23 at 18:36
  

  
  
  
  

add a comment | 

12

$begingroup$

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 $0$.

share|cite|improve this answer

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 23 at 18:24
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnDvorak Absolutely true. These questions can have just about any answer with a "non-default" topology.
  $endgroup$
  – J.G.
  Mar 23 at 18:36
  

  
  
  
  

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 $0$.

share|cite|improve this answer

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

$endgroup$

share|cite|improve this answer

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

answered Mar 23 at 16:19

J.G.J.G.

37.6k23656

1

$begingroup$

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.

share|cite|improve this answer

edited Mar 23 at 18:01

Norrius

1055

answered Mar 23 at 17:15

521124521124

78110

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

edited Mar 23 at 18:01

Norrius

1055

answered Mar 23 at 17:15

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.

share|cite|improve this answer

edited Mar 23 at 18:01

Norrius

1055

answered Mar 23 at 17:15

521124521124

78110

$endgroup$

share|cite|improve this answer

edited Mar 23 at 18:01

Norrius

1055

answered Mar 23 at 17:15

521124521124

78110

edited Mar 23 at 18:01

Norrius

1055

edited Mar 23 at 18:01

Norrius

1055

answered Mar 23 at 17:15

521124521124

78110

answered Mar 23 at 17:15

521124521124

78110

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 23 at 18:45

ArnoArno

1,2381615

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 23 at 18:45

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.

share|cite|improve this answer

answered Mar 23 at 18:45

ArnoArno

1,2381615

$endgroup$

share|cite|improve this answer

answered Mar 23 at 18:45

ArnoArno

1,2381615

answered Mar 23 at 18:45

ArnoArno

1,2381615

answered Mar 23 at 18:45

ArnoArno

1,2381615