nyuuzyou/wikis huggingface dataset:
Event/Examples/Prime Number on 6-Sided Die

== Example of Event ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die.
The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.
Consider the subset $E \subseteq \Omega$ defined as:
:$E = \set {2, 3, 5}$
Then $E$ is the event that the result of $\EE$ is a prime number.
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events
Category:Examples of Events
Event/Examples/Score Divisible by 3 on 6-Sided Die

== Example of Event ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die.
Let $E \subseteq \Omega$ be the event that the result of $\EE$ is divisible by $3$.
The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.
Hence:
:$E = \set {3, 6}$
and so:
:$\map \Pr E = \dfrac 1 3$
== Sources ==
* 
* 
Category:Examples of Events
Event/Examples/Tossing 2 Coins

== Examples of Events ==
Let $\EE$ be the experiment consisting of tossing $2$ coins.
From Tossing $2$ Coins, the sample space of $\EE$ is:
:$\Omega = \set {\tuple {\mathrm H, \mathrm H}, \tuple {\mathrm H, \mathrm T}, \tuple {\mathrm T, \mathrm H}, \tuple {\mathrm T, \mathrm T} }$
where $\mathrm H$ denotes heads and $\mathrm T$ denotes tails.
Let $A$ be the subset of $\Omega$ defined as:
:$A = \set {\tuple {\mathrm H, \mathrm H}, \tuple {\mathrm H, \mathrm T}, \tuple {\mathrm T, \mathrm H} }$
Let $B$ be the subset of $\Omega$ defined as:
:$A = \set {\tuple {\mathrm H, \mathrm H} }$
Then:
:$A$ is the event that there is at least one head
:$B$ is the event that there are two heads.
== Sources ==
* : Chapter $2$: Probability and Discrete Sample Spaces: $2.2$ Sample spaces and events
Category:Examples of Events
Event Independence is Symmetric

== Theorem ==
Let $A$ and $B$ be events in a probability space.
Let $A$ be independent of $B$.
Then $B$ is independent of $A$.
That is, is independent of is a symmetric relation.
== Proof ==
We assume throughout that $\map \Pr A > 0$ and $\map \Pr B > 0$.
Let $A$ be independent of $B$.
Then by definition:
:$\condprob A B = \map \Pr A$
From the definition of conditional probabilities, we have:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
and also:
:$\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
So if $\condprob A B = \map \Pr A$ we have:
So by definition, $B$ is independent of $A$.
== Sources ==
* : $\S 1.7$: Independent Events: Exercise $20$
Category:Independent Events
Event Space/Examples/Arbitrary Event Space on 6-Sided Die

== Example of Event Space ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die.
The sample space of $\EE$ is $\Omega = \set {1, 2, 3, 4, 5, 6}$.
Let $\FF$ be the arbitrary set of subsets of $\Omega$ defined as:
:$\FF = \set {\O, \set {1, 2}, \set {3, 4}, \set {5, 6}, \set {1, 2, 3, 4}, \set {3, 4, 5, 6}, \set {1, 2, 5, 6}, \Omega}$
Then $E$ is an event space of $\EE$.
== Proof ==
;Event Space Axiom $(\text {ES} 1)$:
It is specified that $\O \in \FF$.
Thus axiom $(\text {ES} 1)$ is fulfilled.
;Event Space Axiom $(\text {ES} 2)$:
It is specified that $\Omega \in \FF$.
Thus axiom $(\text {ES} 2)$ is fulfilled.
;Event Space Axiom $(\text {ES} 3)$:
We investigate the unions of elements of $\Omega$.
From Union with Superset is Superset, it is unnecessary to investigate the union of any element of $\Omega$ with a subset of it.
We continue:
All the event space axioms are seen to be fulfilled by $\powerset \Omega$.
Hence the result.
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Example $6$
Category:Examples of Event Spaces
Event Space contains Empty Set

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
The event space $\Sigma$ of $\EE$ has the property that:
:$\O \in \Sigma$
That is, the empty set is in the event space.
== Proof ==
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events
Category:Event Spaces
Event Space contains Sample Space

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
The event space $\Sigma$ of $\EE$ has the property that:
:$\Omega \in \Sigma$
where $\Omega$ is the sample space of $\EE$.
== Proof ==
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events
Category:Event Spaces
Event Space from Single Subset of Sample Space

