nyuuzyou/wikis huggingface dataset:
Banach-Alaoglu Theorem/Proof 3

== Theorem ==
== Proof ==
Let $B_{X^\ast}$ be the closed unit ball in $X^\ast$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
From the definition of the norm of a bounded linear functional, we have: 
:$B_{X^\ast} = \set {f : X \to \GF : \cmod {\map f x} \le \norm x \text { and } f \text { is linear} }$
For each $x \in X$, let: 
:$K_x = \set {\lambda \in \GF : \cmod \lambda \le \norm x}$ 
Let: 
:$\ds K = \prod_{x \mathop \in X} K_x$ 
equipped with the product topology. 
Then if $f \in K$ and $x \in X$ we have $\map f x \in K_x$ $\cmod {\map f x} \le \norm x$. 
Then: 
:$B_{X^\ast} = \set {f \in K : f \text { is linear} }$
Let $\pr_x : K \to \GF$ denote the projection onto the $x$th factor, that is: 
:$\map {\pr_x} f = \map f x$ 
for each $f \in K$.
From the definition of the product topology, the product topology on $K$ is the initial topology induced by $\set {\pr_x : x \in X}$. 
From Subspace Topology on Initial Topology is Initial Topology on Restrictions, the subspace topology on $B_{X^\ast}$ inherited by $K$ is the initial topology on $B_{X^\ast}$ induced by $\set {\pr_x \restriction_{B_{X^\ast} } : x \in X}$.
Applying Subspace Topology on Initial Topology is Initial Topology on Restrictions to $\struct {X^\ast, w^\ast}$, this is precisely the subspace topology on $B_{X^\ast}$ inherited by the weak-$\ast$ topology on $X^\ast$. 
From Tychonoff's Theorem, $K$ is compact. 
From Closed Subspace of Compact Space is Compact, it suffices to show that $B_{X^\ast}$ is closed in $\struct {K$.
We may now write: 
From Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous: 
:$\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y}$ is continuous for each $x, y \in X$ and $\lambda, \mu \in \GF$.
From Continuity Defined from Closed Sets, we have that: 
:$\paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} }^{-1} \sqbrk {\set 0}$ is closed in $K$. 
Since the intersection of closed sets is closed, we have that: 
:$\ds \bigcap_{x, y \in X, \, \lambda, \mu \in \GF} \paren {\lambda \pr_x + \mu \pr_y - \pr_{\lambda x + \mu y} }^{-1} \sqbrk {\set 0}$ is closed in $K$.
So $\struct {B_{X^\ast}, w^\ast}$ is closed in $K$.
From Closed Subspace of Compact Space is Compact, $\struct {B_{X^\ast}, w^\ast}$ is compact.
Category:Banach-Alaoglu Theorem
Banach-Schauder Theorem/Lemma 1

== Lemma ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be Banach spaces.
Let $T : X \to Y$ be a surjective bounded linear transformation.
Let $y \in Y$ and $r > 0$ be such that:
:$\map {B_Y} {y, r} \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
where:
:$\map {B_Y} {y, r}$ denotes the open ball in $Y$ centered at $y$ with radius $r$
:$\map {B_X} {0, m}$ denotes the open ball in $X$ centered at $0 \in X$ with radius $m$
:$\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ denotes the topological closure of $T \sqbrk {\map {B_X} {0, m} }$.
Then: 
:$\map {B_Y} {0, r} \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
== Proof ==
From Open Ball is Convex Set, we have: 
:$\map {B_X} {0, m}$ is convex.
Then, from Image of Convex Set under Linear Transformation is Convex, we have: 
:$T \sqbrk {\map {B_X} {0, m} }$ is convex.
From Closure of Convex Subset in Normed Vector Space is Convex, we have: 
:$\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ is convex.
From Open Ball Centred at Origin in Normed Vector Space is Symmetric:
:$\map {B_X} {0, m}$ is symmetric.
From Image of Symmetric Set under Linear Transformation is Symmetric, we have: 
:$T \sqbrk {\map {B_X} {0, m} }$ is symmetric.
From Closure of Symmetric Subset of Normed Vector Space is Symmetric:
:$\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ is symmetric.
Let $x \in \map {B_Y} {0, r}$.
Then $x$ can be written in the form: 
:$x = r u$ 
for $u$ with $\norm u_Y < 1$, in particular: 
:$\ds u = \frac x r$ 
Since $\map {B_Y} {y, r} \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$, we have: 
:$y + r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
and:
:$y - r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
Since $\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ is symmetric, we have: 
:$-y + r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
Writing:
:$\ds r u = \frac 1 2 \paren {y + r u} + \frac 1 2 \paren {-y + r u}$
we have: 
:$r u \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
since $\paren {T \sqbrk {\map {B_X} {0, m} } }^-$ is convex.
So: 
:$x \in \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
So:
:$\map {B_Y} {0, r} \subseteq \paren {T \sqbrk {\map {B_X} {0, m} } }^-$
Category:Banach-Schauder Theorem
Banach-Steinhaus Theorem/F-Space

