nyuuzyou/wikis huggingface dataset:
Bounds for Prime-Counting Function in terms of Second Chebyshev Function

== Theorem ==
There exists a real function $R : \hointr 2 \infty \to \R$ such that:
:$\ds \frac {\map \psi x} {\ln x} + \map R x \le \map \pi x \le \frac {2 \map \psi x} {\ln x} + \sqrt x$
for all real numbers $x \ge 2$, where:
:$\pi$ is the prime counting function
:$\psi$ is the second Chebyshev function
:$R = \map \OO {\sqrt x \ln x}$ where $\OO$ is big-$\OO$ notation.
== Proof ==
We have, from the definition of the prime counting function: 
:$\ds \map \pi x = \sum_{p \le x} 1$
We can write: 
:$\ds \sum_{p \le x} 1 = \sum_{p \le \sqrt x} 1 + \sum_{\sqrt x < p \le x} 1$
We have that: 
For $1 < \sqrt x < p$, we have from Logarithm is Strictly Increasing:
:$0 < \map \ln {\sqrt x} < \ln p$
So:
:$\ds \frac {\ln p} {\map \ln {\sqrt x} } > 1$
Then:
So:
:$\ds \map \psi x \le \sqrt x + \frac 2 {\ln x} \map \psi x$ 
Now, from Order of Second Chebyshev Function, there exists a real function $r : \hointr 2 \infty \to \R$ such that: 
:$\ds \map \psi x = \sum_{p \le x} \ln p + \map r x$
where:
:$r = \map \OO {\sqrt x \paren {\ln x}^2}$
For $p \le x$, we have from Logarithm is Strictly Increasing:
:$0 < \ln p \le \ln x$
so:
:$\ds \frac {\ln p} {\ln x} < 1$
Then:
From Product of Big-O Estimates, we have: 
:$\dfrac {\map r x} {\ln x} = \map \OO {\sqrt x \ln x}$
So if we take:
:$\map R x = \dfrac {\map r x} {\ln x}$
for each $x \ge 2$, we have: 
:$\ds \map \pi x = \sum_{p \le x} 1 \ge \frac {\map \psi x} {\ln x} + \map R x$
with $R = \map \OO {\sqrt x \ln x}$. 
So we obtain: 
:$\ds \frac {\map \psi x} {\ln x} + \map R x \le \map \pi x \le \frac {2 \map \psi x} {\ln x} + \sqrt x$
with $R = \map \OO {\sqrt x \ln x}$ as required.
Category:Second Chebyshev Function
Category:Prime-Counting Function
Category:Inequalities
Bounds for Rank of Subset

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\rho: \powerset S \to \Z$ be the rank function of $M$.
Let $A \subseteq S$ be subset of $S$.
Then:
:$0 \le \map \rho A \le \size A$
== Proof ==
By definition of the rank function:
From Cardinality of Subset of Finite Set:
:$\forall X \subseteq A : \size X \le \size A$
In particular:
:$\forall X \subseteq A : X \in \mathscr I$ then $\size X \le \size A$
From Max Operation Yields Supremum of Operands:
:$\max \set {\size X : X \subseteq A \land X \in \mathscr I} \le \size A$
From Empty Set is Subset of All Sets:
:$\O \subseteq A$
From Cardinality of Empty Set:
:$\size \O = 0$
By matroid axiom $(\text I 1)$:
:$\O \in \mathscr I$
From Max Operation Yields Supremum of Operands:
:$0 \le \max \set {\size X : X \subseteq A \land X \in \mathscr I}$
It follows that:
:$0 \le \map \rho A \le \size A$
== Sources ==
* Chapter $1.$ $\S 6.$ Properties of the rank function
Category:Matroid Rank Functions
Bounds for Weierstrass Elementary Factors

