nyuuzyou/wikis huggingface dataset:
Hilbert Space Direct Sum of Countably Many Separable Hilbert Spaces is Separable

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
For each $n \in \N$, let $\struct {\HH_n, \innerprod \cdot \cdot_n}$ be a separable Hilbert space over $\GF$. 
Let:
:$\ds \HH = \bigoplus_{n \mathop \in \N} \HH_n$
be the direct sum of $\sequence {\HH_n}_{n \mathop \in \N}$. 
Then $\HH$ is separable.
== Proof ==
For each $n \in \N$, let $\sequence {e_k^{(n)} }_{k \mathop \in \N}$ be a countable everywhere dense subset of $\HH_n$.
Let $\SS$ be the set of $f \in \HH$ for which there exists $N \in \N$ such that:
:$\map f n = e_{\map k n}^{(n)} \in \HH_n$ for some $\map k n \in \N$ if $n \le N$
:$\map f n = {\mathbf 0}_{\HH_n}$ if $n > N$.
To every $f \in \SS$, we can associate a function $\phi_f : \N \to \Z_{\ge 0}$ defined by:
:$\map {\phi_f} n = \map k n$ for $n \le N$
:$\map {\phi_f} n = 0$ for $n > N$.
From Set of Finitely Supported Functions on Integers is Countable and Subset of Countable Set is Countable, $\set {\phi_f : f \in \SS}$ is countable.
Hence $\SS$ is countable.
From Everywhere Dense Subset of Countable Hilbert Space Direct Sum in terms of Everywhere Dense Subsets of Summands, $\SS$ is everywhere dense in $\HH$.
Hence $\HH$ is separable.
Category:Hilbert Space Direct Sums
Hilbert Space Isomorphism is Bijection

== Theorem ==
Let $H, K$ be Hilbert spaces.
Denote by $\innerprod \cdot \cdot_H$ and $\innerprod \cdot \cdot_K$ their respective inner products.
Let $U: H \to K$ be an isomorphism.
Then $U$ is a bijection.
== Proof ==
As $U$ is an isomorphism, it is necessarily surjective.
Suppose now that $g, h \in H$ are such that $\map U g = \map U h$.
Then as $U$ is a linear map, it follows that $\map U {g - h} = \bszero_K$, the zero vector of $K$.
From property $(3)$ of an isomorphism, conclude that:
:$0 = \innerprod {\bszero_K} {\bszero_K}_K = \innerprod {\map U {g - h} } {\map U {g - h} }_K = \innerprod {g - h} {g - h}_H$
Property $(5)$ of an inner product ensures us that $g - h = \bszero_H$, that is $g = h$.
Hence $U$ is injective.
As $U$ is both injective and surjective, it is a bijection.
Category:Hilbert Spaces
Hilbert Space Isomorphism is Equivalence Relation

