nyuuzyou/wikis huggingface dataset:
Baire-Osgood Theorem

== Theorem ==
Let $X$ be a Baire space.
Let $Y$ be a metrizable topological space
Let $f: X \to Y$ be a mapping which is the pointwise limit of a sequence $\sequence {f_n}$ in $\map C {X, Y}$.
Let $\map D f$ be the set of points where $f$ is discontinuous.
Then $\map D f$ is a meager subset of $X$.
== Proof ==
Let $d$ be a metric on $Y$ generating its topology.
Let $\map {\omega_f} x$ denote the oscillation of $f$ at $x$.
We have:
:$\ds \map D f = \bigcup_{n \mathop = 1}^\infty \set {x \in X: \map {\omega_f} x \ge \frac 1 n}$
which is a countable union of closed sets.
Since we have this expression for $\map D f$, the claim follows if we can prove that for all $\epsilon \in \R_{>0}$ the closed set:
:$F_\epsilon = \set {x \in X: \map {\omega_f} x \ge 5 \epsilon}$
is nowhere dense.
Let $\epsilon \in \R_{>0}$ be given and consider the sets:
:$\ds A_n = \bigcap_{i, j \mathop \ge n} \set {x \in X: \map d {\map {f_i} x, \map {f_j} x} \le \epsilon}$
which are closed because $d$ and the $f_i$ are continuous.
Because $\sequence {f_n}$ is pointwise convergent, it is pointwise Cauchy with respect to any metric generating the topology on $Y$, so $\ds \bigcup_{n \mathop = 1}^\infty A_n = X$.
Given a nonempty open $U \subseteq X$ we wish to show that $U \nsubseteq F_\epsilon$.
Consider the sequence $\sequence {A_n \cap U}$ of closed subsets of $U$.
The union of these is all of $U$.
As $U$ is an open subspace of a Baire space, it is a Baire space.
So one of the elements of $\sequence {A_n \cap U}$, say $A_k$, must have an interior point, so there is an open $V \subseteq A_k \cap U$.
Because $U$ is open in $X$, $V$ is open in $X$ as well. 
We will show that:
:$V \subseteq F_\epsilon^c = \set {x \in X: \map {\omega_f} x < 5 \epsilon}$
This will show that:
:$V \nsubseteq F_\epsilon$ and thus that $U \nsubseteq F_\epsilon$
Since $V \subseteq A_k$:
:$\map d {\map {f_i} x, \map {f_j} x} \le \epsilon$
for all $x \in V$ and all $i, j \ge k$.
Pointwise convergence of $\sequence {f_n}$ gives that:
:$\map d {\map f x, \map {f_k} x} \le \epsilon$
for all $x \in V$.
By continuity of $f_k$ we have for every $x_0 \in V$ an open $V_{x_0} \subseteq V$ such that:
:$\map d {\map {f_k} x, \map {f_k} {x_0} } \le \epsilon$
for all $x \in V_{x_0}$.
By the triangle inequality:
:$\map d {\map f x, \map {f_k} {x_0} } \le 2 \epsilon$ 
for all $x \in V_{x_0}$.
Applying the triangle inequality again:
:$\map d {\map f x, \map f y} \le 4 \epsilon$
for all $x, y \in V_{x_0}$.
Thus we have the bound:
:$\map {\omega_f} {x_0} \le \map {\omega_f} {V_{x_0} } \le 4 \epsilon$
showing that $x_0 \notin F_\epsilon$ as desired.
== Sources ==
* 
Category:Topology
Category:Meager Spaces
Category:Baire Spaces
Baire Category Theorem