== Theorem ==
Let $E_p: \C \to \C$ denote the $p$th Weierstrass elementary factor:
$\quad\map {E_p} z = \begin {cases} 1 - z & : p = 0 \\ 
\paren {1 - z} \map \exp {z + \dfrac {z^2} 2 + \cdots + \dfrac {z^p} p} & : \text {otherwise} \end {cases}$
Let $z \in \C$.
=== Some bound ===
Let $\cmod z \le \dfrac 1 2$.
Then:
:$\cmod {\map {E_p} z - 1} \le 3 \cmod z^{p + 1}$
=== Another bound ===
Let $\cmod z \le 1$.
Then:
:$\cmod {\map {E_p} z - 1} \le \cmod z^{p + 1}$
== Proof ==
=== Proof of some bound ===
Let $\cmod z \le \dfrac 1 2$.
We may assume $p \ge 1$.
We have:
:$\map {E_p} z = \map \exp {\map \log {1 - z} + \ds \sum_{k \mathop = 1}^p \frac {z^k} k}$
Then:
Because $p \ge 1$:
:$2 \cmod z^{p + 1} \le \dfrac 1 2$
By Bounds for Complex Exponential:
:$\cmod {\map {E_p} z - 1} \le 3 \cmod z^{p + 1}$
=== Proof of another bound ===
Let $\cmod z \le 1$.
We may assume $p \ge 1$.
For a fixed $p$, let:
:$\map {E_p} z = \paren {1 - z} \map \exp {z + \dfrac {z^2} 2 + \cdots + \dfrac {z^p} p} = 1 + \ds \sum_{k \mathop = 1}^\infty a_k z^k$
be its power series expansion about $z = 0$.
By Derivative of Complex Power Series we obtain:
:$\map {E_p'} z = \ds \sum_{k \mathop = 1}^\infty k a_k z^{k - 1} = -z^p \map \exp {z + \cdots + \dfrac {z^p} p}$
Comparing the two expressions gives two pieces of information about the coefficients $a_k$.
First:
:$a_1 = a_2 = \cdots = a_p = 0$
Second, since the coefficients of the expansion of $\map \exp {z + \cdots + \dfrac {z^p} p}$ are all positive:
:$a_k \le 0$ for $k \ge p + 1$
Thus:
:$\cmod {a_k} = -a_k$ for $k \ge p + 1$
This gives:
:$0 = \map {E_p} 1 = 1 + \ds \sum_{k\mathop = p + 1}^\infty a_k$
or:
:$\ds \sum_{k \mathop = p + 1}^\infty \cmod {a_k} = -\sum_{k \mathop = p + 1}^\infty a_k = 1$
Hence:
== Also see ==
* Weierstrass Factorization Theorem, what this is made for
* Bounds for Complex Exponential
* Bounds for Complex Logarithm
== Sources ==
* : $15$: Zeros of holomorphic functions: The Weierstrass Factorization Theorem: Lemma $15.8$
* $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Lemma $5.11$
Category:Infinite Products
Bounds of GCD for Sum and Difference Congruent Squares

== Theorem ==
Let $x, y, n$ be integers.
Let:
:$x \not \equiv \pm y \pmod n$
and:
:$x^2 \equiv y^2 \pmod n$
where $a \equiv b \pmod n$ denotes that $a$ is congruent to $b$ modulo $n$.
Then:
:$1 < \gcd \set {x - y, n} < n$
and:
:$1 < \gcd \set {x + y, n} < n$
where $\gcd \set {a, b}$ is the GCD of $a$ and $b$.
== Proof ==
But since $x \not \equiv -y \pmod n$, then:
:$n \nmid \paren {x + y}$
and since $x \not \equiv y \pmod n$, then:
:$n \nmid \paren {x - y}$
Therefore:
:$\gcd \set {x - y, n} < n$
and:
:$\gcd \set {x + y, n} < n$
So if $p \divides \paren {x - y}$ then:
:$1 < \gcd \set {x - y, n} < n$
and also there exists $q$ such that:
:$q \divides n$
:$q \divides \paren {x + y}$
:$1 < q \le \gcd \set {x + y, n}$
Likewise if $p \divides \paren {x + y}$ then:
:$1 < \gcd \set {x + y, n} < n$
and also there exists $q$ such that:
:$q \divides n$
:$q \divides \paren {x - y}$
:$1 < q \le \gcd \set {x - y, n}$
Category:Greatest Common Divisor
Bounds of Natural Logarithm

