nyuuzyou/wikis huggingface dataset:
Weak-* Limit in Normed Dual Space is Unique

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$. 
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space for $\struct {X, \norm \cdot}$.
Let $f, g \in X^\ast$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ such that: 
:$f_n \weakstarconv f$
and:
:$f_n \weakstarconv g$
where $\weakstarconv$ denotes weak-$\ast$ convergence.
Then: 
:$f = g$
== Proof ==
By the definition of weak-$\ast$ convergence, we have: 
:$\map {f_n} x \to \map f x$ for each $x \in X$
and:
:$\map {f_n} x \to \map g x$ for each $x \in X$.
Then, from Convergent Complex Sequence has Unique Limit we have: 
:$\map f x = \map g x$ for each $x \in X$.
That is: 
:$f = g$
== Sources ==
* $27.4$: Weak-$\ast$ Convergence
Category:Weak-* Convergence (Normed Vector Spaces)
Weak-* Metrizability of Closed Unit Ball in Normed Dual of Normed Vector Space implies Original Space is Separable

== Theorem ==
Let $X$ be a normed vector space such that: 
:$\struct {B_{X^\ast}^-, w^\ast}$ is metrizable
where $X^\ast$ is the normed dual of $X$ and $B_{X^\ast}^-$ is the closed unit ball of $X^\ast$. 
Then $X$ is separable. 
== Proof ==
Let: 
:$K = \struct {B_{X^\ast}^-, w^\ast}$
From the Banach-Alaoglu Theorem, $K$ is compact.
From Weak-* Topology is Hausdorff, $K$ is Hausdorff.
Since $K$ is metrizable, $\map \CC K$ is separable from Space of Complex-Valued Continuous Functions on Compact Hausdorff Space is Separable iff Space is Metrizable.
Define $T : X \to \map \CC K$ by: 
:$T x = x^\wedge \restriction_{B_{X^\ast}^-}$
Clearly $T$ is a linear transformation.
Further, we have for each $f \in X^\ast$ with $\norm f_{X^\ast} = 1$:
:$\ds \cmod {\map {\paren {T x} } f} = \cmod {\map f x} \le \norm x$
from Fundamental Property of Norm on Bounded Linear Functional.
From Existence of Support Functional, for each $x \in X$ there exists $f \in X^\ast$ with $\norm f_{X^\ast} = 1$ and $\map f x = \norm x$.
Hence we have:
:$\norm {T x}_{\map \CC K} = \norm x$ 
So $T$ is a linear isometry, hence so is $T^{-1} : T \sqbrk X \to X$.
Since $\map \CC K$ is separable, $T \sqbrk X$ is separable.
From Separability of Normed Vector Space preserved under Isometric Isomorphism, $X$ is separable.
== See also ==
* Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable shows that the converse of this result is also true.
== Sources ==
* : Lemma $3.23$
Category:Weak-* Topologies
Category:Separable Spaces
Category:Metrizable Topologies
Weak-* Topology is Hausdorff

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$. 
Let $X^\ast$ be the topological dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Then $\struct {X^\ast, w^\ast}$ is Hausdorff space. 
== Proof ==
From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $w^\ast$ can be induced by: 
:$\PP = \set {p_{x^\wedge} : x \in X}$
where we define $p_{x^\wedge} : X^\ast \to \R_{\ge 0}$ by: 
:$\map {p_{x^\wedge} } f = \cmod {\map {x^\wedge} f} = \cmod {\map f x}$
From Locally Convex Space is Hausdorff iff induces Hausdorff Topology, it suffices to show that $\struct {X^\ast, w^\ast}$ is Hausdorff as a locally convex space.
That is, that $\PP$ is separating.
Let $f \in X^\ast$ be such that $f \ne 0$. 
Then there exists $x \in X$ such that $\map f x \ne 0$. 
Then $\map {p_{x^\wedge} } f = \map {x^\wedge} f \ne 0$. 
So $\PP$ is indeed separating.
Category:Weak-* Topologies
Category:Hausdorff Spaces
Weak Convergence in Hilbert Space