== Theorem ==
Let $M = \struct {A, d}$ be a complete metric space.
Then $M = \struct {A, d}$ is also a Baire space.
=== Corollary ===
== Proof ==
Let $U_n$ be a countable set of open sets of $M$ all of which are everywhere dense.
The strategy of this proof is to show that the intersection $\ds \bigcap U_n$ is everywhere dense.
Let $W \subseteq A$ be a non-empty open set of $M$.
From Open Set Characterization of Denseness, since $U_1$ is everywhere dense:
:$W \cap U_1 \ne \O$
Thus, there is a point $x_1 \in A$ and $\epsilon_1 \in \R_{>0}$ such that:
:$\map {\overline B} {x_1, \epsilon_1} \subset W \cap U_1$
where $\map B {x, \epsilon}$ and $\map {\overline B} {x, \epsilon}$ denote an open $\epsilon$-ball of $x$ and its closure, respectively.
Recall that the $U_n$ are everywhere dense.
Hence we can use in a recursive manner to find a pair of sequences $\sequence {x_n}$ and $\sequence {\epsilon_n} > 0$ such that:
:$\map {\overline B} {x_n, \epsilon_n} \subset \map B {x_{n - 1}, \epsilon_{n - 1} } \cap U_n$
as well as $\epsilon_n < 1/n $.
Since $x_n \in \map B {x_m, \epsilon_m}$ when $n > m$, we have that $x_n$ is a Cauchy Sequence.
Thus $x_n$ converges to some limit $x$ by definition of completeness.
For any $n$, by closedness:
:$x \in \map {\overline B} {x_{n + 1}, \epsilon_{n + 1} } \subset \map B {x_n, \epsilon_n}$
Thus:
:$\forall n \in \N: x \in W \cap U_n$
From Open Set Characterization of Denseness it follows that $\ds \bigcap U_n$ is (everywhere) dense.
So, by definition, $\struct {M, d}$ is a Baire space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
* : Chapter XV, $\S1$. Theorem 1.1 Baire's Theorem
* $22.1$: The Baire Category Theorem
Category:Baire Category Theorem
Category:Complete Metric Spaces
Category:Baire Spaces
Category:Examples of Use of Axiom of Dependent Choice
Baire Category Theorem/Corollary/Proof 1

== Corollary to Baire Category Theorem ==
== Proof ==
From Real Number Line is Complete Metric Space, $\R$ is a complete metric space.
From the Baire Category Theorem, a complete metric space is a Baire space.
The result follows from Baire Space is Non-Meager.
Category:Baire Category Theorem
Baire Function may not be Continuous

== Theorem ==
Let $X \subseteq \R$.
Let $f : X \to \R$ be a Baire function.
Then $f$ is not necessarily continuous.
== Proof ==
Let:
:$X = \closedint 0 1$
For each $n$, define the function $f_n : \closedint 0 1 \to \R$ to have: 
:$\map {f_n} x = x^n$
for each $x \in \closedint 0 1$.
Note that for $0 \le x < 1$, we have: 
:$x^n \to 0$
Note also that: 
:$\map {f_n} 1 = 1$
so:
:$\map {f_n} 1 \to 1$
Define $f : \closedint 0 1 \to \R$ by: 
:$\ds \map f x = \begin{cases}0 & 0 \le x < 1 \\ 1 & x = 1\end{cases}$
for each $x \in \closedint 0 1$.
We have: 
:$\sequence {f_n}$ converges to $f$ pointwise.
Clearly $f$ is not continuous at $1$, since:
:$\size {\map f x - \map f 1} = 1$
for any $x \in \hointr 0 1$.
So $f$ is a Baire function that is not continuous. 
== Also see ==
* Continuous Real Function is Baire Function: this result shows that the converse of that result is false.
* Set of Discontinuities of Baire Function is Meager: quantifies the discontinuity sets of Baire functions more precisely.
Category:Baire Functions
Baire Space iff Open Sets are Non-Meager