== Theorem ==
Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\O \subsetneqq A \subsetneqq \Omega$.
Then $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ is an event space of $\EE$.
== Proof ==
;Event Space Axiom $(\text {ES} 1)$:
From its definition:
:$\Sigma \ne \O$
thus fulfilling axiom $(\text {ES} 1)$.
;Event Space Axiom $(\text {ES} 2)$:
From Set Difference with Empty Set is Self:
:$\Omega \setminus \O = \Omega \in \Sigma$
From Set Difference with Self is Empty Set:
:$\Omega \setminus \Omega = \O \in \Sigma$
By definition:
:$\Omega \setminus A \in \Sigma$
From Relative Complement of Relative Complement:
:$\Omega \setminus \paren {\Omega \setminus A} = A \in \Sigma$
Thus axiom $(\text {ES} 2)$ is fulfilled.
;Event Space Axiom $(\text {ES} 3)$:
From Union with Empty Set:
:$\forall X \in \Sigma: X \cup \O = X \in \Sigma$
From Union with Superset is Superset:
:$\forall X \in \Sigma: X \cup \Sigma = \Sigma \in \Sigma$
From Set Union is Idempotent:
:$\forall X \in \Sigma: X \cup X = X \in \Sigma$
From Union with Relative Complement:
:$A \cup \paren {\Omega \setminus A} = \Sigma \in \Sigma$
It follows that axiom $(\text {ES} 3)$ is fulfilled.
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Example $5$
Category:Probability Theory
Event Space of Experiment with Final Sample Space has Even Cardinality

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $\Omega$ be a finite set.
Then the event space $\Sigma$ consists of an even number of subsets of $\Omega$.
== Proof ==
Let $A \in \Sigma$ be one of the events of $\EE$.
We have by definition that $\Omega$ is itself an events of $\EE$.
Hence by Set Difference of Events is Event, $\Omega \setminus A$ is also an event of $\EE$.
As $A$ is arbitrary, the same applies to all events of $\EE$.
Thus all events of $\EE$ come in pairs: $A$ and $\Omega \setminus A$.
Hence the result.
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $5$
Category:Event Spaces
Event of Stopping Time Equal to Infinity is Measurable in Limit of Filtration/Continuous Time

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\sequence {\FF_t}_{t \ge 0}$ be a continuous-time filtration of $\Sigma$.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\FF_\infty$ be the limit of the filtration $\sequence {\FF_t}_{t \ge 0}$.
Then:
:$\set {\omega \in \Omega : \map T \omega = \infty} \in \FF_\infty$
== Proof ==
Note that we have $\map T \omega < \infty$ $\map T \omega \le N$ for some $N \in \N$. 
So, we have: 
:$\ds \set {\omega \in \Omega : \map T \omega < \infty} = \bigcup_{N \in \N} \set {\omega \in \Omega : \map T \omega \le N}$
Since $T$ is a stopping time, we have: 
:$\set {\omega \in \Omega : \map T \omega \le N} \in \FF_N$
For each $N \in \N$, we have: 
:$\ds \FF_N \subseteq \bigcup_{t \in \hointl 0 \infty} \FF_t$ 
from Set is Subset of Union, so that:
:$\FF_N \subseteq \FF_\infty$
from the definition of the $\sigma$-algebra generated by collection of subsets.
Now, since $\sigma$-algebras are closed under countable union, we have:
:$\ds \bigcup_{N \in \N} \set {\omega \in \Omega : \map T \omega \le N} \in \FF_\infty$ 
So:
:$\set {\omega \in \Omega : \map T \omega < \infty} \in \FF_\infty$
Then since $\FF_\infty$ is closed under relative complement, we have: 
:$\set {\omega \in \Omega : \map T \omega = \infty} \in \FF_\infty$
== Sources ==
* : $3.2$: Stopping Times and Associated $\sigma$-Fields
Category:Event of Stopping Time Equal to Infinity is Measurable in Limit of Filtration
Event of Stopping Time Equal to Infinity is Measurable in Limit of Filtration/Discrete Time