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau_X}$ be an $F$-Space over $\GF$.
Let $\struct {Y, \tau_Y}$ be a topological vector space over $\GF$.
Let $\Gamma$ be a set of continuous linear transformations $X \to Y$ such that for all $x \in X$:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is von Neumann-bounded in $Y$.
Then $\Gamma$ is equicontinuous.
== Proof ==
Let $B$ be the set of $x \in X$ such that:
:$\map \Gamma x = \set {T x : T \in \Gamma}$ is von Neumann-bounded in $Y$.
By hypothesis, $B = X$. 
Let $d$ be a metric inducing $\tau_X$ such that $\struct {X, d}$ is a complete metric space. 
From the Baire Category Theorem, $\struct {X, d}$ is a Baire space.
From Baire Space is Non-Meager, $\struct {X, d}$ is not meager.
Since $B$ is then not meager, we can apply the Banach-Steinhaus Theorem to obtain that $\Gamma$ is equicontinuous.
== Historical Note ==
== Sources ==
* : $2.6$: Theorem
Category:F-Spaces
Category:Banach-Steinhaus Theorem
Banach-Steinhaus Theorem/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$.
Suppose that:
:$\ds \forall x \in X: \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite.
Then:
:$\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha}$ is finite
where $\norm {T_\alpha}$ denotes the norm of the linear transformation $T_\alpha$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also known as ==
== Historical Note ==
== Sources ==
* $22.2$: The Principle of Uniform Boundedness
Category:Linear Transformations on Banach Spaces
Category:Banach-Steinhaus Theorem
Banach-Steinhaus Theorem/Normed Vector Space/Also known as

== Banach-Steinhaus Theorem (Normed Vector Space): Also known as ==
This form of the Banach-Steinhaus Theorem is also known as the Uniform Boundedness Principle.
Some sources give it as the Uniform Bounded Principle, but it is possible that this is a mistake.
== Sources ==
* 
Category:Banach-Steinhaus Theorem
Banach-Steinhaus Theorem/Normed Vector Space/Proof 2

