nyuuzyou/wikis huggingface dataset:
Totally Disconnected Space is Totally Pathwise Disconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is a totally pathwise disconnected space.
== Proof ==
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then by definition $T$ contains no non-degenerate connected sets.
 $T$ is not a totally pathwise disconnected space.
That is, there exist two points $x, y \in S$ such that there exists a path between $x$ and $y$.
That is, $x$ and $y$ are path-connected.
Thus they are in the same path component.
From Path-Connected Space is Connected, $x$ and $y$ are therefore in the same component.
But that contradicts the definition of $T$ having no non-degenerate connected sets.
Hence, by Proof by Contradiction, $T$ is a totally pathwise disconnected space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Totally Pathwise Disconnected Spaces
Category:Totally Disconnected Spaces
Totally Disconnected and Locally Connected Space is Discrete

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is both totally disconnected and locally connected.
Then $T$ is the discrete space on $S$.
== Proof ==
So, let $T = \struct {S, \tau}$ be that topological space which is both totally disconnected and locally connected.
As $T$ is totally disconnected, every point is a component and therefore closed.
As $T$ is locally connected, there exists a basis $\BB$ of $T$ such that every element of $\BB$ is a component of $T$.
In order for $T$ to be covered by $\BB$, every singleton subset of $T$ must be in $\BB$.
From the definition of topology, the union of any number of these singleton sets is an open set of $T$.
That is, every subset of $S$ is open in $T$.
That is, every element of the power set of $S$ is open in $T$.
This is precisely the definition of the discrete space on $S$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Discrete Topology
Category:Locally Connected Spaces
Category:Totally Disconnected Spaces
Totally Disconnected but Connected Set must be Singleton

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both totally disconnected and connected.
Then $H$ is a singleton.
== Proof ==
If $H$ is totally disconnected then its individual points are separated.
If $H$ is connected it can not be represented as the union of two (or more) separated sets.
So $H$ can have only one point in it.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Totally Disconnected Spaces
Category:Connected Sets
Category:Singletons
Totally Ordered Abelian Group Isomorphism

== Theorem ==
Let $\struct {\Z', +', \le'}$ be a totally ordered abelian group.
Let $0'$ be the identity of $\struct {\Z', +', \le'}$.
Let $\N' = \set {x \in \Z': x \ge' 0'}$.
Let $\Z'$ contain at least two elements.
Let $\N'$ be well-ordered for the ordering induced on $\N'$ by $\le'$.
Then the mapping $g: \Z \to \Z'$ defined by:
:$\forall n \in \Z: \map g n = \paren {+'}^n 1'$
is an isomorphism from $\struct {\Z, +, \le}$ onto $\struct {\Z', +', \le'}$, where $1'$ is the smallest element of $\N' \setminus \set {0'}$.
== Proof ==
First we establish that $g$ is a homomorphism.
Suppose $z \in \Z'$ such that $z \ne 0'$.
Then by Ordering of Inverses in Ordered Monoid, either $z >' 0'$ or $-z >' 0'$.
Thus either:
:$z \in \N' \setminus \set {0'}$
or:
:$-z \in \N' \setminus \set {0'}$
and thus $\N' \setminus \set {0'}$ is not empty.
Therefore $\N' \setminus \set {0'}$ has a minimal element.
Call this minimal element $1'$.
It is clear that $\N'$ is an ordered semigroup satisfying:
:
:
:.
Also:
Thus $\N'$ also satisfies .
So $\struct {\N', +', \le'}$ is a naturally ordered semigroup.
So, by Naturally Ordered Semigroup is Unique, the restriction to $\N$ of $g$ is an isomorphism from $\struct {\N, +, \le}$ to $\struct {\N', +', \le'}$.
By Index Law for Sum of Indices, $g$ is a homomorphism from $\struct {\Z, +}$ into $\struct {\Z', +'}$.
Next it is established that $g$ is surjective.
Let $y \in \Z': y <' 0'$.
Therefore $g$ is a surjection.
Now to show that $g$ is a monomorphism, that is, it is injective.
Let $n < m$.
Therefore it can be seen that $g$ is strictly increasing.
It follows from Monomorphism from Total Ordering that $g$ is a monomorphism from $\struct {\Z, +, \le}$ to $\struct {\Z', +', \le'}$.
A surjective monomorphism is an isomorphism, and the result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.14$
Category:Isomorphisms (Abstract Algebra)
Category:Abelian Groups
Totally Ordered Ring Zero Precedes Element or its Inverse

