nyuuzyou/wikis huggingface dataset:
Stirling Number of the Second Kind of Number with Greater/Proof 2

== Theorem ==
Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from Stirling Number of the Second Kind of 0.
So by definition of Kronecker delta:
:$\forall k \in \Z_{\ge 0}: k > 0 \implies \ds {0 \brace k} = 0$
and so $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds k > r \implies {r \brace k} = 0$
from which it is to be shown that:
:$\ds k > r + 1 \implies {r + 1 \brace k} = 0$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$
Category:Stirling Number of the Second Kind of Number with Greater
Stirling Number of the Second Kind of Number with Self

== Theorem ==
:$\ds {n \brace n} = 1$
where $\ds {n \brace n}$ denotes a Stirling number of the second kind.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n} = 1$
=== Basis for the Induction ===
$\map P 0$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brace k} = 1$
from which it is to be shown that:
:$\ds {k + 1 \brace k + 1} = 1$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: {n \brace n} = 1$
== Also see ==
* Unsigned Stirling Number of the First Kind of Number with Self
* Signed Stirling Number of the First Kind of Number with Self
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(48)$
Category:Stirling Numbers
Stirling Number of the Second Kind of n+1 with 0

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brace 0} = 0$
where $\ds {n + 1 \brace 0}$ denotes a Stirling number of the second kind.
== Proof ==
We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brace k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus
Hence the result.
== Also see ==
* Unsigned Stirling Number of the First Kind of n+1 with 0
* Signed Stirling Number of the First Kind of n+1 with 0
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(50)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Number of the Second Kind of n+1 with 1

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brace 1} = 1$
where $\ds {n + 1 \brace 1}$ denotes a Stirling number of the second kind.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brace 1} = 1$
=== Basis for the Induction ===
$\map P 0$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k + 1 \brace 1} = 1$
from which it is to be shown that:
:$\ds {k + 2 \brace 1} = 1$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$
== Also see ==
* Unsigned Stirling Number of the First Kind of n+1 with 1
* Signed Stirling Number of the First Kind of n+1 with 1
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(50)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Number of the Second Kind of n+1 with 2

== Theorem ==
Let $n \in \Z_{\ge 0}$.
Then:
:$\ds {n + 1 \brace 2} = 2^n - 1$
where $\ds {n + 1 \brace 2}$ denotes a Stirling number of the second kind.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brace 2} = 2^n - 1$
=== Basis for the Induction ===
$\map P 0$ is the case:
So the result holds for $\map P 0$.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k + 1 \brace 2} = 2^k - 1$
from which it is to be shown that:
:$\ds {k + 2 \brace 2} = 2^{k + 1} - 1$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0}: {n + 1 \brace 2} = 2^n - 1$
== Also see ==
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(50)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Number of the Second Kind of n with n-1

== Theorem ==
Let $n \in \Z_{> 0}$ be an integer greater than $0$.
Then:
:$\ds {n \brace n - 1} = \binom n 2$
where:
:$\ds {n \brace n - 1}$ denotes a Stirling number of the second kind
:$\dbinom n 2$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n - 1} = \binom n 2$
=== Basis for the Induction ===
$\map P 1$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brace k - 1} = \binom k 2$
from which it is to be shown that:
:$\ds {k + 1 \brace k} = \binom {k + 1} 2$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 1}: {n \brace n - 1} = \binom n 2$
== Also see ==
* Unsigned Stirling Number of the First Kind of n with n-1
* Signed Stirling Number of the First Kind of n with n-1
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(49)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Number of the Second Kind of n with n-2

== Theorem ==
Let $n \in \Z_{\ge 2}$ be an integer greater than or equal to $2$.
Then:
:$\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$
where:
:$\ds {n \brace n - 2}$ denotes an Stirling number of the second kind
:$\dbinom n 4$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
=== Basis for the Induction ===
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$
$\map P 2$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brace k - 2} = \binom {k + 1} 4 + 2 \binom k 4$
from which it is to be shown that:
:$\ds {k + 1 \brace k - 1} = \binom {k + 2} 4 + 2 \binom {k + 1} 4$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 2}: {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$
== Also see ==
* Unsigned Stirling Number of the First Kind of n with n-2
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(57)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Number of the Second Kind of n with n-3