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$. 
Let $x \in X$.
Then:
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
:
:$\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.
=== Corollary ===
== Proof ==
Let $\struct {\HH^\ast, \norm \cdot_{\HH^\ast} }$ be the normed dual space of $\HH$.
=== Necessary Condition ===
Suppose that:
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
For each $y \in \HH$ define the function $f_y : \HH \to \R$ by: 
:$\map {f_y} x = \innerprod x y$
for each $x \in \HH$.
From Inner Product with Vector is Bounded Linear Functional, it follows that $f_y$ is a bounded linear functional.
So, by the definition of the normed dual:
:$f_y \in \HH^\ast$
By hypothesis we have: 
:$\map {f_y} {x_n} \to \map {f_y} x$
That is: 
:$\innerprod {x_n} y \to \innerprod x y$
=== Sufficient Condition ===
Suppose that:
:$\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.
Let: 
:$f \in \HH^\ast$
Then from the Riesz Representation Theorem (Hilbert Spaces), there exists $z \in \HH$ such that: 
:$\map f x = \innerprod x z$
for each $x \in \HH$.
By hypothesis, we have: 
:$\innerprod {x_n} z \to \innerprod x z$
so:
:$\map f {x_n} \to \map f x$ 
Since $f \in \HH^\ast$ was arbitrary, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$.
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Category:Hilbert Spaces
Category:Weak Convergence in Hilbert Space
Weak Convergence in Hilbert Space/Corollary

== Corollary ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$. 
Let $x \in X$.
Then:
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
:
:$\innerprod y {x_n} \to \innerprod y x$ for each $y \in \HH$.
== Proof ==
From Weak Convergence in Hilbert Space, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
:
:$\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.
From Convergence of Complex Conjugate of Convergent Complex Sequence, we have: 
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
:
:$\overline {\innerprod {x_n} y} \to \overline {\innerprod x y}$ for each $y \in \HH$
From conjugate symmetry of the inner product, we have: 
:$\overline {\innerprod {x_n} y} \to \overline {\innerprod x y}$ for each $y \in \HH$
:
:$\innerprod y {x_n} \to \innerprod y x$ for each $y \in \HH$.
Hence the result.
Category:Weak Convergence in Hilbert Space
Weak Convergence in Normed Dual Space of Reflexive Normed Vector Space is Equivalent to Weak-* Convergence

== Theorem ==
 
Let $\mathbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a reflexive normed vector space over $\mathbb F$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $f \in X^\ast$. 
Then:
:$\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$ $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $27.4$: Weak-$\ast$ Convergence
Category:Reflexive Spaces
Category:Weak Convergence (Normed Vector Spaces)
Category:Weak-* Convergence (Normed Vector Spaces)
Weak Convergence in Normed Dual Space of Reflexive Normed Vector Space is Equivalent to Weak-* Convergence/Proof 1

== Theorem ==
== Proof ==
=== Sufficient Condition ===
Suppose that $\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$.
From Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent, we have: 
:$\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.
=== Necessary Condition ===
Suppose that $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.
Then: 
:$\map {f_n} x \to \map f x$
for each $x \in X$.
We aim to show that: 
:$\map F {f_n} \to \map F f$ for all $F \in X^{\ast \ast}$.
Let $F \in X^{\ast \ast}$.
Since $X$ is reflexive, we have: 
:for each $F \in X^{\ast \ast}$ there exists $x_F \in X$ such that $F = x_F^\wedge$
where $x_F^\wedge : X^\ast \to \mathbb F$ is defined by: 
:$\map {x_F^\wedge} f = \map f x$
for each $f \in X^\ast$.
Then, we have: 
:$\map {f_n} {x_F} \to \map f {x_F}$ 
That is: 
:$\map {x_F^\wedge} {f_n} \to \map {x_F^\wedge} f$
Since $F = x_F^\wedge$, we therefore have: 
:$\map F {f_n} \to \map F f$
Since $F \in X^{\ast \ast}$ was arbitrary, we have: 
:$\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$ in $X^{\ast \ast}$.
Weak Countable Compactness is not Preserved under Continuous Maps