== Theorem ==
Let $\struct {R, +, \circ, \preceq}$ be an ordered ring.
From the definition of ordered ring, $\preceq$ is compatible with $+$.
Let $0_R$ be the zero element of $R$.
Let $x \ne 0_R$ be a non-zero element of $R$.
Let $-x$ be the ring negative of $x$.
Then:
:$0_R \prec x \lor 0_R \prec -x$
but not both.
== Proof ==
By the definition of total ordering, $\preceq$ is connected.
As $x \ne 0_R$, one of the following is true, but not both:
:$(1): \quad 0_R \prec x$
:$(2): \quad x \prec 0_R$
If $(2)$, because $\prec$ is compatible with $+$:
Category:Ordered Rings
Totally Ordered Set/Examples/Divisor Relation

== Examples of Totally Ordered Sets ==
Let $\struct {\Z_{>0}, \divides}$ denote the order structure consisting of the (strictly) positive integers under the divisor relation $\divides$.
Then $\struct {\Z_{>0}, \divides}$ is specifically not a totally ordered set.
== Proof ==
;Proof by Counterexample
From Divisor Relation on Positive Integers is Partial Ordering we have that $\struct {\Z_{>0}, \divides}$ is a partially ordered set.
However, note that:
:$3 \nmid 5$
and:
:$5 \nmid 3$
where $x \nmid y$ denotes that $x$ is not a divisor of $y$.
Hence $\divides$ is not a total ordering.
== Sources ==
* 
* 
Category:Examples of Orderings
Totally Ordered Set/Examples/Example Ordering on Integers

== Examples of Totally Ordered Sets ==
Let $\preccurlyeq$ denote the relation on the set of integers $\Z$ defined as:
:$a \preccurlyeq b$ $0 \le a \le b \text { or } b \le a < 0 \text { or } a < 0 \le b$
where $\le$ is the usual ordering on $\Z$.
Then $\struct {\Z, \preccurlyeq}$ is a totally ordered set.
== Proof ==
From Example Ordering on Integers, $\preccurlyeq$ is an ordering on $\Z$.
Let $a, b \in \Z$.
Let $a, b \ge 0$.
Then either $a \le b$ or $b \le a$.
Hence either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Let $a, b < 0$.
Then either $a \le b$ or $b \le a$.
Hence either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Let $a < 0$ but $b \ge 0$.
Then $a \preccurlyeq b$.
Let $b < 0$ but $a \ge 0$.
Then $b \preccurlyeq a$.
In all cases, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $4$
Category:Examples of Orderings
Totally Ordered Set/Examples/Real Numbers under Usual Ordering

== Examples of Totally Ordered Sets ==
Let $\struct {\R, \le}$ denote the real numbers under the usual ordering on $\R$.
Then $\struct {\R, \le}$ is a totally ordered set.
== Proof ==
Demonstrated in Real Numbers form Totally Ordered Field.
== Sources ==
* 
* 
Category:Examples of Orderings
Totally Ordered Set is Lattice

== Theorem ==
Every totally ordered set is a lattice.
== Proof ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Then we have:
:$\forall x, y \in S: x \preceq y \lor y \preceq x$
:$\forall x, y \in S: x \preceq y \implies \sup \set {x, y} = y \land \inf \set {x, y} = x$
:$\forall x, y \in S: y \preceq x \implies \sup \set {x, y} = x \land \inf \set {x, y} = y$
Thus the conditions for $\struct {S, \preceq}$ to be a lattice are fulfilled.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
* 
* 
Category:Total Orderings
Category:Lattice Theory
Totally Ordered Set is Well-Ordered iff Subsets Contain Infima