== Theorem ==
Let $n \in \Z_{\ge 3}$ be an integer greater than or equal to $3$.
Then:
:$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$
where:
:$\ds {n \brace n - 3}$ denotes an Stirling number of the second kind
:$\dbinom n 6$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
=== Basis for the Induction ===
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$
$\map P 3$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brace k - 3} = \binom {k + 2} 6 + 8 \binom {k + 1} 6 + 6 \binom k 6$
from which it is to be shown that:
:$\ds {k + 1 \brace k - 2} = \binom {k + 3} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 1} 6$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 3}: {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$
== Also see ==
* Unsigned Stirling Number of the First Kind of n with n-3
* Particular Values of Stirling Numbers of the Second Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(57)$
Category:Examples of Stirling Numbers of the Second Kind
Stirling Numbers of the First Kind/Examples/5th Falling Factorial

== Example of Stirling Numbers of the First Kind==
:$x^{\underline 5} = x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x$
and so:
:$\dbinom x 5 = \dfrac 1 {120} \paren {x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x}$
== Proof ==
Follows directly from Stirling's triangle of the first kind (unsigned):
Also from Stirling's triangle of the first kind (signed):
By definition of binomial coefficient:
:$\dbinom x 5 = \dfrac {x^{\underline 5} } {5!} = \dfrac {x^{\underline 5} } {120}$
Hence:
:$\dbinom x 5 = \dfrac 1 {120} \paren {x^5 - 10 x^4 + 35 x^3 - 50 x^2 + 24 x}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Falling Factorials
Category:Examples of Stirling Numbers of the First Kind
Stirling Numbers of the Second Kind/Examples/5th Power

== Example of Stirling Numbers of the Second Kind==
:$x^5 = x^{\underline 5} + 10 x^{\underline 4} + 25 x^{\underline 3} + 15 x^{\underline 2} + x^{\underline 1}$
and so:
:$x^5 = 120 \dbinom x 5 + 240 \dbinom x 4 + 150 \dbinom x 3 + 30 \dbinom x 2 + \dbinom x 1$
== Proof ==
From the definition of Stirling numbers of the second kind:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
Reading the values directly from Stirling's triangle of the second kind:
By definition of binomial coefficient:
Hence:
:$x^5 = 120 \dbinom x 5 + 240 \dbinom x 4 + 150 \dbinom x 3 + 30 \dbinom x 2 + \dbinom x 1$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients
Category:Falling Factorials
Category:Examples of Stirling Numbers of the Second Kind
Stochastic/Examples/Queueing System

== Examples of Stochastic ==
A queueing system at a public facility such as a bank or a post office is an example of a system which is stochastic.
Hence the following are stochastic variables:
:the time intervals between the arrival of people requiring the service
:the times taken to service the needs of those people
:the number of people in the queue awaiting service at any point in time.
== Sources ==
* 
* 
Category:Examples of Stochastic
Stokes's Law

== Physical Law ==
Stokes's Law is an empirical law for the drag exerted on a spherical body $\BB$ with a small Reynolds number in a viscous fluid $\FF$.
It is formulated as:
:$\mathbf F_{\rm d} = 6 \pi \mu r \mathbf v$
where:
:$\mathbf F_{\rm d}$ is the Stokes's drag acting on the boundary between $\FF$ and $\BB$ in newtons
:$\mu$ is the dynamic viscosity in pascal-seconds
:$r$ is the radius of $\BB$ in metres
:$\mathbf v$ is the velocity of $\FF$ relative to $\BB$ in metres per second.
== Also known as ==
== Sources ==
* 
* 
Category:Stokes's Law
Category:Hydrodynamics
Stolz-Cesàro Theorem

== Theorem ==
Let $\sequence {a_n}$ be a sequence.
Let $\sequence {b_n}$ be a sequence of (strictly) positive real numbers such that:
:$\ds \sum_{i \mathop = 0}^\infty b_n = \infty$
If:
:$\ds \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = L \in \R$
then also:
:$\ds \lim_{n \mathop \to \infty} \dfrac {a_1 + a_2 + \cdots + a_n} {b_1 + b_2 + \cdots + b_n} = L$
=== Corollary ===
== Proof ==
Define the following sums:
:$\ds A_n = \sum_{i \mathop = 1}^n a_i$
:$\ds B_n = \sum_{i \mathop = 1}^n b_i$
Let $\epsilon > 0$ and $\mu = \dfrac {\epsilon} 2$. 
By the definition of convergent sequences, there exists $k \in \N$ such that:
:$\forall n > k: \paren {L - \mu} b_n < a_n < \paren {L + \mu} b_n$
Rewrite the sum as:
Divide above by $B_n$:
:$\dfrac {A_k - \paren {L - \mu} B_k} {B_n} + \paren {L - \mu} < \dfrac {A_n} {B_n} < \paren {L + \mu} + \dfrac {A_k - \paren {L + \mu} B_k} {B_n}$
Let $k$ be fixed. 
From Reciprocal of Null Sequence and Combination Theorem for Sequences, the sequence $\sequence {\dfrac {A_k - \paren {L \pm \mu} B_k} {B_n} }$ converges to zero.
By the definition of convergent sequences, there exists $N > k > 0$ such that:
:$\size {\dfrac {A_k - \paren {L \pm \mu} B_k} {B_n} } < \mu$ for all $n > N$
Substitute the above into the inequality and obtain:
Hence by the definition of convergent sequences the result follows.
== Remarks ==
* Using the similar proof technique with limits inferior and superior, a more general version of this theorem can be obtained. In that case the limit $L$ can be either a real number or $\pm \infty$.
* By setting $b_n = 1$ the theorem turns into Cesàro Mean for real-valued sequences.
Category:Analysis
Category:Limits of Sequences
Stolz-Cesàro Theorem/Corollary

