Proof that the intersection of any finite number of convex sets is a convex set
$begingroup$
How to prove that the intersection of any finite number of convex sets is a convex set?
I have no idea.
convex-analysis convex-optimization
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
$endgroup$
add a comment |
$begingroup$
How to prove that the intersection of any finite number of convex sets is a convex set?
I have no idea.
convex-analysis convex-optimization
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
$endgroup$
$begingroup$
Well it makes sense, because it's a going to be a subset of one or more of the convex sets,making it convex..I just don't know how to provide a rigorous proof.
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 5:33
add a comment |
$begingroup$
How to prove that the intersection of any finite number of convex sets is a convex set?
I have no idea.
convex-analysis convex-optimization
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
$endgroup$
How to prove that the intersection of any finite number of convex sets is a convex set?
I have no idea.
convex-analysis convex-optimization
convex-analysis convex-optimization
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
edited Nov 11 '14 at 6:25
user147263
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
asked Nov 11 '14 at 5:24
Aggressive Sneeze.Aggressive Sneeze.
3315
3315
$begingroup$
Well it makes sense, because it's a going to be a subset of one or more of the convex sets,making it convex..I just don't know how to provide a rigorous proof.
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 5:33
add a comment |
$begingroup$
Well it makes sense, because it's a going to be a subset of one or more of the convex sets,making it convex..I just don't know how to provide a rigorous proof.
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 5:33
$begingroup$
Well it makes sense, because it's a going to be a subset of one or more of the convex sets,making it convex..I just don't know how to provide a rigorous proof.
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 5:33
$begingroup$
Well it makes sense, because it's a going to be a subset of one or more of the convex sets,making it convex..I just don't know how to provide a rigorous proof.
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 5:33
add a comment |
1 Answer
1
active
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votes
$begingroup$
Let $(S_i)$ be a convex set for $i = 1,2,ldots,n$.
For any $x,y in cap_{i=1}^n S_i$, $t in [0, 1]$, we have:
For $i = 1,2,ldots,n$, $x in S_i$ and $y in S_i$ implies $tx + (1-t)y in S_i$ by convexity of $S_i$.
Hence $tx + (1-t)y in cap_{i=1}^nS_i$.
Therefore $cap_{i=1}^nS_i$ is convex.
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
$endgroup$
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Let $(S_i)$ be a convex set for $i = 1,2,ldots,n$.
For any $x,y in cap_{i=1}^n S_i$, $t in [0, 1]$, we have:
For $i = 1,2,ldots,n$, $x in S_i$ and $y in S_i$ implies $tx + (1-t)y in S_i$ by convexity of $S_i$.
Hence $tx + (1-t)y in cap_{i=1}^nS_i$.
Therefore $cap_{i=1}^nS_i$ is convex.
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
$endgroup$
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
add a comment |
$begingroup$
Let $(S_i)$ be a convex set for $i = 1,2,ldots,n$.
For any $x,y in cap_{i=1}^n S_i$, $t in [0, 1]$, we have:
For $i = 1,2,ldots,n$, $x in S_i$ and $y in S_i$ implies $tx + (1-t)y in S_i$ by convexity of $S_i$.
Hence $tx + (1-t)y in cap_{i=1}^nS_i$.
Therefore $cap_{i=1}^nS_i$ is convex.
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
$endgroup$
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
add a comment |
$begingroup$
Let $(S_i)$ be a convex set for $i = 1,2,ldots,n$.
For any $x,y in cap_{i=1}^n S_i$, $t in [0, 1]$, we have:
For $i = 1,2,ldots,n$, $x in S_i$ and $y in S_i$ implies $tx + (1-t)y in S_i$ by convexity of $S_i$.
Hence $tx + (1-t)y in cap_{i=1}^nS_i$.
Therefore $cap_{i=1}^nS_i$ is convex.
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
$endgroup$
Let $(S_i)$ be a convex set for $i = 1,2,ldots,n$.
For any $x,y in cap_{i=1}^n S_i$, $t in [0, 1]$, we have:
For $i = 1,2,ldots,n$, $x in S_i$ and $y in S_i$ implies $tx + (1-t)y in S_i$ by convexity of $S_i$.
Hence $tx + (1-t)y in cap_{i=1}^nS_i$.
Therefore $cap_{i=1}^nS_i$ is convex.
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
answered Nov 11 '14 at 5:35
EmpiricistEmpiricist
6,79011433
6,79011433
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
add a comment |
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
By S_i in the 3rd line, do you mean some S_i out of the S_i's from 1 to n?
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 6:02
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
$begingroup$
@AggressiveSneeze. No, its valid for all $S_i$
$endgroup$
– Aram
Nov 11 '14 at 6:34
add a comment |
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$begingroup$
Well it makes sense, because it's a going to be a subset of one or more of the convex sets,making it convex..I just don't know how to provide a rigorous proof.
$endgroup$
– Aggressive Sneeze.
Nov 11 '14 at 5:33