== Theorem ==
Hilbert space isomorphism is an equivalence relation.
== Proof ==
Checking the three defining properties of an equivalence relation in turn:
=== Reflexivity ===
For any Hilbert space $H$, we have the identity map $I_H: H \to H$.
Identity Mapping is Linear yields $I_H$ to be a linear map.
Also, Identity Mapping is Bijection ensures that $I_H$ is a surjection.
Lastly, for any $g, h \in H$, we have:
:$\innerprod g h = \innerprod {\map {I_H} g} {\map {I_H} h}$
Therefore, the three defining properties for a Hilbert space isomorphism are satisfied by $I_H$.
Hence, any Hilbert space $H$ is isomorphic to itself.
It follows that Hilbert space isomorphism is reflexive.
=== Symmetry ===
Let $H, K$ be isomorphic Hilbert spaces, and let $U: H \to K$ be a Hilbert space isomorphism.
From Hilbert Space Isomorphism is Bijection and the definition of bijection, it follows that $U$ has a two-sided inverse, $U^{-1}: K \to H$.
Inverse of Linear Map is Linear yields $U^{-1}$ to be a linear map as well.
Bijection iff Inverse is Bijection yields that $U^{-1}$ is also a surjection.
For $k, l \in K$, the computation $\innerprod k l _K = \innerprod {U U^{-1}k} {U U^{-1} l}_K = \innerprod {U^{-1} k} {U^{-1} l}_H$ shows that $U^{-1}$ preserves the inner product (where in the last step, the fact that $U$ is an isomorphism is used).
Hence, $U^{-1}$ is an isomorphism, so $K$ and $H$ are also isomorphic.
It follows that Hilbert space isomorphism is symmetric.
=== Transitivity ===
Let $H, K, L$ be Hilbert spaces, and let $U: H \to K, V: K \to L$ be Hilbert space isomorphisms.
Consider the composition $VU : H \to L$.
From Composition of Linear Maps is Linear, $VU$ is a linear map.
Composite of Surjections is Surjection yields $VU$ to be a surjection.
Lastly, for $g, h \in H$, we have:
:$\innerprod g h_H = \innerprod {U g} {U h}_K = \innerprod {V U g} {V U h}_L$
Hence $VU$ is a Hilbert space isomorphism, and $H$ and $L$ are isomorphic.
It follows that Hilbert space isomorphism is transitive.
Having verified these three conditions, it follows that Hilbert space isomorphism is an equivalence relation, 
== Sources ==
* : $\S \text I.5$
Category:Hilbert Spaces
Category:Examples of Equivalence Relations
Hilbert Space Isomorphism preserves Orthocomplements

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $T : \HH \to \KK$ be a Hilbert space isomorphism.
Let $U \subseteq \HH$.
Then:
:$T \sqbrk {U^\bot} = \paren {T \sqbrk U}^\bot$
== Proof ==
Let $x \in \KK$.
Since $T$ is a Hilbert space isomorphism, there exists $v \in \HH$ such that $x = T v$.
We then have $x \in \paren {T \sqbrk U}^\bot$ :
:$\innerprod x {T u}_\KK = 0$ for each $u \in U$.
That is:
:$\innerprod {T v} {T u}_\KK = 0$
Since $T$ is a Hilbert space isomorphism, this is equivalent to $\innerprod v u_\HH = 0$ for each $u \in U$.
So this is equivalent to $v \in U^\bot$.
Since $x = T v$, this is equivalent to $x \in T \sqbrk {U^\bot}$. 
So we obtain:
:$T \sqbrk {U^\bot} = \paren {T \sqbrk U}^\bot$
Category:Orthocomplements
Category:Hilbert Spaces
Hilbert Space Projection is *-Algebra Projection

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B \HH$ be the space of bounded linear operators on $\HH$ understood as a $\text C^\ast$-algebra.
Let $P : \HH \to \HH$ be a Hilbert space projection.
Then $P$ is a $\ast$-algebra projection.
== Proof ==
From Characterization of Projections, $P$ is Hermitian.
By definition, $P$ is an idempotent operator so that $P^2 = P$.
So $P$ is a $\ast$-algebra projection.
Category:Projections (*-Algebras)
Hilbert Space Projections with Zero Product are Pointwise Orthogonal

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $P, Q : \HH \to \HH$ be projections (in the Hilbert space sense) such that:
:$P Q = 0$ 
Then:
:$\innerprod {Q x} {P y} = 0$ for each $x, y \in \HH$.
== Proof ==
We have, for each $x \in \HH$:
:$\map {\paren {P Q} } x = {\mathbf 0}_\HH$
Hence:
:$\map P {Q x} = {\mathbf 0}_\HH$
So we have $Q x \in \ker P$.
Since $x \in \HH$ was arbitrary we have $\Rng Q \subseteq \ker P$.
From the definition of a projections (in the Hilbert space sense), we have:
:$\ker P = \paren {\Rng Q}^\bot$
Hence:
:$\Rng Q \subseteq \paren {\Rng P}^\bot$
Hence for each $x \in \HH$, we have $Q x \in \paren {\Rng Q}^\bot$.
Hence for each $y \in \HH$ we have $Q x \perp P y$ from the definition of orthocomplement.
That is, $\innerprod {Q x} {P y} = 0$. 
Category:Orthogonal Projections
Hilbert Space is Hausdorff Topological Vector Space