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\sequence {\FF_n}_{n \ge 0}$ be a discrete-time filtration of $\Sigma$.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\FF_\infty$ be the limit of the filtration $\sequence {\FF_n}_{n \ge 0}$.
Then:
:$\set {\omega \in \Omega : \map T \omega = \infty} \in \FF_\infty$
== Proof ==
Note that we have: 
From the definition of a stopping time, we have:
:$\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$
for each $t \in \Z_{\ge 0}$.
For each $t \in \Z_{\ge 0}$, we have:
:$\ds \FF_t \subseteq \bigcup_{s \in \Z_{\ge 0} } \FF_s$
from Set is Subset of Union, so that:
:$\FF_t \subseteq \FF_\infty$ 
from the definition of the $\sigma$-algebra generated by collection of subsets.
Then we have: 
:$\set {\omega \in \Omega : \map T \omega = t} \in \FF_\infty$
for each $t \in \Z_{\ge 0}$.
Now, since $\sigma$-algebras are closed under countable union, we have:
:$\ds \bigcup_{t \in \Z_{\ge 0} } \set {\omega \in \Omega : \map T \omega = t} \in \FF_\infty$ 
So: 
:$\set {\omega \in \Omega : \map T \omega < \infty} \in \FF_\infty$
Then since $\FF_\infty$ is closed under relative complement, we have: 
:$\set {\omega \in \Omega : \map T \omega = \infty} \in \FF_\infty$
Category:Event of Stopping Time Equal to Infinity is Measurable in Limit of Filtration
Events One of Which equals Intersection

== Theorem ==
Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.
Let $A$ and $B$ be such that:
:$A \cap B = A$
Then whenever $A$ occurs, it is always the case that $B$ occurs as well.
== Proof ==
From Intersection with Subset is Subset:
:$A \cap B = A \iff A \subseteq B$
Let $A$ occur.
Let $\omega$ be the outcome of $\EE$.
Let $\omega \in A$.
That is, by definition of occurrence of event, $A$ occurs.
Then by definition of subset:
:$\omega \in B$
Thus by definition of occurrence of event, $B$ occurs.
Hence the result.
== Examples ==
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $2$
Category:Intersections of Events
Category:Events One of Which equals Intersection
Events One of Which equals Intersection/Examples/Target of Concentric Circles

== Examples of Use of Events One of Which equals Union ==
Let $T$ be a target which consists of $10$ concentric circles $C_1$ to $C_{10}$ whose radii are respectively $r_k$ for $k = 1, 2, \ldots, 10$.
Let $r_k < r_{k + 1}$ for all $k = 1, 2, \ldots, 9$.
That is, let $C_1$ be the innermost and $C_{10}$ be the outermost.
Let $A_k$ denote the event of hitting $T$ inside the circle of radius $r_k$.
Let $C$ denote the event:
:$C = \ds \bigcap_{k \mathop = 5}^{10} A_k$
Then $C$ is the event of hitting $T$ inside circle $C_5$.
== Proof ==
By the geometry of the situation:
:$C_5 \subseteq C_6 \subseteq \cdots \subseteq C_{10}$
By Events One of Which equals Intersection:
The result follows.
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $3$
::Note that the question does not state whether $r_k < r_{k + 1}$ for all $k = 1, 2, \ldots, 9$ or $r_k > r_{k + 1}$ for all $k = 1, 2, \ldots, 9$. The interpretation made here is the one which provides the correct answer.
Category:Events One of Which equals Union
Events One of Which equals Union

== Theorem ==
Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.
Let $A$ and $B$ be such that:
:$A \cup B = A$
Then whenever $B$ occurs, it is always the case that $A$ occurs as well.
== Proof ==
From Union with Superset is Superset:
:$A \cup B = A \iff B \subseteq A$
Let $B$ occur.
Let $\omega$ be the outcome of $\EE$.
Let $\omega \in B$.
That is, by definition of occurrence of event, $B$ occurs.
Then by definition of subset:
:$\omega \in A$
Thus by definition of occurrence of event, $A$ occurs.
Hence the result.
== Examples ==
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Example $1.1$
Category:Unions of Events
Category:Events One of Which equals Union
Events One of Which equals Union/Examples/Target of Concentric Circles