== Theorem ==
Let $\ln x$ be the natural logarithm of $x$ where $x \in \R_{>0}$.
Then $\ln$ satisfies the compound inequality:
:$1 - \dfrac 1 x \le \ln x \le x - 1$
== Proof ==
From Upper Bound of Natural Logarithm:
:$\ln x \le x - 1$
From Lower Bound of Natural Logarithm:
:$1 - \dfrac 1 x \le \ln x$
== Also see ==
* Bounds for Complex Logarithm
Category:Analysis
Category:Logarithms
Category:Inequalities
Bounds of Riemannian Metric by Euclidean Metric on Euclidean Space

== Theorem ==
Let $W \subseteq \R^n$ be a subset of an $n$-dimensional Riemannian manifold.
Let $g$ be a Riemannian metric on $W$.
Let $\tilde g$ be a Euclidean metric on $W$.
Let $K \subseteq W$ be a compact subset.
Let $c, C \in \R_{\mathop > 0}$ be positive constants.
Let $T_x \R^n$ be the tangent space of $\R^n$ at $x \in \R^n$.
Let $\size {\, \cdot \,}_g$ be the Riemannian inner product norm.
Then:
:$\forall K \subseteq W : \exists c, C \in \R_{\mathop > 0} : \forall x \in K : \forall v \in T_x \R^n : c \size v_{\tilde g} \le \size v_g \le C \size v_{\tilde g}$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Riemannian Geometry
Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering on $A$.
Let $a \in A$ be Hermitian.
Then we have $\map {\sigma_A} x \subseteq \closedint \alpha \beta$ :
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
In particular:
:$-\norm a {\mathbf 1}_A a \le_A \norm a {\mathbf 1}_A$
== Proof ==
=== Necessary Condition ===
Since $a$ is Hermitian, and ${\mathbf 1}_A$ is Hermitian from Identity Element in Unital *-Algebra is Hermitian, we have:
:$a - \lambda {\mathbf 1}_A$ is Hermitian for each $\lambda \in \R$.
From Spectral Mapping Theorem for Polynomials, we have:
:$\map {\sigma_A} {a - \alpha {\mathbf 1}_A} = \set {t - \alpha : t \in \map {\sigma_A} a} \subseteq \closedint \alpha \beta - \alpha$
We have $\closedint \alpha \beta - \alpha = \closedint 0 {\beta - \alpha} \subseteq \hointr 0 \infty$. 
Hence $a - \alpha {\mathbf 1}_A$ is positive, giving $\alpha {\mathbf 1}_A \le a$. 
From Spectral Mapping Theorem for Polynomials, we also have:
:$\map {\sigma_A} {\beta {\mathbf 1}_A - a} = \set {\beta - t : t \in \map {\sigma_A} a} \subseteq \beta - \closedint \alpha \beta$
Again, we have $\beta - \closedint \alpha \beta = \closedint 0 {\beta - \alpha} \subseteq \hointr 0 \infty$, so:
:$\beta {\mathbf 1}_A - a$ is positive, giving $a \le \beta {\mathbf 1}_A$.
Hence we have:
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
=== Sufficient Condition ===
Suppose that:
:$\alpha {\mathbf 1}_A \le_A a \le_A \beta {\mathbf 1}_A$
So that:
:$a - \alpha {\mathbf 1}_A \ge_A {\mathbf 0}_A$
and:
:$\beta {\mathbf 1}_A - a \ge_A {\mathbf 0}_A$
Then by the Spectral Mapping Theorem for Polynomials, we have that:
:$\map {\sigma_A} {a - \alpha {\mathbf 1}_A} = \set {t - \alpha : t \in \map {\sigma_A} a} \subseteq \hointr 0 \infty$
Hence $\map {\sigma_A} a \subseteq \hointr \alpha \infty$.
Similarly by the Spectral Mapping Theorem for Polynomials, we have that:
:$\map {\sigma_A} {\beta {\mathbf 1}_A - a} = \set {\beta - t : t \in \map {\sigma_A} a} \subseteq \hointl {-\infty} \beta$
Hence we have:
:$\map {\sigma_A} a \subseteq \hointr \alpha \infty \cap \hointl {-\infty} \beta = \closedint \alpha \beta$
=== Particular bound ===
From Spectrum of Element of Banach Algebra is Bounded, we have:
:$\cmod \lambda \le \norm a$ for each $a \in \map {\sigma_A} x$.
Since $\map {\sigma_A} x \subseteq \R$, we have $-\norm a \le \lambda \le \norm a$ for each $a \in \map {\sigma_A} x$.
Hence we can take $\alpha = -\norm a$ and $\beta = \norm a$ to obtain:
:$-\norm a {\mathbf 1}_A \le a \le \norm a {\mathbf 1}_A$
Category:Canonical Preorderings of C*-Algebras
Category:Hermitian Elements of *-Algebras
Bounds on Number of Odd Terms in Pascal's Triangle