== Corollary to Stolz-Cesàro Theorem ==
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$ such that $\sequence {b_n}$ is strictly increasing and $\ds \lim_{n \mathop \to \infty} b_n = \infty$.
If:
:$\ds \lim_{n \mathop \to \infty} \frac {a_n - a_{n - 1} } {b_n - b_{n - 1} } = L \in \R$
then also:
:$\ds \lim_{n \mathop \to \infty} \frac {a_n} {b_n} = L$
== Proof ==
Define the following sequences:
:$x_1 = a_1$, $x_n = a_n - a_{n - 1}$
:$y_1 = b_1$, $y_n = b_n - b_{n - 1}$
It follows that:
:$\ds \sum_{i \mathop = 1}^n x_i = a_n$
and:
:$\ds \sum_{i \mathop = 1}^n y_i = b_n$
From above follows:
:$\ds \lim_{n \mathop \to \infty} \frac {x_n} {y_n} = \lim_{n \mathop \to \infty} \frac {a_n - a_{n - 1} } {b_n - b_{n - 1} } = L$
From the definition of divergent sequences there exists $N \in \N$ such that 
:$b_n$ is positive for all $n > N$
From the general Stolz-Cesàro Theorem follows that
:$\ds \lim_{n \mathop \to \infty} \frac {a_n} {b_n} = \lim_{n \mathop \to \infty} \frac {\sum_{i \mathop = 1}^n x_i} {\sum_{i \mathop = 1}^n y_i} = L$
Hence the result.
== Remarks ==
* The theorem also holds if $\sequence {b_n}$ is strictly monotone and $\ds \lim_{n \mathop \to \infty} b_n = \pm \infty$.
* Just as the general Stolz-Cesàro Theorem, the corollary also a version using limit inferior and limit superior. In that case the limit $L$ can be a real number or $\pm \infty$.
* This form of Stolz-Cesàro Theorem is most commonly found in books.
* The theorem can be considered as a special case of L'Hopital's Rule for sequences.
Category:Real Analysis
Category:Limits of Sequences
Stone-Weierstrass Theorem/Compact Space

== Theorem ==
Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\map C {X, \R}$ be the set of real-valued continuous functions on $T$.
Let $\times$ be the pointwise multiplication on $\map C {X, \R}$.
Let $\struct {\map C {X, \R}, \times}$ be the Banach algebra with respect to $\norm \cdot_\infty$.
Let $\AA$ be a unital subalgebra of $\map C {X, \R}$.
Suppose that $\AA$ separates points of $X$, that is:
:for distinct $p, q \in X$, there exists $h_{p q} \in \AA$ such that $\map {h_{p q} } p \ne \map {h_{p q} } q$.
Then the closure $\overline \AA$ of $\AA$ is equal to $\map C {X, \R}$.
=== Complex-Valued ===
== Proof ==
Let $\struct {C', \norm {\,\cdot\,}_{C'} }$ be the dual space of $\struct {\map C {X, \R}, \norm {\,\cdot\,}_\infty }$.
From Spanning Criterion of Normed Vector Space, it suffices to show that:
:$\forall \ell \in C' : \ell \restriction_\AA = 0 \implies \ell = 0$
Let $B' \subseteq C'$ be the closed unit ball, that is:
:$B' := \set {\ell \in C' : \norm \ell_{C'} \le 1}$
Let:
:$U := \set {\ell \in B' : \ell \restriction_\AA = 0}$
Let $\map E U$ be the set of extreme points of $U$.
=== Lemma ===
Let $w^\ast$ be the weak-$\star$ topology on $C'$.
Recall that $B'$ is $w^\ast$-compact by Banach-Alaoglu Theorem.
Thus $U$ is a $w^\ast$-compact convex subset of $C'$.
By Weak-* Topology is Hausdorff, $U$ is Hausdorff.
Thus by Krein-Milman Theorem we have:
:$\map E U \ne \O$
Thus:
:$\map E U = \set 0$
Therefore:
:$U = \set 0$
== Also presented as ==
== Also see ==
* Weierstrass Approximation Theorem, of which the Stone-Weierstrass Theorem is a generalization.
== Sources ==
* : $13.2$: The Krein-Milman Theorem
Category:Stone-Weierstrass Theorem
Category:Compact Spaces
Category:Banach Algebras
Category:Hausdorff Spaces
Stone-Weierstrass Theorem/Compact Space/Also presented as