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ be a topological space which is weakly countably compact.
Let $T_B = \struct {S_B, \tau_B}$ be another topological space.
Let $\phi: T_A \to T_B$ be a continuous mapping.
Then $T_B$ is not necessarily weakly countably compact.
== Proof ==
Let $\Z_{>0}$ be the strictly positive integers:
:$\Z_{>0} = \set {1, 2, 3, \ldots}$
Let $T_A = \struct {\Z_{>0}, \tau_A}$ be the odd-even topology.
Let $T_B = \struct {\Z_{>0}, \tau_B}$ be the discrete topology on $\Z_{>0}$.
Let $\phi: T_A \to T_B$ be the mapping:
:$\map \phi {2 k} = k, \map \phi {2 k - 1} = k$
Then:
:$\map {\phi^{-1} } k = \set {2 k, 2 k - 1} \in \tau_A$
demonstrating that $\phi$ is continuous.
Now we have that the Odd-Even Topology is Weakly Countably Compact.
But we also have that a Countable Discrete Space is not Weakly Countably Compact.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $6$. Odd-Even Topology: $4$
Category:Weakly Countably Compact Spaces
Weak Existence of Matrix Logarithm

== Theorem ==
Let $T$ be a square matrix of order $n$.
Let $\norm {T - I} < 1$ in the norm on bounded linear operators, where $I$ the identity matrix.
Then there is a square matrix $S$ such that:
:$e^S = T$
where $e^S$ is the matrix exponential.
== Proof ==
Define:
:$\ds S = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n$
$S$ converges since $\norm {T - I} < 1$.
We have that $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \norm {T - I}^n$ is the Newton-Mercator Series.
This converges since $\norm {T - I} < 1$.
Hence the series for $S$ converges absolutely, and so $S$ is well defined.
Using the series definition for the matrix exponential: 
 
If $c_i = 0$ for $i \ge 2$, then $e^S = T$, and the result is shown.
The Newton-Mercator Series is a Taylor expansion for $\map \ln {1 + x}$.
When combined with the Power Series Expansion for Exponential Function, it gives: 
But $e^{\map \ln {1 + x} } = 1 + x$.
Thus:
:$1 + x = 1 + x + c_2 x^2 + c_3 x^3 + \cdots \implies c_i = 0$
for $i \ge 2$.
== Also see ==
* Existence of Matrix Logarithm
== Sources ==
* Ch. $5$ $\S 3$, Problem $14$.
Category:Matrix Algebra
Category:Matrix Logarithms
Weak Inequality of Integers iff Strict Inequality with Integer plus One

== Theorem ==
Let $a, b \in \Z$ be integers.
:$(1): \quad a \le b$
:$(2): \quad a < b + 1$
where:
:$\le$ is the ordering on the integers
:$<$ is the strict ordering on the integers.
== Proof ==
Category:Orderings on Integers
Weak Limit in Normed Vector Space is Unique

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x, y \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that: 
:$x_n \weakconv x$
and:
:$x_n \weakconv y$
where $\weakconv$ denotes weak convergence.
Then: 
:$x = y$
== Proof ==
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Since: 
:$x_n \weakconv x$
we have: 
:$\map f {x_n} \to \map f x$ for each $f \in X^\ast$.
Since: 
:$x_n \weakconv x$
we have: 
:$\map f {x_n} \to \map f y$ for each $f \in X^\ast$.
From Convergent Complex Sequence has Unique Limit, we have: 
:$\map f x = \map f y$ for each $f \in X^\ast$. 
From Normed Dual Space Separates Points, we therefore have: 
:$x = y$
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Weak Local Compactness is Preserved under Open Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous mapping which is also an open mapping and a surjection.
If $T_A$ is weakly locally compact, then $T_B$ is also weakly locally compact.
== Proof ==
Let $\phi$ be a mapping which is surjective, continuous and open.
Let $T_A$ be weakly locally compact.
Take $b \in S_B$.
Let $V$ be a neighbourhood of $b$.
Since $\phi$ is surjective:
:$\forall y \in S_B: \exists x \in S_A: x \in \map {\phi^{-1} } y$
From the weak local compactness of $T_A$ and the continuity of $\phi$, there exists a compact neighbourhood $K$ of $x$ such that $\phi \sqbrk K \subseteq V$.
Since $K$ is a neighbourhood of $x$, then $x \in K^\circ$ and $y \in \phi \sqbrk {K^\circ} \subseteq \phi \sqbrk K$, where $K^\circ$ is the interior of $K$.
$\phi$ is an open mapping and $K^\circ$ is an open set, so $\phi \sqbrk {K^\circ}$ is also open.
Finally we get that $y \in \phi \sqbrk K \subseteq V$, where $\phi \sqbrk K$ is a compact neighbourhood.
Thus $T_B$ is weakly locally compact.
== Also see ==
* Local Compactness is Preserved under Open Continuous Surjection
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Weakly Locally Compact Spaces
Category:Open Mappings
Category:Continuous Mappings
Weak Lower Closure in Restricted Ordering