== Theorem ==
Let $\struct {H, \innerprod \cdot \cdot}$ be a Hilbert space over $\mathbb F \in \set {\R, \C}$.
Let $d: H \times H \to \R_{\ge 0}$ be the metric induced by the inner product norm $\norm {\,\cdot\,}$ on $H$.
Let $\tau$ be the the topology on $H$ induced by the metric $d$.
Then $\struct {H, \tau}$ is a Hausdorff topological vector space.
== Proof ==
If $\mathbb F = \R$, then Real Numbers form Field shows that $\mathbb F$ is a field.
From Euclidean Space is Normed Vector Space, it follows for $r \in \mathbb F$ that:
:$\size r = \sqrt {r^2}$
is a norm on $\mathbb F$.
If $\mathbb F = \C$, then Complex Numbers form Field shows that $\mathbb F$ is a field.
From Complex Modulus is Norm, it follows for $z \in \mathbb F$ that:
:$\cmod z$
is a norm on $\mathbb F$.
In both cases, it follows from definition of valued field that $\mathbb F$ is a normed division ring.
By definition of normed vector space, $\struct {H, \norm {\,\cdot\,} }$ is a normed vector space.
The result follows from Normed Vector Space is Hausdorff Topological Vector Space.
Category:Hilbert Spaces
Category:Hausdorff Topological Vector Spaces
Hinge Theorem

== Theorem ==
If two triangles have two pairs of sides which are the same length, the triangle with the larger included angle also has the larger third side.
== Proof ==
:
Let $\triangle ABC$ and $DEF$ be two triangles in which $AB = DE$, $AC = DF$, and $\angle CAB > \angle FDE$.
Construct $\angle EDG$ on $DE$ at point $D$.
Place $G$ so that $DG = AC$.
Join $EG$ and $FG$.
Since $AB = DE$, $\angle BAC = \angle EDG$, and $AC = DG$, by Triangle Side-Angle-Side Congruence:
:$BC = GE$
By Euclid's first common notion:
:$DG = AC = DF$
Thus by Isosceles Triangle has Two Equal Angles:
:$\angle DGF = \angle DFG$
So by Euclid's fifth common notion:
:$\angle EFG \, > \, \angle DFG = \angle DGF \, > \, \angle EGF$
Since $\angle EFG > \angle EGF$, by Greater Angle of Triangle Subtended by Greater Side:
:$EG > EF$
Therefore, because $EG = BC$, $BC > EF$.
== Also known as ==
This theorem is also known as the side-angle side inequality theorem, or SAS inequality theorem.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.19$
Category:Triangles
Category:Named Theorems
Hodge Conjecture

== Conjecture ==
It is conjectured that:
:For projective algebraic varieties, Hodge cycles are rational linear combinations of algebraic cycles.
== Progress ==
== Sources ==
* 
* 
* : Appendix $18$: Millennium Prize problems: $2$.
* : Appendix $23$: Millennium Prize problems: $2$.
Category:Millennium Problems
Category:Open Questions
Holomorphic 1-forms are closed

==Theorem==
Holomorphic 1-forms are closed. 
==Proof==
Let $\omega$ be a holomorphic 1-form.
By wedge product of differential forms is antisymmetric
:$\rd z \wedge\rd z=0$
By Cauchy-Riemann Equations
:$\frac{\partial f}{\partial \bar{z}}=0$
Then:
:$\rd \omega=\rd(f(z)\rd z)=\frac{\partial f}{\partial z}\rd z \wedge\rd z+\frac{\partial f}{\partial \bar{z}}\rd \bar{z} \wedge\rd z=0+0=0$
so $\omega$ is closed. 
== Also see ==
Category:1-Forms
==Source==
* 2. Basic Notions
Holomorphic Function is Analytic