== Theorem ==
Let $P_n$ be the number of odd elements in the first $n$ rows of Pascal's triangle.
Then:
:$0 \cdotp 812 \ldots < \dfrac {P_n} {n^{\lg 3} } < 1$
where $\lg 3$ denotes logarithm base $2$ of $3$.
The lower bound $0 \cdotp 812 \ldots$ is known as the Stolarsky-Harborth constant.
== Proof ==
== Also see ==
* Definition:Stolarsky-Harborth Constant
== Sources ==
* 
* 
* : $0,81256 6 \ldots$
* 
Category:Analytic Number Theory
Bounds on Number of Odd Terms in Pascal's Triangle/Mistake

== Source Work ==
:
:Thème et variations
::$0,81256 6 \ldots$
== Mistake ==
* Soit $P_n$ le nombre de termes impairs dans le $n$ premieres lignes du triangle de Pascal.
:Alors $0,812 \ldots < P_n / n^{\Log 2 / \Log 3} < 1$.
That is, in English:
* Let $P_n$ be the number of odd terms in the first $n$ rows of Pascal's triangle.
:Then $0 \cdotp 812 \ldots < P_n / n^{\ln 2 / \ln 3} < 1$.
== Correction ==
The expression ought to read:
:$P_n / n^{\Log 3 / \Log 2}$
that is:
:$P_n / n^{\ln 3 / \ln 2}$
== Also see ==
* Definition:Stolarsky-Harborth Constant
== Sources ==
* : $0,81256 6 \ldots$
Category:Mistakes/Les Nombres Remarquables
Bounds on Projection in Unital C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering of $A$.
Let $p$ be a projection on $A$.
Then we have:
:${\mathbf 0}_A \le_A p \le_A {\mathbf 1}_A$
== Proof ==
Since $p$ is a projection, it is Hermitian.
From Spectrum of Projection in *-Algebra: Corollary, we have:
:$\map {\sigma_A} p \subseteq \set {0, 1}$
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:${\mathbf 0}_A \le_A p \le_A {\mathbf 1}_A$
Category:Projections (*-Algebras)
Category:Canonical Preorderings of C*-Algebras
Bourbaki-Witt Fixed Point Theorem