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Then $\struct {S, \preccurlyeq}$ is a well-ordered set every non-empty subset of $T \subseteq S$ has an infimum such that $\map \inf T \in T$.
== Proof ==
=== Sufficient Condition ===
Let $T \subseteq S$ such that $m := \map \inf T \in T$.
By definition, $m$ is a lower bound of $T$ and so:
:$\forall x \in T: m \preccurlyeq x$
That is:
:$\neg \exists x \in T: x \prec m$
Thus by definition $x$ is a minimal element of $T$.
Thus by definition $S$ is well-founded and so is a well-ordered set.
=== Necessary Condition ===
Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Let $T \subseteq S$.
Then from Subset of Well-Ordered Set is Well-Ordered, $\struct {T, \preccurlyeq}$ is also a well-ordered set.
Hence, by definition, $\struct {T, \preccurlyeq}$, has a smallest element $m$.
From Smallest Element is Infimum, $m$ is an infimum.
But $m \in T$.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Total Orderings
Category:Well-Orderings
Totally Separated Space is Completely Hausdorff and Urysohn

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is totally separated.
Then $T$ is completely Hausdorff and Urysohn.
== Proof ==
Let $T = \struct {S, \tau}$ be a totally separated space.
Then for every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$.
Consider a mapping $f: S \to \closedint 0 1$ such that:
:$\Img {f {\restriction_U} } = \set 0$
:$\Img {f {\restriction_V} } = \set 1$
where:
:$f {\restriction_U}$ and $f {\restriction_V}$ denote the restrictions of $f$ to $U$ and $V$ respectively
:$\Img g$ denotes the image set of a mapping $g$.
From Constant Mapping is Continuous, both $f {\restriction_U}$ and $f {\restriction_V}$ are continuous on $U$ and $V$ respectively.
From Continuous Mapping of Separation, it follows that $f$ is continuous on $S$ itself.
By definition, $f$ is then an Urysohn function on $x$ and $y$.
Hence $T$ is Urysohn.
Having proved that $T$ is an Urysohn space, it follows from Urysohn Space is Completely Hausdorff Space that $T$ is also completely Hausdorff.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Totally Separated Spaces
Category:Completely Hausdorff Spaces
Category:Urysohn Spaces
Totally Separated Space is Totally Disconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is totally separated.
Then $T$ is totally disconnected.
== Proof ==
Let $T = \struct {S, \tau}$ be a totally separated space.
Then for every $x, y \in S: x \ne y$ there exists a partition $U \mid V$ of $T$ such that $x \in U, y \in V$.
Suppose $T$ were not totally disconnected.
Then $\exists H \subseteq S: x, y \in U$ such that $H$ is connected.
But by definition of connected, there can be no partition $U \mid V$ of $T$ such that $x \in U, y \in V$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Totally Separated Spaces
Category:Totally Disconnected Spaces
Totally Skew Symmetric Function/Examples/Arbitrary Example 1

== Examples of Totally Skew Symmetric Functions ==
Let $f: \R^3 \to \R$ be the real-valued function defined as:
:$\forall \tuple {x, y, z} \in \R^3: \map f {x, y, z} = \paren {x - y} \paren {y - z} \paren {z - x}$
Then $f$ is a totally skew symmetric function.
== Sources ==
* 
* 
Category:Examples of Totally Skew Symmetric Functions
Touching Circles have Different Centers

== Theorem ==
== Proof ==
If the two circles are outside one another, the result is trivial.
This proof will focus on the situation where one circle is inside the other one.
:
Let $ABC$ and $CDE$ be circles which touch one another at $C$, such that $CDE$ is inside $ABC$
 they had the same center $F$.
Join $FC$ and let $FB$ be drawn at random through $E$.
As $F$ is the center of $ABC$, by , we have that $FB = FC$.
Similarly, as $F$ is also the center of $CDE$, we have that $FC = FE$.
But they are clearly unequal by the method of construction.
So from this contradiction, the two circles can not have the same center.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Tower Law for Subgroups