== Theorem == 
Let $a \in \C$ be a complex number.
Let $r > 0$ be a real number.
Let $f$ be a function holomorphic on some open ball $D = \map B {a, r}$. 
Then $f$ is complex analytic on $D$.
== Proof == 
Let $z \in D$.
Then: 
Note that for all $t \in \partial D$, we have:
:$\cmod {z - a} < \cmod {t - a} = r$.
Therefore:
:$\cmod {\dfrac {z - a} {t - a} } < 1$
so Sum of Infinite Geometric Sequence may be applied.
This gives:
From Continuous Function on Compact Space is Bounded, there exists a real $M \ge 0$, such that:
:$\forall t \in \partial D: \cmod {\map f t} \le M$
As $\cmod {\dfrac {z - a} {t - a} } < 1$, there exists a real $0 \le N < 1$ such that:
:$\cmod {\dfrac {z - a} {t - a} } \le N$
Then: 
:$\cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \map f t} = \cmod {\dfrac 1 {t - a} } \cmod {\dfrac {\paren {z - a}^n} {\paren {t - a}^n} } \cmod {\map f t} \le \dfrac 1 r M N^n$ 
As $N < 1$, we have that $\ds \frac 1 r \sum_{n \mathop = 0}^\infty M N^n$ converges by Sum of Infinite Geometric Sequence. 
Therefore by the Weierstrass M-Test, we have that: 
:$\ds \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} }$ 
converges uniformly on $D$.
Therefore: 
:$\ds \frac 1 {2 \pi i} \int_{\partial D} \sum_{n \mathop = 0}^\infty \map f t \frac {\paren {z - a}^n} {\paren {t - a}^{n + 1} } \rd t = \sum_{n \mathop = 0}^\infty \paren {z - a}^n \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t}$
Let: 
:$\ds c_n = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t$
Then: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n$ 
so $f$ is complex analytic on $D$. 
== Sources ==
* 
Category:Holomorphic Functions
Category:Analytic Functions
Holomorphic Function is Continuously Differentiable

== Theorem ==
Let $U \subseteq \C$ be an open set.
Let $f : U \to \C$ be a holomorphic function.
Then $f$ is continuously differentiable in $U$.
== Proof ==
Let $z \in U$.
The derivatives $\map {f'} z$ and $\map {f} z$ exist by Cauchy's Integral Formula for Derivatives.
Therefore $f'$ is continuous at $z$ by Complex-Differentiable Function is Continuous.
== Sources ==
* : $\S 4.2$
Category:Holomorphic Functions
Holomorphic Function is Identified by Real or Imaginary Part

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f, g : D \to \C$ be holomorphic functions such that either:
:$\map \Re f = \map \Re g$
or:
:$\map \Im f = \map \Im g$
Then $f = g$. 
== Proof ==
From Combination Theorem for Complex Derivatives: Combined Sum Rule, we have:
:$f - g$ is holomorphic.
Suppose first that:
:$\map \Re f = \map \Re g$
Then:
:$\map \Re {f - g} = 0$
Hence from , we have $f = g$. 
Suppose also that:
:$\map \Im f = \map \Im g$
Then:
:$\map \Im {f - g} = 0$
Hence from Holomorphic Function with Constant Imaginary Part is Constant, we have $f = g$. 
Category:Holomorphic Functions
Holomorphic Function with Constant Imaginary Part is Constant

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a holomorphic function such that $\map \Im f$ is constant.
Then $f$ is constant.
== Proof ==
Define $u : \R^2 \to \R$ and $v : \R^2 \to \R$ such that:
:$\map f {x + i y} = \map u {x, y} + i \map v {x, y}$ 
for each $x, y \in \R$.
By hypothesis, $v$ is constant.
Hence from Derivative of Constant, we have:
:$\ds \frac {\partial v} {\partial x} = 0$
and:
:$\ds \frac {\partial v} {\partial y} = 0$
Hence by the Cauchy-Riemann Equations, we have:
:$\ds \frac {\partial u} {\partial x} = 0$
and:
:$\ds \frac {\partial u} {\partial y} = 0$
Applying Zero Derivative implies Constant Function to the function $x \mapsto \map u {x, y}$ for fixed $y \in D$, we obtain:
:$\map u {x, y} = \map F y$ for some $\map F y \in \C$
from the first equation.
We can apply Zero Derivative implies Constant Function to the function $y \mapsto \map u {x, y}$ for fixed $x \in D$ to obtain:
:$\map u {x, y} = \map G x$ for some $\map G x \in \C$
from the second equation. 
We have $\map F y = \map G x$ for each $x, y \in D$. 
Fixing $x_0 \in D$ and setting $c = \map G {x_0}$, we have $\map F y = c$ for all $y \in \C$, and so $u$ is constant. 
Since both $u$ and $v$ are constant, $f$ is constant.
Category:Holomorphic Functions
Category:Holomorphic Function with Constant Imaginary Part is Constant
Holomorphic Function with Constant Imaginary Part is Constant/Corollary