== Theorem == 
Let $\struct {X, \le}$ be a non-empty chain complete ordered set (that is, an ordered set in which every chain has a supremum).
Let $f: X \to X$ be an inflationary mapping, that is, so that $\map f x \ge x$.
Then for every $x \in X$ there exists $y \in X$ where $y \ge x$ such that $\map f y = y$.
== Proof ==
Let $\gamma$ be the Hartogs number of $X$.
Suppose that $x \in X$.
Define $g : \gamma \to X$ by transfinite recursion as follows: 
:$\map g 0 = x$
:$\map g {\alpha + 1} = \map f {\map g \alpha}$
:$\map g \alpha = \sup \set {\map g \beta: \beta < \alpha}$ when $\alpha$ is a limit ordinal.
That $f$ is inflationary guarantees both that:
:$(1) \quad \set {\map g \beta: \beta < \alpha}$ is a chain for each $\alpha < \gamma$
:$(2) \quad g$ is increasing.
If $g$ is strictly increasing, then $g$ is strictly monotone and, by Strictly Monotone Mapping with Totally Ordered Domain is Injective, an injection.
But by definition of Hartogs number, $\gamma$ is the least ordinal such that there is no injection from $\gamma$ to $X$.
Therefore there must be some $\alpha < \beta < \gamma$ with $\map g \alpha = \map g \beta$.
Then:
:$\alpha + 1 \le \beta$
so:
:$\map g \alpha \le \map g {\alpha + 1} \le \map g \beta = \map g \alpha$
whence:
:$\map f {\map g \alpha} = \map g \alpha$
Because $x = \map g 0$:
:$\exists y \in \map g \gamma \subseteq \set {y \in X : x \le y}: \map f y = y$
== Sources ==
* 
* 
Category:Set Theory
Category:Fixed Point Theorems
Box-and-Whisker Diagram/Examples/Arbitrary Example 1

== Example of Box-and-Whisker Diagram ==
:
The diagram above shows two such box-and-whisker diagrams, illustrating samples with the five-number summaries:
:$(1): \quad 2 \quad 17 \quad 23 \quad 49 \quad 71$
:$(2): \quad 6 \quad 11 \quad 25 \quad 56 \quad 93$
== Sources ==
* 
* 
Category:Box-and-Whisker Diagrams
Box Topology contains Product Topology

== Theorem ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the product topology on $X$.
Let $\tau'$ be the box topology on $X$.
Then:
:$\tau \subseteq \tau'$
== Proof ==
From Natural Basis of Product Topology and Basis for Box Topology, it follows immediately that the natural basis $\BB$ for the product topology is contained in the basis $\BB'$ for the box topology.
From Corollary of Basis Condition for Coarser Topology, it follows that $\tau \subseteq \tau'$.
Category:Product Topology
Category:Box Topology
Box Topology may not be Coarsest Topology such that Projections are Continuous

== Theorem ==
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Then $\tau$ may not be the coarsest topology on $X$ for which the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.
== Proof ==
From Projection from Box Topology is Continuous, the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.
From Product Topology is Coarsest Topology such that Projections are Continuous, the product topology is the coarsest topology such that the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous.
Therefore, it is sufficient to show that in some cases the product topology is strictly coarser than the box topology.
That is, it is sufficient to show that in some cases the product topology is not equal to the box topology.
=== Lemma ===
Let $\struct{\R, \tau}$ denote the real numbers with Euclidean topology.
Let $\epsilon \in \R_{>0}$.
Let $U$ be the open interval $\openint {-\epsilon} {\epsilon}$.
From the Lemma:
:$U^\N$ is open in the box topology on $\R^\N$
:$U^\N$ is not open in the product topology on $\R^\N$
where $\R^\N$ is the countable-dimensional real Cartesian space.
Category:Box Topology
Box Topology may not be Coarsest Topology such that Projections are Continuous/Lemma

