Prove point is in closure of set
Consider the set $M={ (1,x,x^2):x in mathbb{R}}$.
We define $text{cone}(M)$ as the conic hull of $M$, which admits conic combinations of its points, i.e. $sum_i lambda_i(1,x_i,x_i^2)in M$ where $lambda_i>=0 forall i$.
We can apparently see that $(0,0,1)intext{cl(cone(}M))$, the closure of the conic hull of $M$. The proof is simple:
$$
begin{align}
(0,0,1)&= lim_{xrightarrow infty}Big(dfrac{1}{x^2},dfrac{1}{x},1Big)\
&=lim_{xrightarrow infty}dfrac{1}{x^2}(1,x,x^2)
end{align}.
$$
But I don't understand it. Can someone explain why the limit of $dfrac{1}{x^2}times(text{an element of M})$ gives an element in the closure of $text{cone}(M)$?
general-topology convex-analysis convex-optimization
asked Dec 1 '18 at 21:02
Dan
276
add a comment |
Consider the set $M={ (1,x,x^2):x in mathbb{R}}$.
We define $text{cone}(M)$ as the conic hull of $M$, which admits conic combinations of its points, i.e. $sum_i lambda_i(1,x_i,x_i^2)in M$ where $lambda_i>=0 forall i$.
We can apparently see that $(0,0,1)intext{cl(cone(}M))$, the closure of the conic hull of $M$. The proof is simple:
$$
begin{align}
(0,0,1)&= lim_{xrightarrow infty}Big(dfrac{1}{x^2},dfrac{1}{x},1Big)\
&=lim_{xrightarrow infty}dfrac{1}{x^2}(1,x,x^2)
end{align}.
$$
But I don't understand it. Can someone explain why the limit of $dfrac{1}{x^2}times(text{an element of M})$ gives an element in the closure of $text{cone}(M)$?
general-topology convex-analysis convex-optimization
asked Dec 1 '18 at 21:02
Dan
276
Well what you have is a sequence of elements of $Cone(M)$. In $mathbb R^n$ that implies that the limit will be a in the cosure for sure.
– Dog_69
Dec 1 '18 at 21:16
add a comment |
Consider the set $M={ (1,x,x^2):x in mathbb{R}}$.
We define $text{cone}(M)$ as the conic hull of $M$, which admits conic combinations of its points, i.e. $sum_i lambda_i(1,x_i,x_i^2)in M$ where $lambda_i>=0 forall i$.
We can apparently see that $(0,0,1)intext{cl(cone(}M))$, the closure of the conic hull of $M$. The proof is simple:
$$
begin{align}
(0,0,1)&= lim_{xrightarrow infty}Big(dfrac{1}{x^2},dfrac{1}{x},1Big)\
&=lim_{xrightarrow infty}dfrac{1}{x^2}(1,x,x^2)
end{align}.
$$
But I don't understand it. Can someone explain why the limit of $dfrac{1}{x^2}times(text{an element of M})$ gives an element in the closure of $text{cone}(M)$?
general-topology convex-analysis convex-optimization
asked Dec 1 '18 at 21:02
Dan
276
Consider the set $M={ (1,x,x^2):x in mathbb{R}}$.
We define $text{cone}(M)$ as the conic hull of $M$, which admits conic combinations of its points, i.e. $sum_i lambda_i(1,x_i,x_i^2)in M$ where $lambda_i>=0 forall i$.
We can apparently see that $(0,0,1)intext{cl(cone(}M))$, the closure of the conic hull of $M$. The proof is simple:
$$
begin{align}
(0,0,1)&= lim_{xrightarrow infty}Big(dfrac{1}{x^2},dfrac{1}{x},1Big)\
&=lim_{xrightarrow infty}dfrac{1}{x^2}(1,x,x^2)
end{align}.
$$
But I don't understand it. Can someone explain why the limit of $dfrac{1}{x^2}times(text{an element of M})$ gives an element in the closure of $text{cone}(M)$?
general-topology convex-analysis convex-optimization
general-topology convex-analysis convex-optimization
asked Dec 1 '18 at 21:02
Dan
276
asked Dec 1 '18 at 21:02
Dan
276
asked Dec 1 '18 at 21:02
Dan
276
asked Dec 1 '18 at 21:02
Dan
276
asked Dec 1 '18 at 21:02
Dan
276
276
Well what you have is a sequence of elements of $Cone(M)$. In $mathbb R^n$ that implies that the limit will be a in the cosure for sure.