== Theorem ==
Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.
Let $f: D \to \C$ be a holomorphic function such that $\map \Re f$ is constant.
Then $f$ is constant.
== Proof ==
From Combination Theorem for Complex Derivatives: Multiple Rule, $i f$ is holomorphic.
From Imaginary Part of Imaginary Unit times Element of *-Algebra, we have:
:$\map \Im {i f} = \map \Re f$ 
Hence $\map \Im {i f}$ is constant.
From Holomorphic Function with Constant Imaginary Part is Constant, there exists $c \in \C$ such that:
:$i \map f z = c$
for all $z \in D$.
Hence $\map f z = -i c$ for all $z \in D$.
So $f$ is constant.
Category:Holomorphic Function with Constant Imaginary Part is Constant
Homeomorphic Image of Local Basis is Local Basis

== Theorem ==
Let $T_\alpha = \struct{S_\alpha, \tau_\alpha}$ and $T_\beta = \struct{S_\beta, \tau_\beta}$ be topological spaces.
Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.
Let $s \in S_\alpha$.
Let $\BB$ be a local basis of $s$ in $T_\alpha$.
Then:
:$\BB' = \set{ \phi \sqbrk B : B \in \BB}$ is a local basis of $\map \phi s$ in $T_\beta$
== Proof ==
By definition of homeomorphism:
:$\forall U \in \tau_\alpha : \phi \sqbrk U \in \tau_\beta$
Hence:
:$\BB'$ is a set of open sets in $T_\beta$ containing $\map \phi s$
Let $U \in \tau_\beta$ containing $\map \phi s$.
By definition of homeomorphism:
:$\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$
By definition of local basis:
:$\exists V \in \BB: s \in V \subseteq \phi^{-1} \sqbrk U$
By definition of image of subset:
:$\map \phi s \in \phi \sqbrk V$
From Subset Maps to Subset:
:$\phi \sqbrk V \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$
From Image of Preimage under Mapping:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$
From Subset Relation is Transitive]:
:$\phi \sqbrk V \subseteq U$
Hence:
:$\exists \phi \sqbrk V \in \BB': \map \phi s \in \phi \sqbrk V \subseteq U$
== Also see ==
* Homeomorphic Image of Neighborhood Basis is Neighborhood Basis
Category:Homeomorphisms (Topological Spaces)
Category:Local Bases
Homeomorphic Image of Meager Set is Meager

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A \subseteq X$ be meager in $X$.
Then $f \sqbrk A$ is meager in $Y$. 
== Proof ==
Since $A$ is meager in $X$, there exists a countable set $\family {A_n}_{n \in \N}$ of nowhere dense sets in $X$ such that:
:$\ds A = \bigcup_{n \mathop = 1}^\infty A_n$
So, we have:
:$\ds f \sqbrk A = f \sqbrk {\bigcup_{n \mathop = 1}^\infty A_n}$
So, from Image of Union under Mapping:
:$\ds f \sqbrk A = \bigcup_{n \mathop = 1}^\infty f \sqbrk {A_n}$ 
From Homeomorphic Image of Nowhere Dense Set is Nowhere Dense, $f \sqbrk {A_n}$ is nowhere dense in $Y$ for each $n \in \N$. 
Hence $f \sqbrk A$ is the countable union of nowhere dense sets in $Y$.
So $f \sqbrk A$ is meager in $Y$. 
Category:Meager Spaces
Homeomorphic Image of Neighborhood Basis is Neighborhood Basis