== Theorem ==
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\preccurlyeq T} = T \cap t^{\preccurlyeq S}$
where:
:$t^{\preccurlyeq T}$ is the weak lower closure of $t$ in $\left({T, \preccurlyeq \restriction_T}\right)$
:$t^{\preccurlyeq S}$ is the weak lower closure of $t$ in $\left({S, \preccurlyeq}\right)$.
== Proof ==
Let $t \in T$, and suppose that $t' \in t^{\preccurlyeq T}$.
By definition of weak lower closure $t^{\preccurlyeq T}$, this is equivalent to:
:$t' \preccurlyeq \restriction_T t$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
:$t' \preccurlyeq t \land t' \in T$
as it is assumed that $t \in T$.
The first conjunct precisely expresses that $t' \in t^{\preccurlyeq S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\preccurlyeq S}$
 $t' \in T$ and $t' \in t^{\preccurlyeq S}$.
Thus, it follows that the following are equivalent:
:$t' \in t^{\preccurlyeq T}$
:$t' \in T \cap t^{\preccurlyeq S}$
and hence the result follows, by definition of set equality.
Category:Lower Closures
Weak Nullstellensatz

== Theorem ==
Let $K$ be an algebraically closed field.
Let $n \ge 0$ be an natural number.
Let $K \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Let $I \subseteq K \sqbrk {x_1,\ldots, x_n}$ be an ideal.
# $I$ is the unit ideal: $I = (1)$.
# Its zero-locus is empty set: $\map V I = \O$.
== Proof ==
Category:Algebraic Geometry
Weak Solution to Dx u + 3yu = 0 with Heaviside Step Function Boundary Condition

== Theorem ==
Consider the boundary value problem:
:$\begin{cases}
\dfrac {\partial u} {\partial x} + 3 y u = 0 & : x \in \R_{>0},~ y \in \R \\
& \\
\map u {0, y} = \map H y & : y \in \R \\
\end{cases}$
Then it has a weak solution of the form:
:$u = e^{-3 y x} \map H y$
== Proof ==
Let $u = e^{-3 y x} \map H y$
We have that:
:Heaviside Step Function is Locally Integrable
:Locally Integrable Function defines Distribution
:Multiplication of Distribution induced by Locally Integrable Function by Smooth Function
Hence, we can define a distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$.
Then in the distributional sense we have that:
That is:
Let $\phi \in \map \DD {\R^2}$ be a test function.
Then:
Hence, in the distributional sense we have that:
:$\dfrac \partial {\partial x} \map H y = \mathbf 0$
where $\mathbf 0$ is the zero distribution.
Therefore:
Moreover:
== Sources ==
* : Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions
Category:Examples of Weak Solutions
Weak Solution to Dx u = Heaviside Step Function

== Theorem ==
Let $H: \R \to \closedint 0 1$ be the Heaviside step function.
Let $u : \R \to \R$ be such that:
:$\map u x = \begin{cases}
c & : x < 0 \\
x + c & : x > 0 
\end{cases}$ 
where $c \in \R$.
Let $T_u$ be the distribution associated with $u$.
Then $u$ is a weak solution of:
:$u' = H$
That is, in the distributional sense it holds that:
:$\dfrac \d {\d x} T_u = T_H$
== Proof ==
$u$ is continuous on $\R$ and continously differentiable on $\R \setminus \set 0$.
For $x < 0$ we have $\map {u'} x = 0$.
For $x > 0$ we have $\map {u'} x = 1$.
That is:
:$\map {u'} x = \map H x$
Furthermore:
:$\ds \lim_{x \mathop \to 0^-} = 0$
:$\ds \lim_{x \mathop \to 0^+} = 1$
By the jump rule:
== Proof of no classical solutions ==
 $u$ is a differentiable function such that $u' = H$, then
:$\ds \lim_{x \mathop \to 0-} \map {u'} x = 0$
:$\ds \lim_{x \mathop \to 0+} \map {u'} x = 1$
contradicts the Intermediate Value Theorem for Derivatives.
== Sources ==
* : Chapter $\S 6.3$: A glimpse of distribution theory. Weak solutions
Category:Examples of Weak Solutions
Weak Solution to Dx u = u

