Property of a n-simplex?
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Let $A subseteq mathbb{R}^n$ a open set. Then there exists a subset $B subseteq A$ where $B$ is a $n-$ simplex?.
Is the condition that $A$ is open necessary?
general-topology convex-analysis simplex
edited Dec 3 at 4:38
D_S
13.2k51551
asked Dec 3 at 4:17
Juan Daniel V.F
526
add a comment |
up vote
0
down vote
favorite
Let $A subseteq mathbb{R}^n$ a open set. Then there exists a subset $B subseteq A$ where $B$ is a $n-$ simplex?.
Is the condition that $A$ is open necessary?
general-topology convex-analysis simplex
edited Dec 3 at 4:38
D_S
13.2k51551
asked Dec 3 at 4:17
Juan Daniel V.F
526
Can you explain more carefully? I don't understand what you're asking at all
– D_S
Dec 3 at 4:25
I want to know if it is true that: Let $A subseteq mathbb{R}^n$ a open set. Then exist $B$ a $n-$simplex such that $B subseteq A$.
– Juan Daniel V.F
Dec 3 at 4:29
You can assume that $A$ is an open ball.
– Lord Shark the Unknown
Dec 3 at 4:37
1
The condition that $A$ is open is not necessary. What is necessary (and sufficient), however, is that $A$ contains an open ball (which is implied by $A$ being open). This is the case since every open ball contains an $n$-simplex and every $n$-simplex contains an open ball.
– munchhausen
Dec 3 at 4:52
There is exactly one counterexample to the claim: The empty set is open but does not contain a simplex. Of course, this might be irrelevant for the context of the question.
– Ingix
Dec 3 at 15:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A subseteq mathbb{R}^n$ a open set. Then there exists a subset $B subseteq A$ where $B$ is a $n-$ simplex?.
Is the condition that $A$ is open necessary?
general-topology convex-analysis simplex
edited Dec 3 at 4:38
D_S
13.2k51551
asked Dec 3 at 4:17
Juan Daniel V.F
526
Let $A subseteq mathbb{R}^n$ a open set. Then there exists a subset $B subseteq A$ where $B$ is a $n-$ simplex?.
Is the condition that $A$ is open necessary?
general-topology convex-analysis simplex
general-topology convex-analysis simplex
edited Dec 3 at 4:38
D_S
13.2k51551
asked Dec 3 at 4:17
Juan Daniel V.F
526
edited Dec 3 at 4:38
D_S
13.2k51551
asked Dec 3 at 4:17
Juan Daniel V.F
526
edited Dec 3 at 4:38
D_S
13.2k51551
edited Dec 3 at 4:38
D_S
13.2k51551
edited Dec 3 at 4:38
D_S
13.2k51551
13.2k51551
asked Dec 3 at 4:17
Juan Daniel V.F
526
asked Dec 3 at 4:17
Juan Daniel V.F
526
asked Dec 3 at 4:17
Juan Daniel V.F
526
526
Can you explain more carefully? I don't understand what you're asking at all
– D_S
Dec 3 at 4:25
I want to know if it is true that: Let $A subseteq mathbb{R}^n$ a open set. Then exist $B$ a $n-$simplex such that $B subseteq A$.
– Juan Daniel V.F
Dec 3 at 4:29
You can assume that $A$ is an open ball.
– Lord Shark the Unknown
Dec 3 at 4:37
1
The condition that $A$ is open is not necessary. What is necessary (and sufficient), however, is that $A$ contains an open ball (which is implied by $A$ being open). This is the case since every open ball contains an $n$-simplex and every $n$-simplex contains an open ball.
– munchhausen
Dec 3 at 4:52
There is exactly one counterexample to the claim: The empty set is open but does not contain a simplex. Of course, this might be irrelevant for the context of the question.
– Ingix
Dec 3 at 15:58
add a comment |
Can you explain more carefully? I don't understand what you're asking at all
– D_S
Dec 3 at 4:25
I want to know if it is true that: Let $A subseteq mathbb{R}^n$ a open set. Then exist $B$ a $n-$simplex such that $B subseteq A$.
– Juan Daniel V.F
Dec 3 at 4:29
You can assume that $A$ is an open ball.
– Lord Shark the Unknown
Dec 3 at 4:37
1
The condition that $A$ is open is not necessary. What is necessary (and sufficient), however, is that $A$ contains an open ball (which is implied by $A$ being open). This is the case since every open ball contains an $n$-simplex and every $n$-simplex contains an open ball.
– munchhausen
Dec 3 at 4:52
There is exactly one counterexample to the claim: The empty set is open but does not contain a simplex. Of course, this might be irrelevant for the context of the question.
– Ingix
Dec 3 at 15:58
Can you explain more carefully? I don't understand what you're asking at all
– D_S
Dec 3 at 4:25
Can you explain more carefully? I don't understand what you're asking at all
– D_S
Dec 3 at 4:25
I want to know if it is true that: Let $A subseteq mathbb{R}^n$ a open set. Then exist $B$ a $n-$simplex such that $B subseteq A$.
– Juan Daniel V.F
Dec 3 at 4:29
I want to know if it is true that: Let $A subseteq mathbb{R}^n$ a open set. Then exist $B$ a $n-$simplex such that $B subseteq A$.
– Juan Daniel V.F
Dec 3 at 4:29
You can assume that $A$ is an open ball.
– Lord Shark the Unknown
Dec 3 at 4:37
You can assume that $A$ is an open ball.
– Lord Shark the Unknown
Dec 3 at 4:37
1
1
The condition that $A$ is open is not necessary. What is necessary (and sufficient), however, is that $A$ contains an open ball (which is implied by $A$ being open). This is the case since every open ball contains an $n$-simplex and every $n$-simplex contains an open ball.
– munchhausen
Dec 3 at 4:52
The condition that $A$ is open is not necessary. What is necessary (and sufficient), however, is that $A$ contains an open ball (which is implied by $A$ being open). This is the case since every open ball contains an $n$-simplex and every $n$-simplex contains an open ball.
– munchhausen
Dec 3 at 4:52
There is exactly one counterexample to the claim: The empty set is open but does not contain a simplex. Of course, this might be irrelevant for the context of the question.
– Ingix
Dec 3 at 15:58
There is exactly one counterexample to the claim: The empty set is open but does not contain a simplex. Of course, this might be irrelevant for the context of the question.
– Ingix
Dec 3 at 15:58
add a comment |
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Can you explain more carefully? I don't understand what you're asking at all
– D_S
Dec 3 at 4:25
I want to know if it is true that: Let $A subseteq mathbb{R}^n$ a open set. Then exist $B$ a $n-$simplex such that $B subseteq A$.
– Juan Daniel V.F
Dec 3 at 4:29
You can assume that $A$ is an open ball.
– Lord Shark the Unknown
Dec 3 at 4:37
1
The condition that $A$ is open is not necessary. What is necessary (and sufficient), however, is that $A$ contains an open ball (which is implied by $A$ being open). This is the case since every open ball contains an $n$-simplex and every $n$-simplex contains an open ball.
– munchhausen
Dec 3 at 4:52
There is exactly one counterexample to the claim: The empty set is open but does not contain a simplex. Of course, this might be irrelevant for the context of the question.
– Ingix
Dec 3 at 15:58