Property of a n-simplex?
0
$begingroup$
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 '18 at 4:38
D_S
13.5k51551
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
$endgroup$
add a comment |
0
$begingroup$
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 '18 at 4:38
D_S
13.5k51551
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
$endgroup$
$begingroup$
Can you explain more carefully? I don't understand what you're asking at all
$endgroup$
– D_S
Dec 3 '18 at 4:25
$begingroup$
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 $B subseteq A$.
$endgroup$
– Juan Daniel V.F
Dec 3 '18 at 4:29
$begingroup$
You can assume that $A$ is an open ball.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 4:37
1
$begingroup$
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.
$endgroup$
– munchhausen
Dec 3 '18 at 4:52
1
$begingroup$
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.
$endgroup$
– Ingix
Dec 3 '18 at 15:58
add a comment |
0
0
0
$begingroup$
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 '18 at 4:38
D_S
13.5k51551
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
$endgroup$
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 '18 at 4:38
D_S
13.5k51551
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
edited Dec 3 '18 at 4:38
D_S
13.5k51551
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
edited Dec 3 '18 at 4:38
D_S
13.5k51551
edited Dec 3 '18 at 4:38
D_S
13.5k51551
edited Dec 3 '18 at 4:38
D_S
13.5k51551
13.5k51551
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
asked Dec 3 '18 at 4:17
Juan Daniel V.FJuan Daniel V.F
526
526
$begingroup$
Can you explain more carefully? I don't understand what you're asking at all
$endgroup$
– D_S
Dec 3 '18 at 4:25
$begingroup$
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 $B subseteq A$.
$endgroup$
– Juan Daniel V.F
Dec 3 '18 at 4:29
$begingroup$
You can assume that $A$ is an open ball.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 4:37
1
$begingroup$
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.
$endgroup$
– munchhausen
Dec 3 '18 at 4:52
1
$begingroup$
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.
$endgroup$
– Ingix
Dec 3 '18 at 15:58
add a comment |
$begingroup$
Can you explain more carefully? I don't understand what you're asking at all
$endgroup$
– D_S
Dec 3 '18 at 4:25
$begingroup$
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 $B subseteq A$.
$endgroup$
– Juan Daniel V.F
Dec 3 '18 at 4:29
$begingroup$
You can assume that $A$ is an open ball.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 4:37
1
$begingroup$
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.
$endgroup$
– munchhausen
Dec 3 '18 at 4:52
1
$begingroup$
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.
$endgroup$
– Ingix
Dec 3 '18 at 15:58
$begingroup$
Can you explain more carefully? I don't understand what you're asking at all
$endgroup$
– D_S
Dec 3 '18 at 4:25
$begingroup$
Can you explain more carefully? I don't understand what you're asking at all
$endgroup$
– D_S
Dec 3 '18 at 4:25
$begingroup$
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 $B subseteq A$.
$endgroup$
– Juan Daniel V.F
Dec 3 '18 at 4:29
$begingroup$
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 $B subseteq A$.
$endgroup$
– Juan Daniel V.F
Dec 3 '18 at 4:29
$begingroup$
You can assume that $A$ is an open ball.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 4:37
$begingroup$
You can assume that $A$ is an open ball.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 4:37
1
1
$begingroup$
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.
$endgroup$
– munchhausen
Dec 3 '18 at 4:52
$begingroup$
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.
$endgroup$
– munchhausen
Dec 3 '18 at 4:52
1
1
$begingroup$
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.
$endgroup$
– Ingix
Dec 3 '18 at 15:58
$begingroup$
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.
$endgroup$
– Ingix
Dec 3 '18 at 15:58
add a comment |
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$begingroup$
Can you explain more carefully? I don't understand what you're asking at all
$endgroup$
– D_S
Dec 3 '18 at 4:25
$begingroup$
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 $B subseteq A$.
$endgroup$
– Juan Daniel V.F
Dec 3 '18 at 4:29
$begingroup$
You can assume that $A$ is an open ball.
$endgroup$
– Lord Shark the Unknown
Dec 3 '18 at 4:37
1
$begingroup$
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.
$endgroup$
– munchhausen
Dec 3 '18 at 4:52
1
$begingroup$
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.
$endgroup$
– Ingix
Dec 3 '18 at 15:58