Post History
#4: Post edited
by
Incnis Mrsi
·
2021-03-30T18:08:26Z (over 3 years ago)
grammar tweaks
First of all, kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.- Secondly, your $\{\omega_i\}$ obviously are linearly independent. Although if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$ (as explained above), your $f$ is anyway a monomorphism and you *can* pull the topology back to $V$ this way.
Thirdly, **the topology is basis-dependent** unless dimension is finite. It follows from the fact that a hyperplane $\ker\phi$ (where $\phi\in V^\ast$) is closed* in $V$ if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again. Note that Ī̲ require {0} to be closed in $K$ to disqualify trivial topology.- ----
- * With my intuition it would be more convenient to check half-spaces: whether is $\phi > 0$ open and $\phi \le 0$ is closed. But $K$ isn’t an ordered field, alas.
- First of all, the kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.
- Secondly, your $\{\omega_i\}$ obviously are linearly independent. Although if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$ (as explained above), your $f$ is anyway a monomorphism and you *can* pull the topology back to $V$ this way.
- Thirdly, **the topology is basis-dependent** unless the dimension is finite. It follows from the fact that a hyperplane $\ker\phi$ (where $\phi\in V^\ast$) is closed* in $V$ if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again. Note that Ī̲ require {0} to be closed in $K$ to disqualify trivial topology.
- ----
- * With my intuition it would be more convenient to check half-spaces: whether is $\phi > 0$ open and $\phi \le 0$ is closed. But $K$ isn’t an ordered field, alas.
#3: Post edited
by
Incnis Mrsi
·
2021-03-30T04:09:48Z (over 3 years ago)
{0} must be closed in $K$
- First of all, kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.
- Secondly, your $\{\omega_i\}$ obviously are linearly independent. Although if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$ (as explained above), your $f$ is anyway a monomorphism and you *can* pull the topology back to $V$ this way.
Thirdly, **the topology is basis-dependent** unless dimension is finite. It follows from the fact that a hyperplane $\ker\phi$ (where $\phi\in V^\ast$) is closed* in $V$ if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again.- ----
* With my intuition it would be more convenient to ask whether is $\phi > 0$ open (or, equivalently, $\phi \le 0$ closed), but $K$ isn’t an ordered field, alas.
- First of all, kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.
- Secondly, your $\{\omega_i\}$ obviously are linearly independent. Although if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$ (as explained above), your $f$ is anyway a monomorphism and you *can* pull the topology back to $V$ this way.
- Thirdly, **the topology is basis-dependent** unless dimension is finite. It follows from the fact that a hyperplane $\ker\phi$ (where $\phi\in V^\ast$) is closed* in $V$ if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again. Note that Ī̲ require {0} to be closed in $K$ to disqualify trivial topology.
- ----
- * With my intuition it would be more convenient to check half-spaces: whether is $\phi > 0$ open and $\phi \le 0$ is closed. But $K$ isn’t an ordered field, alas.
#2: Post edited
by
Incnis Mrsi
·
2021-03-29T20:21:46Z (over 3 years ago)
$K$ isn’t an ordered field!
- First of all, kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.
Secondly, your $\{\omega_i\}$ obviously are linearly independent, but if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$; see comment in the first paragraph. Anyway your $f$ is a monomorphism and you *can* pull the topology back to $V$ this way.Thirdly, **the topology is basis-dependent** unless dimension is finite. It follows from the fact that a half-space $\phi > 0$ of $V$ (where $\phi\in V^\ast$) is open if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again.
- First of all, kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.
- Secondly, your $\{\omega_i\}$ obviously are linearly independent. Although if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$ (as explained above), your $f$ is anyway a monomorphism and you *can* pull the topology back to $V$ this way.
- Thirdly, **the topology is basis-dependent** unless dimension is finite. It follows from the fact that a hyperplane $\ker\phi$ (where $\phi\in V^\ast$) is closed* in $V$ if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again.
- ----
- * With my intuition it would be more convenient to ask whether is $\phi > 0$ open (or, equivalently, $\phi \le 0$ closed), but $K$ isn’t an ordered field, alas.
#1: Initial revision
by
Incnis Mrsi
·
2021-03-29T17:48:25Z (over 3 years ago)
First of all, kernel of any $\phi\in V^\ast$ has codimension at most one; more precisely, $\ker\phi = V \Leftrightarrow \phi = 0$ and $\operatorname{codim}\ker\phi = 1$ if equalities are false. Hence $V^{[\ast]} = V^\ast$. Ī̲ suspect that by $V^{[\ast]}$ you really assume a basis-dependent thing, namely finite linear combinations of $\{\omega_i\}$ mentioned below. Such $\psi\in V^\ast$ that $\forall i\in I: \psi(b_i) = 1$ cannot be expressed in omegas unless $I$ is finite, but (looking at algebra alone) $\psi$ is not essentially different from any non-zero covector.
Secondly, your $\{\omega_i\}$ obviously are linearly independent, but if $I$ is infinite, then $\{\omega_i:i\in I\}$ *will not* constitute a basis of $V^\ast$; see comment in the first paragraph. Anyway your $f$ is a monomorphism and you *can* pull the topology back to $V$ this way.
Thirdly, **the topology is basis-dependent** unless dimension is finite. It follows from the fact that a half-space $\phi > 0$ of $V$ (where $\phi\in V^\ast$) is open if and only if $\phi$ is generated by $\{\omega_i:i\in I\}$; see the first paragraph again.