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then $T$ is a Baire space every non-empty open set of $T$ is non-meager in $T$.
== Proof ==
We prove the contrapositive:
:$T$ is not a Baire space there exists a non-empty open set of $T$ which is meager in $T$.
We have by definition of Baire space:
:$T$ is a Baire space the interior of the union of any countable set of closed sets of $T$ which are nowhere dense is empty.
Therefore:
:$T$ is not a Baire space the interior of the union of some countable set of closed sets of $T$ which are nowhere dense is non-empty.
We have by definition of meager:
:$A$ is meager in $T$ it is a countable union of subsets of $S$ which are nowhere dense in $T$.
Here $I$ denotes a countable indexing set.
=== Sufficient Condition ===
Suppose $T$ is not a Baire space.
Then there is a countable set of closed sets of $T$ which are nowhere dense with the interior of its union non-empty.
Let $\FF = \set {F_i: i \in I}$ be such a set.
Then $\paren {\bigcup \FF}^\circ$ is a non-empty open set.
Consider the set $\FF' = \set {F_i \cap \paren {\bigcup \FF}^\circ: i \in I}$.
From Intersection is Subset:
:$\forall i \in I: F_i \cap \paren {\bigcup \FF}^\circ \subseteq F_i$
Since $F_i$ are nowhere dense, so is $F_i \cap \paren {\bigcup \FF}^\circ$.
We have:
Therefore $\bigcup \FF'$ is meager in $T$.
Hence we have shown the existence of a non-empty open set of $T$ which is meager in $T$.
=== Necessary Condition ===
Suppose there exists a non-empty open set of $T$ which is meager in $T$.
Let $\ds A = \bigcup_{i \mathop \in I} A_i$ be such a set.
Then $A$ is a non-empty open set, and $A_i$ are nowhere dense for every $i \in I$.
Thus for every $i$, $\paren {A_i^-}^\circ = \O$.
By Closure of Topological Closure equals Closure:
:$\paren {\paren {A_i^-}^-}^\circ = \paren {A_i^-}^\circ = \O$.
By Set is Closed iff Equals Topological Closure, $A_i^-$ are closed sets of $T$ which are nowhere dense.
We also have:
Hence there is a countable set of closed sets of $T$ which are nowhere dense with the interior of its union non-empty.
Therefore $T$ is not a Baire space.
== Historical Note ==
Category:Baire Spaces
Category:Non-Meager Spaces
Baire Space is Non-Meager

== Theorem ==
Let $T = \struct {S, \tau}$ be a Baire space (in the context of topology).
Then $T$ is non-meager in $T$.
== Proof ==
From Baire Space iff Open Sets are Non-Meager, all open sets of $T$ are non-meager in $T$.
But $T$ itself is an open set of $T$ by definition of topological space.
Hence the result.
== Sources ==
* $22.1$: The Baire Category Theorem
Category:Baire Spaces
Category:Non-Meager Spaces
Balance a Stick on its End on a Finger

== Problem ==
How can a stick be made to balance securely on the tip of a finger?
== Solution ==
Push the tips of $2$ knives into the side of the stick, near the tip, opposite each other, so that the handles of the knives hang down well below the finger.
The center of gravity of the arrangement will be below the fingertip and will be stable.
== Historical Note ==
== Sources ==
* 
* : Henry van Etten: $113$
Category:Physics
Category:Classic Problems
Balance a Stick on its End on a Finger/Mistake

== Source Work ==
: 
:The Puzzles:
::Henry van Etten
:::$113$
== Mistake ==
:How can a stick be made to balance securely on the tip of a finger?
The solution is given as:
:Force the tips of three knives into the stick, so that the knives hang well below the finger.
:The centre of gravity of the entire arrangement will then be below the fingertip and will be stable.
== Correction ==
You do not actually need as many as $3$ knives.
$2$ will do just as well.
In fact, any number of knives, as long as the points at which they are forced into the stick form a regular polygon, can be used.
== Sources ==
* : Henry van Etten: $113$
Category:Mistakes/Curious and Interesting Puzzles
Balanced Incomplete Block Design/Examples/Arbitrary Example 1

== Example of Balanced Incomplete Block Design ==
Let there be $4$ treatments $A, B, C, D$.
Let there be $6$ blocks of $2$ units each.
Then we can arrange the $4$ treatments into the $6$ blocks as:
:$AB, AC, AD, BC, BD, CD$
For a valid analysis, the pairs of treatments must be allocated to blocks at random.
Similarly, within each block, the treatments must be allocated to units at random.
== Sources ==
* 
* 
Category:Examples of Balanced Incomplete Block Designs
Balanced Set in Topological Vector Space is Connected