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ with finite index.
Let $K$ be a subgroup of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the index of $H$ in $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $6$: Cosets: Exercise $6$
* : $1$: Subgroups: $1.\text{T}.1$
* : $0$: Some Conventions and some Basic Facts: Exercise $11 \ \text{(i)}$
* 
Category:Subgroups
Category:Tower Law for Subgroups
Category:Named Theorems
Category:Index of Subgroups
Tower Law for Subgroups/Proof 1

== Theorem ==
== Proof ==
Let $p = \index G H$, $q = \index H K$.
By hypothesis these numbers are finite.
Therefore, there exist $g_1, \ldots, g_p \in G$ such that $G$ is a disjoint union:
$\ds G = \bigsqcup_{i \mathop = 1}^p g_i H$
Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a disjoint union:
$\ds H = \bigsqcup_{j \mathop = 1}^q h_j K$
Thus:
This expression for $G$ is the disjoint union of $p q$ cosets.
Therefore the number of elements of the coset space is:
:$\index G K = p q = \index G H \index H K$
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 39$
Category:Tower Law for Subgroups
Tower Law for Subgroups/Proof 2

== Theorem ==
== Proof ==
Assume $G$ is finite.
Then:
Since $K \le H$, from Lagrange's Theorem we have that $\dfrac {\order H} {\order K} = \index H K$.
Hence the result.
== Sources ==
* : $\S 43$. Lagrange's theorem: Worked Example $2$
* : Chapter $5$: Cosets and Lagrange's Theorem: Corollary $5.13$
Category:Tower Law for Subgroups
Tower Property of Conditional Expectation

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\HH \subseteq \GG$ be sub-$\sigma$-algebras of $\Sigma$.
Let $X$ be a integrable random variable. 
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Let $\expect {X \mid \HH}$ be a version of the conditional expectation of $X$ given $\HH$.
 
Let $\expect {\expect {X \mid \GG} \mid \HH}$ be a a version of the conditional expectation of $\expect {X \mid \GG}$ given $\HH$.
Then:
:$\expect {\expect {X \mid \GG} \mid \HH} = \expect {X \mid \HH}$ almost everywhere.
== Proof ==
We show that $\expect {X \mid \HH}$ is a version of $\expect {\expect {X \mid \GG} \mid \HH}$.
Let $A \in \HH$. 
Then:
Since $\expect {X \mid \HH}$ is $\HH$-measurable, we have that $\expect {X \mid \HH}$ is a version of $\expect {\expect {X \mid \GG} \mid \HH}$.
So by Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra:
:$\expect {\expect {X \mid \GG} \mid \HH} = \expect {X \mid \HH}$ almost everywhere.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Tower is Proper Subtower or all of Set

== Lemma ==
Let $X$ be a non-empty set.
Let $\struct {T, \preccurlyeq}$ be a tower in $X$.
Then if $T \ne X$, $\struct {T, \preccurlyeq}$ is a proper subtower in $X$.
== Proof ==
Suppose $T \ne X$.
Then there is an element $x \in X \setminus T$.
The singleton $\set x$ can trivially be given a well-ordering, as there are are no elements in $\set x$ other than $x$ itself.
Consider the order sum on $T \cup \set x$ defined by setting $t \prec x$ for all $t \in T$.
By Order Sum of Totally Ordered Sets is Totally Ordered, this gives a total ordering on $T \cup \set x$.
There is a smallest element of any subset $T \cup \set x$, either the one given by $\preccurlyeq$, or $x$ if the subset considered is the singleton $\set x$.
Extend the choice function $c$ defining $\struct {T, \preccurlyeq}$ by assigning $\map c {X \setminus S_x} = x$.
Then $T$ is a proper subset of $\struct {T \cup \set x, \text {extension of} \preccurlyeq}$.
So $T$ is a proper subtower in $X$.
== Sources ==
* $\S 1.11$ Supplementary Exercise $7$
Category:Well-Orderings
Tower of Hanoi