== Examples of Use of Events One of Which equals Union ==
Let $T$ be a target which consists of $10$ concentric circles $C_1$ to $C_{10}$ whose radii are respectively $r_k$ for $k = 1, 2, \ldots, 10$.
Let $r_k < r_{k + 1}$ for all $k = 1, 2, \ldots, 9$.
That is, let $C_1$ be the innermost and $C_{10}$ be the outermost.
Let $A_k$ denote the event of hitting $T$ inside the circle of radius $r_k$.
Let $B$ denote the event:
:$B = \displaystyle \bigcup_{k \mathop = 1}^6 A_k$
Then $B$ is the event of hitting $T$ inside circle $C_6$.
== Proof ==
By the geometry of the situation:
:$C_1 \subseteq C_2 \subseteq \cdots \subseteq C_6$
By Events One of Which equals Union:
The result follows.
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $3$
::Note that the question does not state whether $r_k < r_{k + 1}$ for all $k = 1, 2, \ldots, 9$ or $r_k > r_{k + 1}$ for all $k = 1, 2, \ldots, 9$. The interpretation made here is the one which provides the correct answer.
Category:Events One of Which equals Union
Events One of Which equals Union/Examples/Target of Concentric Circles/Mistake

== Source Work ==
:
:$\text I$ Random Events
::$1$. Relations among Random Events
:::Problem $3$
== Mistake ==
:A target consists of $10$ concentric circles of radius $r_k (k = 1, 2, 3, \ldots, 10)$. An event $A_k$ means hitting the interior of a circle of radius $r_k (k = 1, 2, \ldots, 10)$. What do the following events mean?
::$\ds B = \bigcup_{k \mathop = 1}^6 A_k, \qquad C = \prod_{k \mathop = 5}^{10} A_k$?
Note that in the above, $\ds C = \prod_{k \mathop = 5}^{10} A_k$ is the notation that uses for what we on would write $\ds C = \bigcap_{k \mathop = 5}^{10} A_k$.
== Correction ==
The question fails to state whether the circle radius $r_1$ or the circle radius $r_{10}$ is the innermost.
It is not obvious: it is feasible for the circles to have been numbered according to the score that the archer would achieve, in which case $r_{10}$ would be innermost.
However, that arrangement would be inconsistent with the answer given in the back of the book, and so it is apparent that:
:$r_k < r_{k + 1}$
for $k = 1, 2, \ldots, 9$. 
== Sources ==
* : $\text I$: Random Events: $1$. Relations among Random Events: Problem $3$
Category:Mistakes/Problems in Probability Theory, Mathematical Statistics and Theory of Random Functions
Eventually Constant Sequence Converges to Constant

== Theorem == 
Let $\struct {R, \norm{\,\cdot\,}}$ be a normed division ring.
Let $\lambda \in R$.
Let $\sequence {x_n}$ be a sequence in $R$ such that:
:$\exists N \in \R_{>0} : \forall n \ge N: x_n = \lambda$
Then:
:$\ds \lim_{n \mathop \to \infty} x_n = \lambda$
== Proof ==
Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n} }$ defined as:
:$\forall n: y_n = x_{N + n}$
The $\sequence {y_n}$ is the constant sequence $\tuple {\lambda, \lambda, \lambda, \dotsc}$.
Then:
Category:Sequences
Category:Normed Division Rings
Every Element is Directed and Every Two Elements are Included in Third Element implies Union is Directed

== Theorem ==
Let $P = \struct {S, \preceq}$ be an ordered set.
Let $A$ be a set of subsets of $S$.
Let
:$\forall X \in A: X$ is directed.
Let
:$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$
Then $\bigcup A$ is directed.
== Proof ==
Let $x, y \in \bigcup A$.
By definition of union:
:$\exists X \in A: x \in X$
and
:$\exists Y \in A: y \in Y$
By assumption:
:$\exists Z \in A: X \cup Y \subseteq Z$
By definition of union:
:$x, y \in X \cup Y$
By definition of subset:
:$x, y \in Z$
By assumption:
:$Z$ is directed.
By definition of directed subset:
:$\exists z \in Z: x \preceq z \land y \preceq z$
Thus by definition of union:
:$z \in \bigcup A$
Thus
:$x \preceq z \land y \preceq z$
Hence $\bigcup A$ is directed.
== Sources ==
* 
Category:Order Theory
Every Element is Lower implies Union is Lower