== Theorem ==
== Proof ==
From Banach Space is F-Space, $\struct {X, \norm {\, \cdot \,}_X}$ can be considered as an $F$-Space.
From Normed Vector Space is Hausdorff Topological Vector Space, $\struct {Y, \norm {\, \cdot \,}_Y}$ can be considered as a topological vector space.
Let $\Gamma = \set {T_\alpha : \alpha \in A}$ and:
:$\map \Gamma x = \set {T_\alpha x : \alpha \in A}$
for each $x \in X$. 
By hypothesis, we have:
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x}_Y < \infty$ for each $x \in X$. 
From Characterization of von Neumann-Boundedness in Normed Vector Space, we have that $\map \Gamma x$ is von Neumann-bounded in $\struct {Y, \norm {\, \cdot \,}_Y}$ for each $x \in X$. 
From Banach-Steinhaus Theorem: $F$-Space, $\Gamma$ is equicontinuous. 
So there exists an open neighborhood $U$ of ${\mathbf 0}_X$ such that:
:$T_\alpha \sqbrk U \subseteq \map {B_1^Y} { {\mathbf 0}_Y} \subseteq B^-_Y$ for each $\alpha \in A$.
where:
:$\map {B_1^Y} { {\mathbf 0}_Y}$ is the open ball of center ${\mathbf 0}_Y$ and radius $1$ in $\struct {Y, \norm {\, \cdot \,}_Y}$
:$B_Y^-$ is the closed unit ball of $\struct {Y, \norm {\, \cdot \,}_Y}$.
From the definition of an open set in a normed vector space, there exists $\delta > 0$ such that:
:$\map {B_\delta^X} { {\mathbf 0}_X} \subseteq U$
and so:
:$T_\alpha \sqbrk {\map {B_\epsilon^X} { {\mathbf 0}_X} } \subseteq \map {B_1^Y} { {\mathbf 0}_Y}$ for each $\alpha \in A$.
We then have:
:$\paren {\delta/2} B_X^- \subseteq \map {B_\delta^X} { {\mathbf 0}_X}$
and:
:$\map {B_1^Y} { {\mathbf 0}_Y} \subseteq B_Y^-$
where $B_X^-$ denotes the closed unit ball of $\struct {X, \norm {\, \cdot \,}_X}$. 
Hence by Image of Dilation of Set under Linear Transformation is Dilation of Image, we have: 
:$T_\alpha \sqbrk {B_X^-} \subseteq \paren {2/\delta} B_Y^-$ for each $\alpha \in A$. 
From Norm of Bounded Linear Transformation in terms of Closed Unit Ball, we have:
:$\ds \norm {T_\alpha} \le \frac 2 \delta$ for each $\alpha \in A$.
That is:
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} \le \frac 2 \delta$
as required.
== Sources ==
* : $2.6$: Theorem
Category:Banach-Steinhaus Theorem
Banach-Steinhaus Theorem/Normed Vector Space/Proof 3

== Theorem ==
== Proof ==
It suffices to show that there exist an $x_0 \in X$ and an $r \in \R_{>0}$ such that:
:$\ds K : = \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y$ is finite
where $\map {B_r} {x_0}$ is the open $r$-ball of $x_0$.
Indeed, then we have for all $x \in X \setminus \set 0$:
That is:
:$\ds \norm {T_\alpha} = \sup_{x \mathop \in X \setminus \set 0} \frac {\norm {T_\alpha x}_Y}{\norm x_X} \le \frac {4 K} r$
 that for all $x_0 \in X$ and $r \in \R_{>0}$:
:$\ds \sup_{x \mathop \in \map {B_r} {x_0} } \sup_{\alpha \mathop \in A} \norm {T_\alpha x}_Y = + \infty$
Then we can choose:
:$\map {B_{r_1} } {x_1} \supseteq \map {B_{r_2} } {x_2} \supseteq \cdots$
and:
:$\alpha_1, \alpha_2, \ldots \in A$
such that:
:$\ds \inf _{x \mathop \in \map {B_{r_k} } {x_k} } \norm {T_{\alpha_k} x}_Y \ge k$
and:
:$0 < r_k < \frac 1 k$
Indeed, let $r_0 \in \R_{>0}$ and $x_0 \in X$ be fixed.
Then, for $k = 1, 2, \ldots$, we have:
:$\exists x_k \in \map {B_{r_{k-1} } } {x_{k-1} } \exists \alpha_k \in A : \norm {T_{\alpha_k} {x_k} }_Y > k$
Furthermore, as $T_{\alpha_k}$ is continuous, $\exists r_k \in \openint 0 {\frac 1 k}$ such that:
:$\map {B_{r_{k-1} } } {x_{k-1} } \supseteq \map {B_{r_k} } {x_k}$
and:
:$\ds \inf _{x \mathop \in \map {B_{r_k} } {x_k} } \norm {T_{\alpha_k} x}_Y \ge k$
Then $\sequence {x_k}$ is a Cauchy sequence in $X$, since for each $N \in \N_{>0}$:
:$k, l \ge N \implies \norm {x_k - x_l}_X \le \norm {\underbrace {x_k - x_N}_{\in \map {B_{r_N} } {x_N} } }_X + \norm {\underbrace {x_l - x_N}_{\in \map {B_{r_N} } {x_N} } }_X\le 2 \, r_N \le \frac 2 N$
As $X$ is a Banach space, there exists:
:$\ds z := \lim_{k \mathop \to +\infty} x_k$
Observe that:
:$\ds z \in \bigcap_{k \in \N_{>0} } \map {\overline {B_{r_k} } } {x_k}$
where $\map {\overline {B_{r_k} } } {x_k}$ is the $r_k$-closed ball of $x_k$.
In particular:
:$\forall k \in \N_{>0} : \norm {T_{\alpha_k} z}_Y \ge k$
Thus:
:$\ds \sup_{\alpha \mathop \in A} \norm {T_\alpha z}_Y = + \infty$
This is a contradiction.
Category:Banach-Steinhaus Theorem
Banach-Tarski Paradox