== Stone-Weierstrass Theorem: Also presented as ==
Some sources specifically define $T = \struct {X, \tau}$ to be a Hausdorff space.
However, we note that $\AA$ is such that it separates points.
Hence from Topological Space Separated by Mappings is Hausdorff it follows that $T$ is a Hausdorff space.
Category:Stone-Weierstrass Theorem
Stone-Weierstrass Theorem/Compact Space/Lemma

== Lemma ==
Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\map C {X, \R}$ be the set of real-valued continuous functions on $T$.
Let $\times$ be the pointwise multiplication on $\map C {X, \R}$.
Let $\struct {\map C {X, \R}, \times}$ be the Banach algebra with respect to $\norm \cdot_\infty$.
Let $\AA$ be a unital subalgebra of $\map C {X, \R}$.
Suppose that $\AA$ separates points of $X$, that is:
:for distinct $p, q \in X$, there exists $h_{p q} \in \AA$ such that $\map {h_{p q} } p \ne \map {h_{p q} } q$.
Let $\struct {C', \norm {\,\cdot\,}_{C'} }$ be the dual space of $\struct {\map C {X, \R}, \norm {\,\cdot\,}_\infty }$.
Let $B' \subseteq C'$ be the closed unit ball, that is:
:$B' := \set {\ell \in C' : \norm \ell_{C'} \le 1}$
Let:
:$U := \set {\ell \in B' : \ell \restriction_\AA = 0}$
Let $\map E U$ be the set of extreme points of $U$.
Then:
:$\map E U \setminus \set 0 = \O$
== Proof ==
 there is an $\ell \in \map E U$ such that:
:$(1):\quad \ell \ne 0$
Then:
:$\norm \ell_{C'} = 1$
since:
:$\ell = \norm \ell_{C'} \underbrace { \norm \ell_{C'}^{-1} \ell}_{\in U} + \paren {1 - \norm \ell_{C'} } \underbrace {0}_{\in U}$
Note that $X$ is Hausdorff by Topological Space Separated by Mappings is Hausdorff.
Thus we can apply Riesz-Kakutani Representation Theorem.
That is, there exists a signed measure $\mu$ on $X$ such that:
:$\ds \forall f \in \map C {X, \R} : \map \ell f = \int_X f \rd \mu$
and:
:$\map {\size \mu} X = 1$
Let $\map \supp \mu$ denote the support of $\mu$.
Consider any $g \in \AA$ such that:
:$\forall x \in X : 0 < \map g x < 1$
Then we have:
:$g \mu \in U$
Let:
:$\ds a := \int_X g \rd \size {\mu}$
and:
:$\ds b := \int_X \paren {1 - g} \rd \size {\mu}$
so that:
:$a + b = 1$
As:
:$\ds \mu = a \underbrace{\frac {g \mu} a}_{\in U} + b \underbrace {\frac {\paren {1 - g} \mu} b}_{\in U}$
we have:
:$\ds \mu = \frac {g \mu} a$
Especially:
:$\forall x \in \map \supp \mu : \map g x = a$
We claim that:
:$\exists x_0 \in X : \map \supp \mu = \set {x_0}$
 there are distinct $p, q \in \map \supp \mu$.
Then there exists $h_{p q} \in \AA$ such that:
:$\map {h_{p q} } p \ne \map {h_{p q} } q$
Let $C,D > 0$ such that:
:$C > \norm {h_{p q} }_\infty $
and:
:$D > \norm {h_{p q} }_\infty + C$
Then:
:$\ds g := \frac {h_{p q} + C} D$
must be constant on $\map \supp \mu$.
This is a contradiction.
Finally, we have:
That is $\ell = 0$.
This is a contradiction to $(1)$.
Category:Stone-Weierstrass Theorem
Stone-Weierstrass Theorem/Locally Compact Hausdorff Space/Complex-Valued