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$.
Let $B \subseteq X$ be a balanced set. 
Then $B$ is connected.
== Proof ==
We first show that $B$ is path connected.
Let $x \in B$. 
Define $p : \closedint 0 1 \to X$ by: 
:$\map p t = t x$
We clearly have: 
:$\map p 0 = 0$
and:
:$\map p 1 = x$ 
Since $B$ is balanced, we have: 
:$t x \in B$
for each $t \in \closedint 0 1$.
It remains to show that $p$ is continuous.
Define $m : \closedint 0 1 \times X$ by:
:$\map m {\lambda, x} = \lambda x$ 
for each $\tuple {\lambda, x} \in \closedint 0 1 \times X$.
From Restriction of Continuous Mapping is Continuous, we have that $m$ is continuous. 
We can then see that $p$ is the $x$-horizontal section of $m$. 
From Horizontal Section of Continuous Function is Continuous, we have that $p$ is continuous.
So every $x \in B$ connects to $0 \in B$.
So from Path-Connected iff Path-Connected to Point, we have that $B$ is path-connected.
From Path-Connected Space is Connected, $B$ is connected.
Category:Connected Spaces
Category:Balanced Sets
Category:Topological Vector Spaces
Balanced Set in Vector Space is Symmetric

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $C \subseteq X$ be a balanced set. 
Then $C$ is symmetric.
== Proof ==
Since $C$ is balanced, we have: 
:$s C \subseteq C$ for all $s \in \C$ with $\cmod s \le 1$.
So in particular, setting $s = -1$:
:$-C \subseteq C$ 
So $C$ is symmetric.
Category:Symmetric Subsets of Vector Spaces
Category:Balanced Sets
Balanced Subset of Complex Plane is Bounded or Entire Space

== Theorem ==
Consider $\C$ as a vector space over $\C$.
Let $E$ be a balanced subset of $\C$.
Then $E$ is bounded, or $E = \C$.
== Proof ==
Suppose that $E$ is not bounded.
Then, for each $R > 0$ there exists some $z_R \in E$ such that $\size {z_R} > R$.
We show that: 
:$\map {B_R} 0 \subseteq E$ for each $R > 0$
where $\map {B_R} 0$ is the open ball centered at $0$ with radius $R$.
Let:
:$w \in \map {B_R} 0$
Then: 
:$\ds \cmod {\frac w {z_R} } < 1$
So, since $E$ is balanced, we have: 
:$\ds z_R \cdot \paren {\frac w {z_R} } = w \in E$
giving:
:$\map {B_R} 0 \subseteq E$ for each $R > 0$.
From Union of Subsets is Subset, we have: 
:$\ds \bigcup_{R > 0} \map {B_R} 0 \subseteq E$
giving:
:$\C \subseteq E$
Since we also have $E \subseteq \C$, we obtain $E = \C$.
Category:Balanced Sets
Balanced Subset of Real Numbers is Bounded or Entire Space

== Theorem ==
Consider $\R$ as a vector space over $\R$.
Let $E$ be a balanced subset of $\R$.
Then $E$ is bounded, or $E = \R$.
== Proof ==
Suppose that $E$ is not bounded.
Then, for each $M > 0$ there exists some $x_M \in E$ such that $\size {x_M} > M$.
We show that: 
:$\closedint {-M} M \subseteq E$ for each $M > 0$.
Let:
:$t \in \closedint {-M} M$
Then, we have: 
:$\ds \size {\frac t {x_M} } < 1$
So, since $E$ is balanced, we have: 
:$\ds x_M \cdot \paren {\frac t {x_M} } = t \in E$
So:
:$\closedint {-M} M \subseteq E$ for each $M > 0$.
From Union of Subsets is Subset, we then have: 
:$\ds \bigcup_{M > 0} \closedint {-M} M \subseteq E$
So:
:$\R \subseteq E$
Since we also have $E \subseteq \R$, we obtain $E = \R$.
CAtegory:Balanced Sets
Banach-Alaoglu Theorem