== Theorem ==
Let $\mathbb D^3 \subset \R^3$ be a unit ball in real Euclidean space of $3$ dimensions.
Then $\mathbb D^3$ is equidecomposable to the union of two such unit balls.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Theorem $1.2$
* 
* 
* 
* 
* 
Category:Banach-Tarski Paradox
Category:Equidecomposable Sets
Category:Veridical Paradoxes
Category:Topology
Banach-Tarski Paradox/Historical Note

== Historical Note on Banach-Tarski Paradox ==
Ever since and raised this question in a collaborative paper in $1924$, whether the Banach-Tarski Paradox is a veridical paradox or an antinomy is being hotly discussed to the present day.
At one point there were even Facebook groups on the subject both pro and anti, but Facebook has since dumbed down and these groups no longer seem to exist.
If you feel really adventurous, check out 's novel .
Category:Historical Notes
Banach-Tarski Paradox/Lemma 1

== Lemma for Banach-Tarski Paradox ==
=== Relation Definition ===
$\approx$ is an equivalence relation.
== Proof ==
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.5$
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Lemma 2

== Lemma for Banach-Tarski Paradox ==
=== Relation Definition ===
Let $X$ and $Y$ be disjoint unions of $X_1, X_2$ and $Y_1, Y_2$ respectively.
Let $X_i \approx Y_i$ for each $i \in \set {1, 2}$.
Then:
:$X \approx Y$
== Proof ==
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.5$
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Lemma 3

== Lemma for Banach-Tarski Paradox ==
=== Relation Definition ===
Let $X_1 \subseteq Y \subseteq X$.
Let$X \approx X_1$.
Then:
:$X \approx Y$
== Proof ==
Recall:
=== Lemma 2 ===
Let:
such that $X^i$ is congruent to $X_1^i$ for each $i \in \set {a, 2, \ldots, n}$.
Let us choose a congruence:
:$f^i: X^i \to X_1^i$
for each $i \in \set {a, 2, \ldots, n}$.
Let $f$ be the bijection of $X$ to $X_i$ which agrees with $f^i$ on each $X^i$.
Now let:
and:
Let:
:$Z = \ds \bigcup_{n \mathop = 0}^\infty \paren {X_n \setminus Y_n}$
Then:
:$f \sqbrk Z$ and $X \setminus Z$ are disjoint, and:
and by :
:$X \approx Y$
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.5$
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Lemmata

== Lemmata for Banach-Tarski Paradox ==
=== Relation Definition ===
Then the following lemmata hold:
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.5$
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Lemmata/Relation Definition