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $A$ be a set of subsets of $S$.
Let
:$\forall X \in A: X$ is a lower section.
Then $\bigcup A$ is a lower section.
== Proof ==
Let $x \in \bigcup A, y \in S$ such that:
:$y \preceq x$
By definition of union:
:$\exists X \in A: x \in X$
By assumption:
:$X$ is a lower section.
By definition of lower section:
:$y \in X$
Thus by definition of union:
:$y \in \bigcup A$
== Sources ==
* 
Category:Lower Sections
Every Filter has Limit Point implies Every Ultrafilter Converges

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let each filter on $S$ have a limit point in $S$.
Then each ultrafilter on $S$ converges to a point in $S$.
== Proof ==
Let $T = \struct {S, \tau}$ be such that each filter on $S$ has a limit point in $S$.
Let $\FF$ be an ultrafilter on $S$.
By hypothesis, $\FF$ has a limit point $x \in S$.
By Limit Point iff Superfilter Converges, there exists a filter $\FF'$ on $S$ which converges to $x$ satisfying $\FF \subseteq \FF'$.
By :
:$\FF$ is an ultrafilter whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$
Therefore:
:$\FF = \FF'$.
Thus $\FF$ converges to $x$.
== Also see ==
* Equivalence of Definitions of Compact Topological Space
Category:Topology
Every Operation is Distributive over Left Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\gets$ denote the left operation on $S$.
Then $\circ$ is distributive over $\gets$.
== Proof ==
Hence the result by definition of distributive.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(a)}$
Category:Left Operation
Category:Distributive Operations
Every Operation is Distributive over Right Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\to$ denote the right operation on $S$.
Then $\circ$ is distributive over $\to$.
== Proof ==
Hence the result by definition of distributive.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(a)}$
Category:Right Operation
Category:Distributive Operations
Every Point except Endpoint in Connected Linearly Ordered Space is Cut Point

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A \subseteq S$ be a connected space.
Let $p \in A$ be a point of $A$ which is not an endpoint of $A$.
Then $p$ is a cut point of $A$.
== Proof ==
We have that $A \setminus \set p$ is separated by $\set {x \in A: x \prec p}$ and $\set {x \in A: p \prec x}$.
If $p$ is an endpoint of $A$, then either:
:$\set {x \in A: x \prec p} = \O$
or:
:$\set {x \in A: p \prec x} = \O$
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $9$
Category:Linearly Ordered Spaces
Category:Connected Spaces
Category:Cut Points
Every Pseudoprime Element is Prime implies Lattice is Arithmetic

== Theorem ==
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below algebraic distributive lattice.
Assume that
:for every element $p$ of $S$ if $p$ is pseudoprime element, then $p$ is prime element.
Then $L$ is arithmetic.
== Proof ==
By If Every Element Pseudoprime is Prime then Way Below Relation is Multiplicative:
:$\ll$ is a multiplicative relation.
where $\ll$ denotes the way below relation.
Thus by Arithmetic iff Way Below Relation is Multiplicative in Algebraic Lattice:
:the result holds.
== Sources ==
* 
* 
Category:Prime Elements
Category:Continuous Lattices
Every Ultrafilter Converges implies Every Filter has Limit Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let every ultrafilter on $S$ be convergent.
Then every filter on $S$ has a limit point.
== Proof ==
Let $\FF$ be a filter on $S$.
By the Ultrafilter Lemma, there exists an ultrafilter $\FF'$ such that $\FF \subseteq \FF'$.
By hypothesis, $\FF'$ converges to some $x \in S$.
This, by Limit Point iff Superfilter Converges, implies that $x$ is a limit point of $\FF$.
== Also see ==
* Equivalence of Definitions of Compact Topological Space
Category:Filter Theory
Everywhere Dense Set determines Adjoint Linear Transformation