== Theorem ==
Let $X$ be a separable normed vector space.
Then the closed unit ball in its normed dual $X^*$ is sequentially compact with respect to the weak-$\ast$ topology.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Sources ==
* 
* 
* : $\S 15.1$
* : $3.3$: The weak* topology
Category:Functional Analysis
Category:Banach-Alaoglu Theorem
Category:Weak-* Topologies
Banach-Alaoglu Theorem/Lemma 1

== Lemma for Banach-Alaoglu Theorem ==
Let $X$ be a separable normed vector space.
Let $X^*$ be the dual space of $X$.
Let $\sequence {l_n}_{n \mathop \in \N}$ be a bounded sequence in $X^*$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable dense subset of $X$.
Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that:
:$\forall j \in \N: \map {l_k} {x_j} \to a_j =: \map l {x_j}$
as $k \to \infty$, $k \in \Lambda_j$.
Let $\Lambda$ be the diagonal sequence.
$l$ can be extended to an element of $X^*$.
== Proof ==
$l$ can be extended in the obvious way to a linear function on $M = \span \set {x_j}_{j \mathop \in \N}$.
We extend it to a functional in $X^*$ by pointwise limit (notice that $M$ is dense in $X$).
We have:
:$\ds \size {\map l x} = \lim_{k \mathop \to \infty} \size {\map l {x_k} } \le \limsup_{k \mathop \to \infty} \norm {l_k}_{X^*} \norm x_X$
where $x_k \to x$ as $k \to \infty$.
Since $\set {l_k}_k \mathop \in \N$ was bounded, $l$ is bounded and thus continuous.
Category:Banach-Alaoglu Theorem
Banach-Alaoglu Theorem/Lemma 2

== Lemma for Banach-Alaoglu Theorem ==
Let $X$ be a separable normed vector space.
Let $X^*$ be the dual space of $X$.
Let $\sequence {l_n}_{n \mathop \in \N}$ be a bounded sequence in $X^*$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable dense subset of $X$.
Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that:
:$\forall j \in \N: \map {l_k} {x_j} \to a_j =: \map l {x_j}$
as $k \to \infty$, $k \in \Lambda_j$.
Let $\Lambda$ be the diagonal sequence.
Then:
$l_k \stackrel {\omega^*} {\to} l$ as $k \to \infty$, $k \in \Lambda$.
== Proof ==
Let:
:$\ds X \ni x = \lim_{\substack {j \mathop \to \infty \\ j \mathop \in J} } x_j$
where $J$ is some subset of $\N$.
We have then, for every $j \in J$:
Now given $\epsilon >0$, we find a $j \in J$ such that the first term is less than $\epsilon / 2$. 
For fixed $j$, we have by construction of $l$ that $\map {l_k} {x_j}$ converges to $\map l {x_j}$.
Therefore, we can find a $k_\epsilon$ such that for $k \ge k_\epsilon$ we have:
:$\size {\map {l_k} x - \map l x} < \epsilon$ 
as requested.
Category:Banach-Alaoglu Theorem
Banach-Alaoglu Theorem/Lemma 3