== Theorem ==
Let $H$ be the Heaviside step function.
Let $\map u {x, t} = \map H t e^x$
Then $u$ is a weak solution of the partial differential equation $\ds \dfrac {\partial u} {\partial x} = u$.
That is, for the distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$ in the distributional sense it holds that:
:$\ds \dfrac {\partial T_u} {\partial x} = T_u$ 
== Proof ==
Let $\phi \in \map \DD {\R^2}$ be a test function.
Then in the distributional sense we have that:
== Sources ==
* : Chapter $\S 6.3$: A glimpse of distribution theory. Weak solutions
Category:Examples of Distributional Solutions
Category:Examples of Weak Solutions
Weak Topology on Infinite Dimensional Normed Vector Space is not Metrizable

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite dimensional normed vector space over $\GF$.
Let $w$ be the weak topology on $X$.
Then $w$ is not metrizable. 
== Proof ==
Let $X^\ast$ be the normed dual space of $X$. 
From Metric Space is First-Countable, it suffices to show that $\struct {X, w}$ is not first-countable.
 that $\struct {X, w}$ is first-countable.
Let $\sequence {U_n}_{n \mathop \in \N}$ be a local basis for $\mathbf 0_X$.
From Open Sets in Weak Topology of Topological Vector Space, for each $n \in \N$ we can pick $\epsilon_n > 0$ and a finite set $F_n \subseteq X$ such that: 
:$V_n = \set {x \in X : \cmod {\map f x} < \epsilon_n \text { for all } f \in F_n} \subseteq U_n$
Then $\sequence {V_n}_{n \mathop \in \N}$ is still a local basis for $\mathbf 0_X$.
Let $g \in X^\ast$ and consider: 
:$U = \set {x \in X : \cmod {\map g x} < 1}$ 
Then there exists $\map n g$ such that: 
:$U_{\map n g} \subseteq U$ 
Set $n = \map n g$. 
Let: 
:$\ds x \in \bigcap_{f \mathop \in F_n} \ker f$
Then $\map f x = 0$ for each $f \in F_n$. 
Then for each $\lambda \in \GF$ and $f \in F_n$ we have $\map f {\lambda x} = 0$.
So we have $\lambda x \in U$ for each $\lambda \in \GF$.
That is, $\cmod {\map g {\lambda x} } < 1$ for each $\lambda \in \GF$.
So for each $\lambda > 0$ we have: 
:$\ds \cmod {\map g x} < \frac 1 \lambda$ 
So we have $\map g x = 0$.
We conclude that: 
:$\ds \bigcap_{f \mathop \in F_n} \ker f \subseteq \ker g$
From Condition for Linear Dependence of Linear Functionals in terms of Kernel, we then have: 
:$g \in \map \span {F_n} = \map \span {F_{\map n g} }$
Since $g \in X^\ast$ was arbitrary, we have that: 
:$\ds X^\ast = \bigcup_{n \mathop = 1}^\infty \map \span {F_n}$
Then: 
:$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is a generator for $X$. 
We have that $X$ has infinite dimension.
Hence, from Normed Dual Space of Infinite-Dimensional Normed Vector Space is Infinite-Dimensional, $X^\ast$ also has infinite dimension. 
Hence: 
:$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is infinite.
From Countable Union of Finite Sets is Countable, it follows that: 
:$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is countable.
From Generator of Vector Space Contains Basis, it follows that $X^\ast$ either has countable or finite dimension.
We have already established that it does not have finite dimension, hence it must have countable dimension.
From Normed Dual Space is Banach Space, $X^\ast$ is a Banach space of countable dimension.
This contradicts Infinite-Dimensional Banach Space has Uncountable Dimension. 
So $\struct {X, w}$ is not first-countable and hence is not metrizable.
Category:Weak Topologies on Topological Vector Spaces
Category:Metrizable Topologies
Weak Topology on Topological Vector Space over Hausdorff Topological Field is Hausdorff