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \innerprod \cdot \cdot_X}$ and $\struct {Y, \innerprod \cdot \cdot_Y}$ be Hilbert spaces.
Let $\DD$ be an everywhere dense subset of $Y$. 
Let $T : X \to Y$ and $S : Y \to X$ be bounded linear transformations such that:
:$\innerprod {T x} y_Y = \innerprod x {S y}_X$ for each $x \in X$ and $y \in \DD$. 
Then $S = T^\ast$. 
== Proof ==
By the definition of the adjoint, we have:
:$\innerprod x {S y}_X = \innerprod {T x} y_Y = \innerprod x {T^\ast y}_X$ for each $x \in X$ and $y \in \DD$.
Let $y \in Y$.
Since $\DD$ is everywhere dense in $Y$, there exists a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $\DD$ such that $y_n \to y$.
Then:
:$\innerprod x {S y_n}_X = \innerprod x {T^\ast y_n}$ for each $x \in X$.
Taking $n \to \infty$ and using Inner Product is Continuous, we have:
:$\innerprod x {S y}_X = \innerprod x {T^\ast y}$ for each $x \in X$ and $y \in Y$.
So:
:$\innerprod x {S y - T^\ast y}_X = 0$ for each $x \in X$ and $y \in Y$. 
Hence $S y - T^\ast y \in X^\bot$. 
From Linear Subspace Dense iff Zero Orthocomplement, we have $X^\bot = \set { {\mathbf 0}_Y}$. 
Hence $S y - T^\ast y = {\mathbf 0}_Y$.
So $S y = T^\ast y$ for each $y \in Y$. 
Hence $S = T^\ast$.
Category:Adjoints
Everywhere Dense Subset of Countable Hilbert Space Direct Sum in terms of Everywhere Dense Subsets of Summands

== Theorem ==
Let $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$ be a sequence of Hilbert spaces over $\C$.
For each $n \in \N$, let $\family {e^{(n)}_\alpha}_{\alpha \in \map J n}$ be an everywhere dense subset of $\HH_n$.
Let $\tuple {\HH, \innerprod \cdot \cdot}$ be the Hilbert space direct sum of $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$. 
Let $\SS$ be the set of $f \in \HH$ such that:
:there exists $N \in \N$ such that:
::$\map f n = e^{(n)}_{\map \alpha n}$ for some $\map \alpha n \in \map J n$ and for all $n \le N$
::$\map f n = {\mathbf 0}_{\HH_n}$ for $n > N$. 
Then $\SS$ is everywhere dense in $\HH$. 
== Proof ==
Let $f \in \HH$ and $\epsilon > 0$. 
We need to show that there exists $g \in \SS$ with $\norm {f - g} < \epsilon$.
Then by definition of the Hilbert space direct sum and Net Convergence Equivalent to Absolute Convergence:
:$\ds \sum_{n \mathop = 1}^\infty \norm {\map f n}_{\HH_n}^2 < \infty$
Hence there exists $N \in \N$ such that:
:$\ds \sum_{n \mathop = N + 1}^\infty \norm {\map f n}_{\HH_n}^2 < \frac {\epsilon^2} 2$
For $n \le N$, pick $\map \alpha n \in \map J n$ such that:
:$\norm {\map f n - e_{\map \alpha n}^{(n)} }_{\HH_n}^2 < \dfrac {\epsilon^2} {2 N}$
Define:
:$\map g n = e_{\map \alpha n}^{(n)} \in \HH_n$ for $n \le N$
:$\map g n = {\mathbf 0}_{\HH_n}$ for $n > N$.
We have:
:$\ds \sum_{n \mathop = 1}^\infty \norm {\map g n}_{\HH_n}^2 = \sum_{n \mathop = 1}^N \norm {\map g n}_{\HH_n}^2 < \infty$
since the sum on the is finite.
Hence $g \in \HH$.
We then have:
hence:
:$\norm {f - g}_\HH < \epsilon$
as required.
Category:Hilbert Space Direct Sums
Everywhere Dense iff Interior of Complement is Empty

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subset S$.
Then $A$ is everywhere dense :
:$\paren {\relcomp S A}^\circ = \O$
where $A^\circ$ is the interior of $A$.
== Proof ==
By definition of everywhere dense, $A$ is everywhere dense :
:$A^- = S$
where $A^-$ is the closure of $A$.
That happens :
Hence the result.
Category:Denseness
Category:Set Interiors
Evolute/Examples/Parabola

== Example of Evolute ==
Let $P$ be a parabola embedded in the Cartesian plane whose equation is given by:
:$y^2 = 4 a x + 8 a^2$
The evolute of $P$ is the semicubical parabola whose equation is given by:
:$4 x^3 = 27 a y^2$
== Proof ==
== Sources ==
* 
* 
Category:Parabolas
Category:Semicubical Parabola
Category:Examples of Evolutes