== Lemmata for Banach-Tarski Paradox: Definition of Relation ==
Let $\approx$ denote the relation between sets in Euclidean space of $3$ dimensions defined as follows:
:$X \approx Y$
:
:there exists a partition of $X$ into disjoint sets:
::$X = X_1 \cup X_2 \cup \cdots \cup X_m$
:and a partition of $Y$ into the same number of disjoint sets:
::$Y = Y_1 \cup Y_2 \cup \cdots \cup Y_m$
such that $X_i$ is congruent to $Y_i$ for each $i \in \set {1, 2, \ldots, m}$.
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.5$
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Proof 1

== Theorem ==
== Proof ==
Let $\mathbb D^3$ be centered at the origin, and $D^3$ be some other unit ball in $\R^3$ such that $\mathbb D^3 \cap D^3 = \O$.
Let $\mathbb S^2 = \partial \mathbb D^3$.
By the Hausdorff Paradox, there exists a decomposition of $ \mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C$ and $B \cup C$ are congruent, and $Q$ is countable.
For $r \in \R_{>0}$, define a function $r^*: \R^3 \to \R^3$ as $\map {r^*} {\mathbf x} = r \mathbf x$, and define the sets:
Let $T = W \cup Z \cup \set {\mathbf 0}$.
$W$ and $X \cup Y$ are clearly congruent by the congruency of $A$ with $B \cup C$, hence $W$ and $X \cup Y$ are equidecomposable.
Since $X$ and $Y$ are congruent, and $W$ and $X$ are congruent, $X \cup Y$ and $W \cup X$ are equidecomposable. 
$W$ and $X \cup Y$ as well as $X$ and $W$ are congruent, so $W \cup X$ and $W \cup X \cup Y$ are equidecomposable. 
Hence $W$ and $W \cup X \cup Y$ are equidecomposable, by Equidecomposability is Equivalence Relation.
So $T$ and $\mathbb D^3$ are equidecomposable, from Equidecomposability Unaffected by Union.
Similarly we find $X$, $Y$, and $W \cup X \cup Y$ are equidecomposable.
Since $Q$ is only countable, but $\map {\operatorname {SO} } 3$ is not, we have:
:$\exists \phi \in \map {\operatorname {SO} } 3: \map \phi Q \subset A \cup B \cup C$
so that $I = \map \phi Q \subset W \cup X \cup Y$.
Since $X$ and $W \cup X \cup Y$ are equidecomposable, by Subsets of Equidecomposable Subsets are Equidecomposable, $\exists H \subseteq X$ such that $H$ and $I$ are equidecomposable.
Finally, let $p \in X - H$ be a point and define $S = Y \cup H \cup \set p$.
Since:
:$Y$ and $W \cup X \cup Y$
:$H$ and $Z$
:$\set 0$ and $\set p$
are all equidecomposable in pairs, $S$ and $\mathbb B^3$ are equidecomposable by Equidecomposability Unaffected by Union.
Since $D^3$ and $\mathbb D^3$ are congruent, $D^3$ and $S$ are equidecomposable, from Equidecomposability is Equivalence Relation. 
By Equidecomposability Unaffected by Union, $T \cup S$ and $\mathbb D^3 \cup D^3$ are equidecomposable.
Hence $T \cup S \subseteq \mathbb D^3 \subset \mathbb D^3 \cup D^3$ are equidecomposable and so, by the chain property of equidecomposability, $\mathbb D^3$ and $\mathbb D^3 \cup D^3$ are equidecomposable.
== Also known as ==
== Historical Note ==
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Proof 2