== Problem ==
There is a tower of $n$ disks, stacked in decreasing size on one of $3$ pegs.
:
The object of the exercise is to move the disks onto a different peg, obeying the rules:
:$(1): \quad$ Only one disk can be moved at a time
:$(2): \quad$ No disk may be placed on a peg with a smaller disk beneath it.
How many moves are needed to move all the disks onto a different peg?
=== Variant ===
== Solution ==
For a tower of $n$ disks, it takes $2^n - 1$ moves.
== Proof ==
=== Setting up a Recurrence Rule ===
Let $T_n$ be the number of moves needed to transfer $n$ disks from one peg to another.
Clearly we have:
: $T_0 = 0$
: $T_1 = 1$
Now, we note that in order to move a tower of $n$ disks, we need to do the following:
:$(1): \quad$ Move the tower of $n - 1$ disks from off the top of the $n$th disk onto another of the pegs
:$(2): \quad$ Move the $n$th disk to the destination peg
:$(3): \quad$ Move the tower of $n - 1$ disks from where it was put temporarily onto the top of the $n$th disk.
It is clear that steps $1$ and $3$ above each take $T_{n - 1}$ moves, while step $2$ takes one move.
Hence we have:
:$T_n \le 2 T_{n - 1} + 1$
The inequality applies here because we have established that although we know we can always get the job done in no more than $2 T_{n-1} + 1$ moves, we are not at this stage certain that there is not a better strategy which needs fewer moves.
So: is there a way of moving the disks that uses fewer moves?
Consider that at some point we need to move disk $n$.
When we do this, we must have moved the $n - 1$ disks above it onto a vacant peg. That must have taken $T_{n - 1}$ moves to do, whatever $T_{n - 1}$ happens to be.
Having moved that final disk for the first (and hopefully last) timeIn it is pointed out that: "We might move the largest disk more than once, if we're not too alert.", we then need to transfer the $n - 1$ smaller disks (which all need to be on a single peg, otherwise there won't be a spare peg to move disk $n$ onto) back onto disk $n$. This also takes $T_{n - 1}$ moves to do.
So now we see that there is not a better way to move the disks, i.e.:
:$T_n \ge 2 T_{n - 1} + 1$
Thus we arrive at our recurrence rule:
:$T_n = 2 T_{n - 1} + 1$
=== Solution of Recurrence ===
Using induction, we show that $T_n = 2^n - 1$.
The base case is straightforward:
: $T_0 = 0 = 2^0 - 1$
: $T_1 = 1 = 2^1 - 1$
Now assume the induction hypothesis:
:$T_k = 2^k - 1$
and try to show:
:$T_{k + 1} = 2^{k + 1} - 1$
Hence the induction step:
Hence the result by induction.
== References ==
== Historical Note ==
== Sources ==
* : $31$
* : $18,446,744,073,709,551,615$
* : $\S 1.1$
* : $31$
* : $18,446,744,073,709,551,615$
* 
Category:Classic Problems
Category:Tower of Hanoi
Tower of Hanoi/Historical Note

== Historical Note on Tower of Hanoi ==
The Tower of Hanoi was invented by in $1893$, under the name M. Claus.
He backed this up by inventing the romantic story about the Tower of Brahma, as follows:
:In the great temple of Benares, beneath the dome which marks the centre of the world, rests a brass plate in which there are fixed three diamond needles, each a cubit high and as thick as the body of a bee. On one of these needles, at the creation, God placed $64$ discs of pure gold, the largest disc resting on the brass plate, and the others getting smaller and smaller up to the top one. This is the Tower of Bramah. Day and night unceasingly the priests transfer the discs from one diamond needle to another according to the fixed and immutable laws of Bramah, which require that the priest on duty must not move more than one disc at a time and that he must place this disc on a needle so that there is no smaller disc below it. When the $64$ discs have been thus transferred from the needle on which at the creation God placed them to one of the other needles, tower, temple, and Brahmins alike will crumble into dust, and with a thunderclap the world will vanish.
It is seen from the solution of the problem that the Brahmin priests would need $2^{64} - 1$ moves:
:$18 \, 446 \, 744 \, 073 \, 709 \, 551 \, 615$
to complete their task.
At $1$ move per second, that is nearly $600 \, 000 \, 000 \, 000$ years.
The world may well have (figuratively) crumbled to dust long before that time.
== Sources ==
* : $31$
* : $18,446,744,073,709,551,615$
* : $\S 1.1$
* : $31$
* : $18,446,744,073,709,551,615$
* 
Category:Historical Notes
Tower of Hanoi/Variant