== Lemma for Banach-Alaoglu Theorem ==
Let $X$ be a normed vector space.
Denote by $B$ the closed unit ball in $X$.
Let $X^*$ be the dual of $X$.
Denote by $B^*$ the closed unit ball in $X^*$.
Let $\map \FF B = \closedint {-1} 1^B$ be the topological space of functions from $B$ to $\closedint {-1} 1$.
By Tychonoff's Theorem, $\map \FF B$ is compact with respect to the product topology.
We define the restriction map:
:$R: B^* \to \map \FF B$
by:
:$\map R \psi = \psi \restriction_B$
$R \sqbrk {B^*}$ is a closed subset of $\map \FF B$.
== Proof ==
Below, we will show that $f \in R \sqbrk {B^*}$ :
First, we see how the result follows from the above claim.
That is, we show that $\map \FF B \setminus R \sqbrk {B^*}$ is open.
For $u \in \closedint {-1} 1$ and $r \in \R_{>0}$, let $\map {I_r} u$ denote the open $r$-ball of $u$ in $\closedint {-1} 1$, i.e.:
:$\map {I_r} u := \openint {u - r} {u + r} \cap \closedint {-1} 1$
Let $f \in \map \FF B \setminus R \sqbrk {B^*}$
It suffices to find an open set $U_f$ in $\map \FF B$ such that:
:$f \in U_f \subseteq \map \FF B \setminus R \sqbrk {B^*}$
In view of the claim above, either $(1)$ or $(2)$ must fail for $f$.
Suppose $(1)$ fails.
That is, there exist $x, y \in B$ such that $x + y \in B$ satisfying:
:$3 \epsilon := \size {\map f x + \map f y - \map f {x + y} } > 0$
Thus we can choose:
:$U_f := \pr_x^{-1} \sqbrk { \map {I_\epsilon} {\map f x} } \cap \pr_y^{-1} \sqbrk { \map {I_\epsilon} {\map f y} } \cap \pr_{x+y}^{-1} \sqbrk { \map {I_\epsilon} {\map f {x+y} } }$
Suppose $(2)$ fails.
That is, there exist $x \in B$ and $\alpha \in \R_{>0}$ such that $\alpha x \in B$ satisfying:
:$2 \epsilon := \size {\map f {\alpha x} - \alpha \map f x} > 0$
Thus we can choose:
:$U_f := \pr_{\alpha x}^{-1} \sqbrk { \map {I_\epsilon} { \map f {\alpha x} } } \cap \pr_x^{-1} \sqbrk { \map {I_{\epsilon / \alpha } } {\map f x} }$
Now, we show the above claim.
Let $f \in R \sqbrk {B^*}$.
That is, there is a $\psi \in B^\ast$ such that $f = \map R \psi$.
$\psi$ satisfies both $(1)$ and $(2)$, as $\psi \in X^\ast$, especially.
Thus $f$ also satisfies them, because it is a restriction of $\psi$.
Conversely, let $f \in \map \FF B$ be satisfying $(1)$ and $(2)$.
Then, define a mapping $\phi : X \to \R$ by:
:$\forall x \in X \setminus \set 0 : \map \psi x := \norm x \map f {\dfrac x {\norm x} }$
and $\map \psi 0 := 0$.
We show that $\psi \in B^\ast$ such that $\map R \psi = f$.
Observe that $\map f 0 = 0$, since $\map f 0 = \map f {0 + 0} = \map f 0 + \map f 0$ by $(1)$.
Thus, for $x \in X$ such that $x \ne 0$:
Furthermore, for all $x,y \in X$ such that $x + y \ne 0$:
Thus for $x,y \in X$:
:$\map \psi {x+y} = \map \psi x + \map \psi y$
In particular, for all $x \in X$:
:$\map \psi {-x} = - \map \psi x$
For $x \in X$ and $\alpha \in \R$ such that $\alpha x \ne 0$:
We also have by Definition of $\psi$:
:$\forall x \in X \setminus \set 0 : \size {\map \psi x} = \norm x \size {\map f {\frac x {\norm x} } } \le \norm x$
Thus $\psi \in B^\ast$.
Finally, we have $\map R \psi = f$, since for all $x \in B \setminus \set 0$:
and $\map \psi 0 = 0 = \map f 0$.
Category:Banach-Alaoglu Theorem
Banach-Alaoglu Theorem/Lemma 4