== Theorem ==
Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.
Let $s \in S_\alpha$.
Let $\NN$ be a neighborhood basis of $s$ in $T_\alpha$.
Then:
:$\NN' = \set {\phi \sqbrk N : N \in \NN}$ is a neighborhood basis of $\map \phi s$ in $T_\beta$.
== Proof ==
Let $N$ be a neighborhood of $s$ in $T_\alpha$.
By definition of neighborhood:
:$\exists V \in \tau_\alpha : s \in V \subseteq N$
By definition of image of subset:
:$\map \phi s \in \phi \sqbrk V$
From Subset Maps to Subset:
:$\phi \sqbrk V \subseteq \phi \sqbrk N$
By definition of homeomorphism:
:$\phi \sqbrk V \in \tau_\beta$
It follows that $\phi \sqbrk N$ is a neighborhood of $\map \phi s$ in $T_\beta$.
Hence:
:$\NN' = \set{ \phi \sqbrk N : N \in \NN}$ is a set of neighborhoods of $\map \phi s$ in $T_\beta$
Let $M$ be a neighborhood of $\map \phi s$ in $T_\beta$.
By definition of neighborhood:
:$\exists U \in \tau_\beta : \map \phi s \in U \subseteq M$
By definition of homeomorphism:
:$\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$
By definition of neighborhood basis:
:$\exists N \in \NN: s \in N \subseteq \phi^{-1} \sqbrk U$
By definition of image of subset:
:$\map \phi s \in \phi \sqbrk N$
From Subset Maps to Subset:
:$\phi \sqbrk N \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$
From Image of Preimage under Mapping:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$
From Subset Relation is Transitive:
:$\phi \sqbrk N \subseteq M$
Hence:
:$\exists \phi \sqbrk N \in \BB': \map \phi s \in \phi \sqbrk N \subseteq M$
The result follows.
== Also see ==
* Homeomorphic Image of Local Basis is Local Basis
Category:Homeomorphisms (Topological Spaces)
Category:Neighborhood Bases
Homeomorphic Image of Nowhere Dense Set is Nowhere Dense

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $f : X \to Y$ be a homeomorphism.
Let $A$ be a nowhere dense subset of $X$.
Then $f \sqbrk A$ is a nowhere dense subset of $Y$.
== Proof ==
Since $A$ is nowhere dense, we have:
:$\paren {A^-}^\circ = \O$
where $A^-$ is the closure of $A$ and $\paren {A^-}^\circ$ is the interior of $A^-$.
From Image of Empty Set is Empty Set, we have:
:$\O = f \sqbrk {\paren {A^-}^\circ}$
So, we have:
Hence $f \sqbrk A$ is a nowhere dense subset of $Y$.
Category:Nowhere Denseness
Category:Homeomorphisms (Topological Spaces)
Homeomorphic Image of Sub-Basis is Sub-Basis

