A property of normed vector spaces equipped with a preorder: inequality between positive vectors implies...
Let $V$ be a normed vector space equipped with a preorder $preceq$. Is there a name for the following property:
$$
forall s, t in S. mathbf{0} preceq s preceq t implies |s| leq |t|tag{*}
$$
For example, if $V = mathbb{R}^n$ for some $n in {1,2,3,dots}$ with the Euclidean norm and with $(v_1,dots,v_n) preceq (w_1,dots,w_n)$ iff $v_i leq w_i$ for every $i in {1,dots,n}$, then $V$ has $(*)$.
terminology normed-spaces order-theory
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
add a comment |
Let $V$ be a normed vector space equipped with a preorder $preceq$. Is there a name for the following property:
$$
forall s, t in S. mathbf{0} preceq s preceq t implies |s| leq |t|tag{*}
$$
For example, if $V = mathbb{R}^n$ for some $n in {1,2,3,dots}$ with the Euclidean norm and with $(v_1,dots,v_n) preceq (w_1,dots,w_n)$ iff $v_i leq w_i$ for every $i in {1,dots,n}$, then $V$ has $(*)$.
terminology normed-spaces order-theory
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
1
I don't know of any source that might justify this answer (hence I am leaving it as a comment), but I would think that the phrase "the preorder is compatible with the norm" would be a way of describing this relation.
– Xander Henderson
Nov 28 '18 at 19:17
add a comment |
Let $V$ be a normed vector space equipped with a preorder $preceq$. Is there a name for the following property:
$$
forall s, t in S. mathbf{0} preceq s preceq t implies |s| leq |t|tag{*}
$$
For example, if $V = mathbb{R}^n$ for some $n in {1,2,3,dots}$ with the Euclidean norm and with $(v_1,dots,v_n) preceq (w_1,dots,w_n)$ iff $v_i leq w_i$ for every $i in {1,dots,n}$, then $V$ has $(*)$.
terminology normed-spaces order-theory
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
Let $V$ be a normed vector space equipped with a preorder $preceq$. Is there a name for the following property:
$$
forall s, t in S. mathbf{0} preceq s preceq t implies |s| leq |t|tag{*}
$$
For example, if $V = mathbb{R}^n$ for some $n in {1,2,3,dots}$ with the Euclidean norm and with $(v_1,dots,v_n) preceq (w_1,dots,w_n)$ iff $v_i leq w_i$ for every $i in {1,dots,n}$, then $V$ has $(*)$.
terminology normed-spaces order-theory
terminology normed-spaces order-theory
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
edited Nov 28 '18 at 11:25
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
asked Nov 28 '18 at 9:20
Evan Aad
5,55411852
5,55411852
1
I don't know of any source that might justify this answer (hence I am leaving it as a comment), but I would think that the phrase "the preorder is compatible with the norm" would be a way of describing this relation.
– Xander Henderson
Nov 28 '18 at 19:17
add a comment |
1
I don't know of any source that might justify this answer (hence I am leaving it as a comment), but I would think that the phrase "the preorder is compatible with the norm" would be a way of describing this relation.
– Xander Henderson
Nov 28 '18 at 19:17
1
1
I don't know of any source that might justify this answer (hence I am leaving it as a comment), but I would think that the phrase "the preorder is compatible with the norm" would be a way of describing this relation.
– Xander Henderson
Nov 28 '18 at 19:17
I don't know of any source that might justify this answer (hence I am leaving it as a comment), but I would think that the phrase "the preorder is compatible with the norm" would be a way of describing this relation.
– Xander Henderson
Nov 28 '18 at 19:17
add a comment |
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1
I don't know of any source that might justify this answer (hence I am leaving it as a comment), but I would think that the phrase "the preorder is compatible with the norm" would be a way of describing this relation.
– Xander Henderson
Nov 28 '18 at 19:17