== Lemma for Banach-Alaoglu Theorem ==
Let $X$ be a normed vector space.
Denote by $B$ the closed unit ball in $X$.
Let $X^*$ be the dual of $X$.
Denote by $B^*$ the closed unit ball in $X^*$.
Let:
:$\map \FF B = \closedint {-1} 1^B$ 
be the topological space of functions from $B$ to $\closedint {-1} 1$
By Tychonoff's Theorem:
:$\map \FF B$ 
is compact with respect to the product topology.
We define the restriction map:
:$R: B^* \to \map \FF B$
by:
:$\map R \psi = \psi \restriction_B$
$R$ is a homeomorphism from $B^*$ with the weak* topology to its image:
:$R \sqbrk {B^*}$ 
seen as a subset of $\map \FF B$ with the product topology.
== Proof ==
Firstly, $R$ is an injection.
Indeed, let $\psi_1, \psi_2 \in B^\ast$ such that $\map R {\psi_1} = \map R {\psi_2}$.
Then, for all $x \in X \setminus \set 0$ we have:
In addition, $\map {\psi_1} 0 = \map {\psi_2} 0 = 0 $.
Thus, $\psi_1 = \psi_2$.
In view of Inverse of Homeomorphism is Homeomorphism, it suffices to show that:
:$R^{-1} : R \sqbrk {B^\ast} \to B^\ast$
is a homeomorphism.
Observe that $X^\ast$ is Hausdorff space by Weak-* Topology is Hausdorff.
Thus $B^\ast$ is also Hausdorff space by Subspace of Hausdorff Space is Hausdorff.
Recall that $\map \FF B$ is compact.
In addition, by Lemma 3, $R \sqbrk {B^\ast}$ is closed.
Thus $R \sqbrk {B^\ast}$ is compact by Closed Subspace of Compact Space is Compact.
Thus, in view of Continuous Bijection from Compact to Hausdorff is Homeomorphism, it suffices to show that:
:$R^{-1} : R \sqbrk {B^\ast} \to B^\ast$
is continuous.
Recall and 
Let $\psi_0 \in X^\ast$, $x \in X$, and $r \in \R_{>0}$ be arbitrary.
Let
:$\map N {\psi_0 ; x ; r } := \set { \psi \in B^\ast : \size {\map \psi x - \map {\psi_0} x } < r } $
Observe, if $x \ne 0$, we have:
:$\map N {\psi_0 ; x ; r } = \map N {\psi_0 ; \frac x {\norm x} ; \frac r {\norm x} }$
, therefore, we may assume $x \in B$.
Now, it suffices to check that $R \sqbrk { \map N {\psi_0 ; x ; r } }$ is open in $R \sqbrk {B^\ast}$.
This can be seen as:
where:
:$\map {I_r} {\map {\psi_0} x} := \openint {\map {\psi_0} x - r} {\map {\psi_0} x + r} \cap \closedint {-1} 1$
Category:Banach-Alaoglu Theorem
Banach-Alaoglu Theorem/Proof 1

== Theorem ==
== Proof ==
The aim of this proof is to show the following:
Given a bounded sequence in $X^*$, there exists a weakly convergent subsequence of that bounded sequence.
Let $\sequence {l_n}_{n \mathop \in \N}$ be a bounded sequence in $X^*$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable dense subset of $X$.
Choose subsequences $\N \supset \Lambda_1 \supset \Lambda_2 \supset \ldots$ such that:
:$\forall j \in \N: \map {l_k} {x_j} \to a_j =: \map l {x_j}$
as $k \to \infty$, $k \in \Lambda_j$.
Let $\Lambda$ be the diagonal sequence.
=== Lemma 1 ===
=== Lemma 2 ===
== Also known as ==
Category:Banach-Alaoglu Theorem
Banach-Alaoglu Theorem/Proof 2

== Theorem ==
== Proof ==
Let $X$ be a normed vector space.
Denote by $B$ the closed unit ball in $X$.
Let $X^*$ be the dual of $X$.
Denote by $B^*$ the closed unit ball in $X^*$.
Let:
:$\map \FF B = \closedint {-1} 1^B$ 
be the topological space of functions from $B$ to $\closedint {-1} 1$.
By Tychonoff's Theorem:
:$\map \FF B$ 
is compact with respect to the product topology.
We define the restriction map:
:$R: B^* \to \map \FF B$
by:
:$\forall \psi \in B^*: \map R \psi = \psi \restriction_B$
=== Lemma 3 ===
=== Lemma 4 ===
Thus by Lemma 4, $B^*$ in the weak* topology is homeomorphic with $R \sqbrk {B^*}$.
This is a closed set of $\map \FF B$ (by Lemma 3) and thus compact.
By the Eberlein-Šmulian Theorem, this is sequentially compact.
== Also known as ==
Category:Banach-Alaoglu Theorem