Proof of the Hyperplane Separation Theorem presented on Wikipedia
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For the proof of the hyperplane separation theorem stated at https://en.wikipedia.org/wiki/Hyperplane_separation_theorem, I'm having trouble understanding the last line. Specifically, when they attempt to cover the case where $v=0$, how do we know that the interior of a convex set can be exhausted by compact convex subsets (like spheres)? How do we know that the sequence of $v_n$ has a convergent subsequence, since I thought that only holds if the sequence is a subset of a compact set, but here we have a sequence of compact sets? I assume they state that the limit is nonzero because $0notin K_n$ for each $K_n$ but I'm not sure. Any explanation on this proof would be very helpful.
real-analysis
asked Nov 28 at 2:20
Alex
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For the proof of the hyperplane separation theorem stated at https://en.wikipedia.org/wiki/Hyperplane_separation_theorem, I'm having trouble understanding the last line. Specifically, when they attempt to cover the case where $v=0$, how do we know that the interior of a convex set can be exhausted by compact convex subsets (like spheres)? How do we know that the sequence of $v_n$ has a convergent subsequence, since I thought that only holds if the sequence is a subset of a compact set, but here we have a sequence of compact sets? I assume they state that the limit is nonzero because $0notin K_n$ for each $K_n$ but I'm not sure. Any explanation on this proof would be very helpful.
real-analysis
asked Nov 28 at 2:20
Alex
1
At the place in the proof where they take a convergent subsequence of $(v_n)$, they've already normalized these vectors to lie on the unit sphere, which is a compact set. And then the limit of that convergent subsequence is also on the unit sphere, hence in particular non-zero.
– Andreas Blass
Nov 28 at 2:41
okay thank you so much, for some reason I wasn't thinking of the unit sphere as the compact set (I was confused). I guess then the argument is that since $(v_n)$ is in the unit sphere (which is compact), it has a subsequence which converges to some $v$ in the unit sphere, then with the continuity of the inner product with one argument fixed we can show that $<x,v>geq 0$, and then with the other argument fixed that this is true for all $x$ not necessarily in the interior of $K$. I'm not sure about the exhausted by compact convex subsets part yet, but I'll think some more. Thank you
– Alex
Nov 28 at 3:19
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up vote
0
down vote
favorite
For the proof of the hyperplane separation theorem stated at https://en.wikipedia.org/wiki/Hyperplane_separation_theorem, I'm having trouble understanding the last line. Specifically, when they attempt to cover the case where $v=0$, how do we know that the interior of a convex set can be exhausted by compact convex subsets (like spheres)? How do we know that the sequence of $v_n$ has a convergent subsequence, since I thought that only holds if the sequence is a subset of a compact set, but here we have a sequence of compact sets? I assume they state that the limit is nonzero because $0notin K_n$ for each $K_n$ but I'm not sure. Any explanation on this proof would be very helpful.
real-analysis
asked Nov 28 at 2:20
Alex
1
For the proof of the hyperplane separation theorem stated at https://en.wikipedia.org/wiki/Hyperplane_separation_theorem, I'm having trouble understanding the last line. Specifically, when they attempt to cover the case where $v=0$, how do we know that the interior of a convex set can be exhausted by compact convex subsets (like spheres)? How do we know that the sequence of $v_n$ has a convergent subsequence, since I thought that only holds if the sequence is a subset of a compact set, but here we have a sequence of compact sets? I assume they state that the limit is nonzero because $0notin K_n$ for each $K_n$ but I'm not sure. Any explanation on this proof would be very helpful.
real-analysis
real-analysis
asked Nov 28 at 2:20
Alex
1
asked Nov 28 at 2:20
Alex
1
asked Nov 28 at 2:20
Alex
1
asked Nov 28 at 2:20
Alex
1
asked Nov 28 at 2:20
Alex
1
1
At the place in the proof where they take a convergent subsequence of $(v_n)$, they've already normalized these vectors to lie on the unit sphere, which is a compact set. And then the limit of that convergent subsequence is also on the unit sphere, hence in particular non-zero.