– Dog_69
Dec 1 '18 at 21:16
add a comment |
Well what you have is a sequence of elements of $Cone(M)$. In $mathbb R^n$ that implies that the limit will be a in the cosure for sure.
– Dog_69
Dec 1 '18 at 21:16
Well what you have is a sequence of elements of $Cone(M)$. In $mathbb R^n$ that implies that the limit will be a in the cosure for sure.
– Dog_69
Dec 1 '18 at 21:16
Well what you have is a sequence of elements of $Cone(M)$. In $mathbb R^n$ that implies that the limit will be a in the cosure for sure.
– Dog_69
Dec 1 '18 at 21:16
add a comment |
1 Answer
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Let x $in$ $mathbb{R}$ and $lambda$ = $frac{1}{x^2}$. Then p = $lambda(1,x,x^2) = (frac{1}{x^2},frac{1}{x},1)$. Note that p $in$ cone(M) $forall$ x $in mathbb{R}$. Now take the limit as x goes to 0. This is a sequence of point in cone(M); thus the limit point is in cl(cone(M)).
answered Dec 1 '18 at 21:18
Joel Pereira
65119
add a comment |
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1 Answer
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1 Answer
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Let x $in$ $mathbb{R}$ and $lambda$ = $frac{1}{x^2}$. Then p = $lambda(1,x,x^2) = (frac{1}{x^2},frac{1}{x},1)$. Note that p $in$ cone(M) $forall$ x $in mathbb{R}$. Now take the limit as x goes to 0. This is a sequence of point in cone(M); thus the limit point is in cl(cone(M)).
answered Dec 1 '18 at 21:18
Joel Pereira
65119
add a comment |
Let x $in$ $mathbb{R}$ and $lambda$ = $frac{1}{x^2}$. Then p = $lambda(1,x,x^2) = (frac{1}{x^2},frac{1}{x},1)$. Note that p $in$ cone(M) $forall$ x $in mathbb{R}$. Now take the limit as x goes to 0. This is a sequence of point in cone(M); thus the limit point is in cl(cone(M)).
answered Dec 1 '18 at 21:18
Joel Pereira
65119
add a comment |
Let x $in$ $mathbb{R}$ and $lambda$ = $frac{1}{x^2}$. Then p = $lambda(1,x,x^2) = (frac{1}{x^2},frac{1}{x},1)$. Note that p $in$ cone(M) $forall$ x $in mathbb{R}$. Now take the limit as x goes to 0. This is a sequence of point in cone(M); thus the limit point is in cl(cone(M)).
answered Dec 1 '18 at 21:18
Joel Pereira
65119
Let x $in$ $mathbb{R}$ and $lambda$ = $frac{1}{x^2}$. Then p = $lambda(1,x,x^2) = (frac{1}{x^2},frac{1}{x},1)$. Note that p $in$ cone(M) $forall$ x $in mathbb{R}$. Now take the limit as x goes to 0. This is a sequence of point in cone(M); thus the limit point is in cl(cone(M)).
answered Dec 1 '18 at 21:18
Joel Pereira
65119
answered Dec 1 '18 at 21:18
Joel Pereira
65119
answered Dec 1 '18 at 21:18
Joel Pereira
65119
answered Dec 1 '18 at 21:18
Joel Pereira
65119
65119
add a comment |
add a comment |
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Well what you have is a sequence of elements of $Cone(M)$. In $mathbb R^n$ that implies that the limit will be a in the cosure for sure.
– Dog_69
Dec 1 '18 at 21:16