== Theorem ==
== Proof ==
=== Lemmata ===
Let $U$ denote the closed ball in real Euclidean space of $3$ dimensions defined as:
:$U = \set {\mathbf x \in \R^3 : \size {\mathbf x - \mathbf c} \le r}$
where:
:$r$ is the radius of $U$
:$\mathbf x$ is the position vector of a point $\tuple {x_1, x_2, x_3} \in \R^3$
:$\mathbf c$ is the position vector of the center $\tuple {c_1, c_2, c_3}$ of $U$
:$\size {\mathbf x - \mathbf c}$ denotes the Euclidean distance:
::$\size {\mathbf x - \mathbf c} = \sqrt { {x_1 - c_1}^2 + {x_2 - c_2}^2 + {x_3 - c_3}^2}$
Let $\Bbb S^2$ denote the sphere which is the boundary of $U$:
:$\Bbb S^2 = \set {\mathbf x \in \R^3 : \size {\mathbf x - \mathbf c} = r}$
Let:
:$(1): \quad \Bbb S^2 = A \cup B \cup C \cup Q$
be the partition of $\Bbb S^2$ from Hausdorff Paradox.
Recall:
:$A$, $B$ and $C$ are disjoint
:$A$, $B$ and $C$ are congruent to each other
:$B \cup C$ is congruent to each of $A$, $B$ and $C$
Hence :
:$A \approx B \approx C \approx B \cup C$
For a given subset $T \subseteq \Bbb S^2$, let $\overline T \subseteq U$ be defined as the set of all $\mathbb x \in U$ apart from $\mathbf c$ whose projection onto $\Bbb S^2$ from $\mathbf c$ is $T$.
Hence using Vector Equation of Straight Line in Space:
:$\overline T = \set {\mathbf y \in U: \mathbf y = \paren {1 - s} \mathbf c + s \mathbf x: \mathbf x \in T, s \in \hointl 0 1}$
That is, for all points $\mathbf x$ in $T$, all the points in $U$ along the radius of $U$ to $\mathbf x$ but not including the center of $U$.
We have that:
:$U = \overline A \cup \overline B \cup \overline C \cup \overline Q \cup \set {\mathbb c}$
Because we have:
:$A \approx B \approx C \approx B \cup C$
it follows that:
:$(2): \quad \overline A \approx \overline B \approx \overline C \approx \overline B \cup \overline C$
Let:
From $(2)$ and :
:$\overline A \approx \overline A \cup \overline B \cup \overline C$
and so:
:$(3): \quad X \approx U$
Let $\alpha$ be a rotation of $U$ such that $\alpha$ is not in the original $G$ that was used to construct $(1)$ with the property that:
:$Q \cap \alpha Q = \O$
Hence using:
:$\overline C \approx \overline A \cup \overline B \cup \overline C$
there exists $S \subseteq C$ such that:
:$\overline S \approx \overline Q$
Let $p$ be an arbitrary point in $\overline C \setminus \overline S$.
We have that:
:$(4): \quad \overline A \cup \overline Q \cup \set {\mathbb c} \approx \overline B \cup \overline S \cup \set p$
Because:
:$\overline B \cup \overline S \cup \set p \subseteq Y \subseteq U$
we can use $(3)$ and $(4)$ along with to get:
:$Y \approx U$
Hence the result.
== Also known as ==
== Historical Note ==
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Proof of Theorem $1.2$
Category:Banach-Tarski Paradox
Banach-Tarski Paradox/Proof 2/Mistake

== Source Work ==
:
:Chapter $1$: Introduction
::$1.3$ A paradoxical decomposition of the sphere:
:::Proof of Theorem $1.2$
== Mistake ==
:Now, it is easy to find some rotation $\alpha$ (not in $G$) such that $Q$ and $Q \cdot \alpha$ are disjoint, and so, using
::$\overline C \approx \overline A \cup \overline B \cup \overline C$,
:there exists $S \subset C$ such that $\overline S \approx \overline Q$. Let $p$ be some point in $\overline S - \overline C$. Obviously, $\ldots$
== Correction ==
If $S \subset C$, as supposed, it follows that $\overline S \subset \overline C$ and so $\overline S - \overline C = \O$.
What is meant is:
:Let $p$ be some point in $\overline C - \overline S$.
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Proof of Theorem $1.2$
Category:Mistakes/The Axiom of Choice
Banach Algebra of Continuous Functions on Compact Hausdorff Space is Banach Algebra