– Andreas Blass
Nov 28 at 2:41
okay thank you so much, for some reason I wasn't thinking of the unit sphere as the compact set (I was confused). I guess then the argument is that since $(v_n)$ is in the unit sphere (which is compact), it has a subsequence which converges to some $v$ in the unit sphere, then with the continuity of the inner product with one argument fixed we can show that $<x,v>geq 0$, and then with the other argument fixed that this is true for all $x$ not necessarily in the interior of $K$. I'm not sure about the exhausted by compact convex subsets part yet, but I'll think some more. Thank you
– Alex
Nov 28 at 3:19
add a comment |
At the place in the proof where they take a convergent subsequence of $(v_n)$, they've already normalized these vectors to lie on the unit sphere, which is a compact set. And then the limit of that convergent subsequence is also on the unit sphere, hence in particular non-zero.
– Andreas Blass
Nov 28 at 2:41
okay thank you so much, for some reason I wasn't thinking of the unit sphere as the compact set (I was confused). I guess then the argument is that since $(v_n)$ is in the unit sphere (which is compact), it has a subsequence which converges to some $v$ in the unit sphere, then with the continuity of the inner product with one argument fixed we can show that $<x,v>geq 0$, and then with the other argument fixed that this is true for all $x$ not necessarily in the interior of $K$. I'm not sure about the exhausted by compact convex subsets part yet, but I'll think some more. Thank you
– Alex
Nov 28 at 3:19
At the place in the proof where they take a convergent subsequence of $(v_n)$, they've already normalized these vectors to lie on the unit sphere, which is a compact set. And then the limit of that convergent subsequence is also on the unit sphere, hence in particular non-zero.
– Andreas Blass
Nov 28 at 2:41
At the place in the proof where they take a convergent subsequence of $(v_n)$, they've already normalized these vectors to lie on the unit sphere, which is a compact set. And then the limit of that convergent subsequence is also on the unit sphere, hence in particular non-zero.
– Andreas Blass
Nov 28 at 2:41
okay thank you so much, for some reason I wasn't thinking of the unit sphere as the compact set (I was confused). I guess then the argument is that since $(v_n)$ is in the unit sphere (which is compact), it has a subsequence which converges to some $v$ in the unit sphere, then with the continuity of the inner product with one argument fixed we can show that $<x,v>geq 0$, and then with the other argument fixed that this is true for all $x$ not necessarily in the interior of $K$. I'm not sure about the exhausted by compact convex subsets part yet, but I'll think some more. Thank you
– Alex
Nov 28 at 3:19
okay thank you so much, for some reason I wasn't thinking of the unit sphere as the compact set (I was confused). I guess then the argument is that since $(v_n)$ is in the unit sphere (which is compact), it has a subsequence which converges to some $v$ in the unit sphere, then with the continuity of the inner product with one argument fixed we can show that $<x,v>geq 0$, and then with the other argument fixed that this is true for all $x$ not necessarily in the interior of $K$. I'm not sure about the exhausted by compact convex subsets part yet, but I'll think some more. Thank you
– Alex
Nov 28 at 3:19
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At the place in the proof where they take a convergent subsequence of $(v_n)$, they've already normalized these vectors to lie on the unit sphere, which is a compact set. And then the limit of that convergent subsequence is also on the unit sphere, hence in particular non-zero.
– Andreas Blass
Nov 28 at 2:41
okay thank you so much, for some reason I wasn't thinking of the unit sphere as the compact set (I was confused). I guess then the argument is that since $(v_n)$ is in the unit sphere (which is compact), it has a subsequence which converges to some $v$ in the unit sphere, then with the continuity of the inner product with one argument fixed we can show that $<x,v>geq 0$, and then with the other argument fixed that this is true for all $x$ not necessarily in the interior of $K$. I'm not sure about the exhausted by compact convex subsets part yet, but I'll think some more. Thank you
– Alex
Nov 28 at 3:19