== Problem ==
There is a tower of $n$ disks, stacked in decreasing size on one of $3$ pegs.
:
The object of the exercise is to move the disks onto a different peg, obeying the rules:
:$(1): \quad$ Only one disk can be moved at a time
:$(2): \quad$ No disk may be placed on a peg with a smaller disk beneath it
:$(3): \quad$ The disks must all be moved from the peg $1$ to peg $3$
:$(4): \quad$ A move consists of transferring a disk from the peg $1$ to peg $2$, or back, or from peg $2$ to peg $3$, or back
:$(5): \quad$ No disk can cross over peg $2$ if it contains a smaller disk.
== Solution ==
For a tower of $n$ disks, it takes $3^n - 1$ moves.
== Proof ==
To move a tower of $n$ disks from the peg $1$ to peg $3$, the following must happen:
:$(1): \quad$ The tower of $n - 1$ disks above the $n$th disk is transferred from peg $1$ to peg $3$.
:$(2): \quad$ The $n$th disk is transferred from peg $1$ to peg $2$.
:$(3): \quad$ The tower of $n - 1$ disks above the $n$th disk is transferred from the peg $3$ to peg $1$.
:$(4): \quad$ The $n$th disk is transferred from the peg $2$ to peg $3$.
:$(5): \quad$ The tower of $n - 1$ disks above the $n$th disk is transferred from the peg $1$ to peg $3$.
Let ${T_n}'$ be the number of moves needed to transfer the $n$-disk tower from peg $1$ to peg $3$.
Then we see that:
:${T_n}' = 3 {T_{n - 1}}' + 2$
A simple proof by induction shows that:
:${T_n}' = 3^n - 1$
== Historical Note ==
== Sources ==
* : $\S 1.1$
Category:Tower of Hanoi
Trace Sigma-Algebra is Sigma-Algebra

== Theorem ==
Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $E \subseteq X$ be a subset of $X$.
Then the trace $\sigma$-algebra $\Sigma_E$ is a $\sigma$-algebra on $E$.
== Proof ==
Verifying the axioms for a $\sigma$-algebra in turn:
=== Axiom $(1)$ ===
From Intersection with Subset is Subset, we have:
:$E = X \cap E$
Hence, as $X \in \Sigma$ by axiom $(1)$:
:$E \in \Sigma_E$
by definition of trace $\sigma$-algebra.
=== Axiom $(2)$ ===
Suppose that:
:$S \in \Sigma_E$.
Then there is some $A \in \Sigma$ such that:
:$S = E \cap A$
by definition of trace $\sigma$-algebra.
Hence:
:$X \setminus A \in \Sigma$
by axiom $(2)$ for $\sigma$-algebras.
So:
:$E \cap \paren {X \setminus A} \in \Sigma_E$
By Intersection with Set Difference is Set Difference with Intersection, this equals:
:$\paren {X \cap E} \setminus A = E \setminus A$.
Now by Set Difference with Intersection is Difference:
:$E \setminus A = E \setminus S$
Hence:
:$E \setminus S \in \Sigma_E$
=== Axiom $(3)$ ===
Let $\sequence {S_i}_{i \mathop \in \N}$ be a sequence in $\Sigma_E$.
For each $i \in \N$, let $A_i \in \Sigma$ such that:
:$S_i = E \cap A_i$
Then observe:
As $\Sigma$ is a $\sigma$-algebra:
:$\ds \bigcup_{i \mathop \in \N} A_i \in \Sigma$
Hence:
:$\ds \bigcup_{i \mathop \in \N} S_i \in \Sigma_E$
by definition of trace $\sigma$-algebra.
== Sources ==
* : $3.3 \ \text{(vi)}$, $\S 3$: Problem $2$
Category:Sigma-Algebras
Category:Trace Sigma-Algebras