== Theorem ==
Let $X$ be a locally compact Hausdorff space.
Let $\struct {\map {\CC_0} {X, \C}, \norm {\, \cdot \,} }$ be the Banach algebra of complex-valued continuous functions vanishing at infinity on $X$.
Let $\AA$ be a $\ast$-subalgebra of $\map {\CC_0} {X, \C}$ such that:
:$(1) \quad$ for each $x, y \in X$ with $x \ne y$ there exists $h_{x y} \in \AA$ such that $\map {h_{x y} } x \ne \map {h_{x y} } y$
:$(2) \quad$ for each $x \in X$ there exists $f_x \in \AA$ such that $\map {f_x} x \ne 0$. 
Then $\AA$ is everywhere dense in $\struct {\map {\CC_0} {X, \C}, \norm {\, \cdot \,} }$.
== Proof ==
This proof is essentially identical to Stone-Weierstrass Theorem: Compact Space: Complex-Valued.
We define:
:$\BB = \set {\map \Re f : f \in \AA} = \set {\map \Im f : f \in \AA}$
Replicating the proof in Stone-Weierstrass Theorem: Compact Space: Complex-Valued, it can be shown:
:$\BB = \set {\map \Re f : f \in \AA} = \set {\map \Im f : f \in \AA}$
It is then shown that $\BB$ separates points.
For $(2)$, for each $x \in X$ there exists $f_x \in \AA$ with $\map {f_x} x \ne 0$.
Then either $\map \Re {\map {f_x} x} \ne 0$ or $\map \Im {\map {f_x} x} \ne 0$. 
We have $\map \Re {f_x}, \map \Im {f_x} \in \BB$, so $(2)$ is satisfied.
Hence we can apply Stone-Weierstrass Theorem: Locally Compact Space to obtain:
:$\BB^- = \map {\CC_0} {X, \R}$
Now let $f \in \map {\CC_0} {X, \C}$. 
Write $f = \map \Re f + i \map \Im f$ with $\map \Re f, \map \Im f \in \map \CC {X, \R}$.
From Real and Imaginary Parts of Complex-Valued Function Vanishing at Infinity Vanish at Infinity, we have $\map \Re f, \map \Im f \in \map {\CC_0} {X, \R}$. 
Let $\epsilon > 0$. 
Since $\BB^- = \map {\CC_0} {X, \R}$, there exists $\phi, \psi \in \BB \subseteq \AA$ such that:
:$\norm {\phi - \map \Re f}_\infty < \dfrac \epsilon 2$
and:
:$\norm {\psi - \map \Im f}_\infty < \dfrac \epsilon 2$
We then have: 
Since $\phi, \psi \in \AA$ and $\AA$ is a subalgebra, we have $\phi + i\psi \in \AA$.
Hence $\AA^- = \map {\CC_0} {X, \C}$.
Category:Stone-Weierstrass Theorem
Stone Space of Boolean Lattice is Hausdorff

== Theorem ==
Let $B = \struct {S, \preceq}$ be a Boolean lattice.
Let $\map S B = \struct {U, \tau}$ be the Stone space of $B$.
Then, $\map S B$ is a Hausdorff space.
== Proof ==
Let $x, y \in U$ be arbitrary, with $x \ne y$.
By definition of set equality, there is some $a \in S$ such that:
:$\neg \paren {a \in x \iff a \in y}$
, assume that $a \in x$ and $a \notin y$.
Now, let:
:$X = \set {z \in U : a \in z}$
:$Y = \set {z \in U : \paren {\neg a} \in z}$
By definition of the Stone space of $B$, each of $X$ and $Y$ is open in $\map S B$.
From the above, it is immediate that $x \in X$.
As $a \notin y$, we have by Ultrafilter Contains Element or Complement that $\paren {\neg a} \in y$.
Hence, $y \in Y$.
If there existed some $z \in X \cap Y$, then $\set {a, \neg a} \subseteq z$.
But then, we would have $\bot = a \wedge \neg a \in z$ by definition of a filter.
Thus, it would hold that $z = S$, since $\bot \preceq c$ for every $c \in S$.
As that contradicts the definition of an ultrafilter, we must have $X \cap Y = \O$.
We have constructed two disjoint open sets $X, Y$ such that:
:$x \in X$
:$y \in Y$
As $x \ne y$ were arbitrary, it follows that $\map S B$ is Hausdorff.
Category:Boolean Lattices
Category:Hausdorff Spaces
Category:Stone Spaces