== Theorem ==
Let $X$ be a compact Hausdorff space.
Let $\map \CC X = \map \CC {X, \C}$ be the vector space of continuous functions on $X$.
Let $\ast$ be pointwise multiplication on $\C^X$. 
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map \CC X$. 
Then $\struct {\map \CC X, \ast, \norm {\, \cdot \,} }$ is a Banach algebra.
== Proof ==
We first show that $\struct {\map \CC X, \ast}$ is an algebra over $\C$.
For this, we show that $\struct {\map \CC X, \ast}$ is a subalgebra of $\C^X$.
As shown in Continuous Functions on Compact Space form Banach Space, $\map \CC X$ is a linear subspace of $\C^X$.
From Product of Continuous Functions on Topological Ring is Continuous, we have $f \ast g \in \map \CC X$. 
Hence $\struct {\map \CC X, \ast}$ is a subalgebra of $\C^X$.
Hence $\struct {\map \CC X, \ast}$ is an algebra over $\C$.
From Continuous Functions on Compact Space form Banach Space, $\map \CC X$ is a Banach space. 
It remains to show that $\norm {\, \cdot \,}_\infty$ is a algebra norm on $\map \CC X$. 
That is: 
:$\norm {f \ast g}_\infty \le \norm f_\infty \norm g_\infty$ for each $f, g \in \map \CC X$.
For each $x \in X$, we have:
Hence $\struct {\map \CC X, \ast, \norm {\, \cdot \,} }$ is a Banach algebra.
Category:Banach Algebras
Banach Fixed-Point Theorem

== Theorem ==
Let $\struct {M, d}$ be a complete metric space.
Let $f: M \to M$ be a contraction.
That is, there exists $q \in \hointr 0 1$ such that for all $x, y \in M$:
:$\map d {\map f x, \map f y} \le q \, \map d {x, y}$
Then there exists a unique fixed point of $f$.
== Proof ==
=== Uniqueness ===
Let $f$ have two fixed points $p_1, p_2 \in M$.
It is to be proved that $p_1 = p_2$.
Then from $\paren {1 - q} > 0$ and $\map d {p_1, p_2} \ge 0$:
:$\map d {p_1, p_2} = 0$
From :
:$p_1 = p_2$
=== Existence ===
The fixed point $p$ will be found from an arbitrary member $a_0$ of $M$ by iteration.
The plan is to obtain $\ds p = \lim_{n \mathop \to \infty} a_n$ with definition $a_{n + 1} = \map f {a_n}$.
The sequence of iterates converges in complete metric space $M$ because it is a Cauchy sequence in $M$, as is proved in the following.
Induction on $n$ applies to obtain the contractive estimate:
:$\map d {a_{n + 1}, a_n} \le q^n \map d {a_1, a_0}$
Induction details $n = 1$:
Assume the contractive estimate for $n = k$.
Induction details for $n = k + 1$:
Next it is to be proved that $\sequence {a_n}$ is a Cauchy sequence in $M$ by showing that $\ds \lim_{m \mathop \to \infty} \map d {a_{n + m}, a_n} = 0$ for $n$ large.
We have that:
:$\ds \lim_{n \mathop \to \infty} q^n = 0$
and:
:$\dfrac {1 - q^m} {1 - q} \le \dfrac 1 {1 - q}$
Then $\sequence {\map d {a_{n + m}, a_n} }$ has limit zero at $m = \infty$ for large $n$.
Sequence $\sequence{a_n}$ is a Cauchy sequence convergent to some $p$ in $M$.
Then:
The has limit zero at $n = \infty$.
Then:
:$\map d {\map f p, p} = 0$
Then by :
:$\map f p = p$
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* : Appendix $\text C$
* 
* 
* 
Category:Banach Fixed-Point Theorem
Category:Metric Spaces
Category:Functional Analysis
Category:Fixed Point Theorems
Banach Fixed-Point Theorem/Also known as

== Banach Fixed-Point Theorem: Also known as ==
The Banach Fixed-Point Theorem is also known as:
:the Contraction Mapping Principle
:the Contraction Mapping Theorem
:the Contraction Theorem
:the Banach Contraction Principle
:the Banach Contraction Theorem
:the Contraction Lemma.
== Sources ==
* 
* 
* 
* 
* 
* 
* 
Category:Banach Fixed-Point Theorem