== Theorem ==
Let $K$ be a Hausdorff topological field. 
Let $X$ be a topological vector space over $K$ with weak topology $w$.
Then $\struct {X, w}$ is Hausdorff.
== Proof ==
From the definition of the weak topology, if $x, y \in X$ have $x \ne y$, then $x - y \ne \mathbf 0_X$: 
:there exists $f \in X^\ast$ such that $\map f {x - y} \ne 0$. 
Since $f$ is linear, we then have $\map f x \ne \map f y$.
Since $K$ is Hausdorff, we therefore obtain that $\struct {X, w}$ is Hausdorff from Initial Topology with respect to Point-Separating Family of Mappings onto Hausdorff Spaces is Hausdorff.
Category:Weak Topologies on Topological Vector Spaces
Category:Hausdorff Spaces
Weak Upper Closure in Restricted Ordering

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\succcurlyeq T} = T \cap t^{\succcurlyeq S}$
where:
:$t^{\succcurlyeq T}$ is the weak upper closure of $t$ in $\struct {T, \preccurlyeq \restriction_T}$
:$t^{\succcurlyeq S}$ is the weak upper closure of $t$ in $\struct {S, \preccurlyeq}$.
== Proof ==
Let $t \in T$.
Suppose that:
:$t' \in t^{\succcurlyeq T}$
By definition of weak upper closure $t^{\succcurlyeq T}$, this is equivalent to:
:$t \preccurlyeq \restriction_T t'$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
:$t \preccurlyeq t' \land t' \in T$
as it is assumed that $t \in T$.
The first conjunct precisely expresses that $t' \in t^{\succcurlyeq S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\succcurlyeq S}$
 $t' \in T$ and $t' \in t^{\succcurlyeq S}$.
Thus it follows that the following are equivalent:
:$t' \in t^{\succcurlyeq T}$
:$t' \in T \cap t^{\succcurlyeq S}$
and hence the result follows, by definition of set equality.
Category:Upper Closures
Weak Whitney Immersion Theorem

== Theorem ==
Every $k$-dimensional manifold $X$ admits a one-to-one immersion in $\R^{2 k + 1}$.
== Proof ==
Let $M > 2 k + 1$ be a natural number such that $f: X \to \R^M$ is an injective immersion.
Define a map $h: X \times X \times \R \to \R^M$ by:
:$\map h {x, y, t} = \map t {\map f x - \map f y}$
Define a map $g: \map T X \to \R^M$ by:
:$\map g {x, v} = \map {\d f_x} v$
where $\map T X$ is the tangent bundle of $X$.
Since $M > 2 k + 1$, the Morse-Sard Theorem implies $\exists a \in \R^M$ such that $a$ is in the image of neither function.
Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$. 
Suppose:
:$\map {\paren {\pi \circ f} } x = \map {\paren {\pi \circ f} } y$
for some $x, y$.
Then:
:$\map f x - \map f y = t a$
for some scalar $t$.
 $x \ne y$.
Then as $f$ is injective:
:$t \ne 0$
But then $\map h {x, y, 1/t} = a$, contradicting the choice of $a$.
By Proof by Contradiction:
:$x = y$
and so $\pi \circ f$ is injective.
Let $v$ be a nonzero vector in $\map {T_x} X$ (the tangent space of $X$ at $x$) for which:
:$\map {\d \paren {\pi \circ f}_x} v = 0$
Since $\pi$ is linear:
:$\map {\d \paren {\pi \circ f}_x} = \pi \circ \d f_x$
Thus:
:$\map {\pi \circ \d f_x} v = 0$
so:
:$\map {\d f_x} v = t a$
for some scalar $t$.
Because $f$ is an immersion, $t \ne 0$.
Hence:
:$\map g {x, 1/t} = a$
contradicting the choice of $a$.
Hence $\pi \circ f: X \to H$ is an immersion. 
$H$ is obviously isomorphic to $\R^{M-1}$.
Thus whenever $M > 2 k + 1$ and $X$ admits of a one-to-one immersion in $\R^M$, it follows that $X$ also admits of a one-to-one immersion in $\R^{M-1}$.
== Also see ==
* Whitney Immersion Theorem, a strictly stronger result: every $k$-manifold admits of a one-to-one immersion in $\R^{2 k}$, and an immersion (not necessarily one-to-one) in $\R^{2 k - 1}$.
Category:Manifolds