== Theorem ==
Let $T_\alpha = \struct{X_\alpha, \tau_\alpha}, T_\beta = \struct{X_\beta, \tau_\beta}$ be topological spaces.
Let $\SS_\alpha \subseteq \tau_\alpha$ be a sub-basis for $\tau_\alpha$.
Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.
Let $\SS_\beta = \set{\phi \sqbrk S : S \in \SS_\alpha}$.
Then:
:$\SS_\beta$ is a sub-basis for $\tau_\beta$ 
== Proof ==
By definition of homeomorphism:
:$\forall U \subseteq X_\alpha : U \in \tau_\alpha \iff \phi \sqbrk U \in \tau_\beta$
By definition of sub-basis:
:$\SS_\alpha \subseteq \tau_\alpha$
Hence:
:$\SS_\beta \subseteq \tau_\beta$
Let $\ds \BB_\alpha = \set {\bigcap \FF: \FF \subseteq \SS_\alpha, \FF \text{ is finite} }$.
Let $\ds \BB_\beta = \set {\bigcap \GG: \GG \subseteq \SS_\beta, \GG \text{ is finite} }$.
Let $B \in \BB_\alpha$.
By definition of $\BB_\alpha$:
:$\exists \FF \subseteq \BB_\alpha : \FF \text{ is finite} : B = \bigcap \set{S : S \in \FF}$
We have:
Let $\GG = \set{\phi \sqbrk S : S \in \FF}$.
Then:
:$\GG \subseteq \BB_\beta : \GG \text{ is finite} : \phi \sqbrk B = \bigcap \GG$
Hence $\phi \sqbrk B \in \BB_\beta$.
We have shown that:
:$(1): \quad \forall B \in \BB_\alpha : \phi \sqbrk B \in \BB_\beta$
Let $V \in \tau_\beta$.
From Inverse of Homeomorphism is Homeomorphism:
:$\phi^{-1}$ is a homeomorphism
By definition of homeomorphism:
:$\phi^{-1} \sqbrk V \in \tau_\alpha$
By definition of sub-basis:
We have:
Let $\AA' = \set{\phi \sqbrk B : B \in \BB_\alpha}$
From $(1)$ above:
:$\AA' \subseteq \BB_\beta$
We have:
:$V = \bigcup \AA'$
Hence by definition, $\SS_\beta$ is a sub-basis for $\tau_\beta$
Category:Homeomorphisms (Topological Spaces)
Category:Topological Bases
Homeomorphic Metric Spaces are not necessarily Isometric

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be homeomorphic.
Then it is not necessarily the case that $M_1$ and $M_2$ are isometric.
== Proof ==
Consider the spaces $\struct {\openint 0 4, d}$ and $\struct {\openint 0 1, d}$, where $d$ is the Euclidean Metric.
By Open Real Intervals are Homeomorphic, they are homeomorphic.
Let $\phi: \openint 0 4 \to \openint 0 1$ be a mapping.
Then:
showing that $\phi$ cannot be an isometry.
Therefore the two spaces above are not isometric.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces
Category:Isometries (Metric Spaces)
Category:Homeomorphisms (Metric Spaces)
Homeomorphic Non-Comparable Particular Point Topologies

== Theorem ==
Let $S$ be a set with at least two elements.
Let $p, q \in S: p \ne q$.
Let $\tau_p$ and $\tau_q$ be the particular point topologies on $S$ by $p$ and $q$ respectively.
Then the topological spaces $T_p = \struct {S, \tau_p}$ and $T_q = \struct {S, \tau_q}$ are homeomorphic.
However, $\tau_p$ and $\tau_q$ are not comparable.
== Proof ==
We can set up the mapping $\phi: S \to S$:
:$\forall x \in S: \map \phi x = \begin {cases}
q & : x = p \\
p & : x = q \\
x & : \text {otherwise} \end {cases}$
It is straightforward to show that $\phi$ is a homeomorphism.
However, we have, for example, that $\set q \notin \tau_p$ and $\set p \notin \tau_q$.
So neither $\tau_p$ nor $\tau_q$ are a subset of the other.
Hence by definition $\tau_p$ and $\tau_q$ are not comparable. 
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $8$
Category:Homeomorphisms (Topological Spaces)
Category:Particular Point Topology
Homeomorphic Topologies on Same Set may not be Identical

== Theorem ==
Let $S$ be a set.
Let $\tau_1$ and $\tau_2$ both be topologies on $S$ such that the topological spaces $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ are homeomorphic.
Then it is not necessarily the case that $\struct {S, \tau_1} = \struct {S, \tau_2}$.
== Proof ==
Let $p, q \in S$ such that $p \ne q$.
Let $\tau_p$ and $\tau_q$ be the particular point topologies on $S$ by $p$ and $q$ respectively.
From Homeomorphic Non-Comparable Particular Point Topologies, $\struct {S, \tau_p}$ and $\struct {S, \tau_q}$ are homeomorphic.
But $\struct {S, \tau_p} \ne \struct {S, \tau_q}$, as $\set p \in \tau_p$ but $\set p \notin \tau_q$, and similarly $\set q \in \tau_q$ but $\set q \notin \tau_p$.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
Category:Homeomorphisms (Topological Spaces)