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $U \in \tau$ such that $U \ne \O$ and $U \ne X$.
Let:
:$Y = \ds \prod_{n \mathop \in \N} X = X \times X \times X \times \ldots$
be the countable Cartesian product of $\family {X}_{n \mathop \in \N}$.
Let $\tau_T$ be the product topology on $Y$.
Let $\tau_b$ be the box topology on $Y$.
Let:
:$V = \ds \prod_{n \mathop \in \N} U = U \times U \times U \times \ldots$
be the countable Cartesian product of $\family {U}_{n \mathop \in \N}$.
Then:
:$V$ is an element of the box topology $\tau_b$
:$V$ is not an element of the product topology $\tau_T$
== Proof ==
By the definition of the box topology:
:$V \in \tau_b$
Let $W \in \tau_T$.
=== Case 1 ===
Let $W \cap V = \O$.
Since $U \ne \O$, there exists $x \in U$.
Then $\family {x}_{n \mathop \in \N} \in V$
So $\family {x}_{n \mathop \in \N} \notin W$
By the definition of set equality:
:$V \ne W$
=== Case 2 ===
Let $W \cap V \ne \O$.
From Natural Basis of Product Topology, $W = \ds \prod_{n \mathop \in \N} W_n$ where:
:for all $n \in \N : W_n \in \tau$
:for all but finitely many indices $n : W_n = X$
Let $m = \max \set {n: W_n \ne X}$
Then:
:$W_{m + 1} = X$
Since $W \cap V \ne \O$, there exists $\family {z_n} \in W \cap V$.
Since $U \ne X$, there exists $y \in X \setminus U$.
Let $\family {y_n}$ be defined as:
:$y_n = \begin{cases}
z_n & : n \ne m + 1 \\
y & : n = m + 1
\end{cases}$
Since $y_{m + 1} = y \in X = W_{m + 1}$, it follows that:
:$\family {y_n} \in W$
Since $y_{m + 1} = y \notin U$, it follows that:
:$\family {y_n} \notin V$
By the definition of set equality:
:$W \ne V$
In either case $W \ne V$.
Since $W$ was arbitrary, it follows that for all $W \in \tau_T$, $W \ne V$.
Thus $V \notin \tau_T$ 
Category:Box Topology
Box Topology may not form Categorical Product in the Category of Topological Spaces

== Theorem ==
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
Then $\tau$ may not be the categorical product in the category of topological spaces.
== Proof ==
Category:Box Topology
Box Topology on Finite Product Space is Product Topology

== Theorem ==
Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $T_k = \struct {X_k, \tau_k}$ be topological spaces.
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the cartesian product of $X_1, \ldots, X_n$.
Then the box topology and the product topology on $X$ are identical.
== Proof ==
Denote the product topology on $X$ as $\tau$, and the box topology on $X$ as $\tau'$.
Suppose that $U \in \tau'$.
Then there exists an index set $I$ such that:
:$\ds U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} }$
where $U_{i, k} \in \tau_k$ for all $i \in I, k \in \set {1, \ldots, n}$.
When $\pr_k: X \to X_k$ denotes the $k$th projection on $X$, we have:
:$\pr_k^{-1} \sqbrk {U_{i, k} } = X_1 \times X_2 \times \cdots \times X_{k - 1} \times U_{i, k} \times X_{k + 1} \times \cdots \times X_n$
It follows that:
:$\ds U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} } = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^{n_i} \pr_k^{-1} \sqbrk {U_{i, k} }$
By definition of product topology, $U \in \tau$, so $\tau' \subseteq \tau$.
Suppose that $U \in \tau$.
Then there exists an index set $I$, so $U$ can be expressed as:
$U = \ds \bigcup_{i \mathop \in I} \bigcap_{l \mathop = 1}^{m_i} \pr_{j_l}^{-1} \sqbrk {U_{i, l} }$
where $m_i \in \N$, $j_1, j_2, \ldots, j_{m_i} \in \set {1, \ldots, n}$, and $U_{i, l} \in \tau_{j_l}$ for all $i \in I, l \in \set {1, \ldots, m_i}$.
Then:
As $\ds \bigcap_{l: k \mathop = j_l} U_{i, l} \in \tau_k$, it follows that:
:$\ds \struct {X_1 \times X_2 \times \cdots \times X_{k - 1} \times \bigcap_{l: k \mathop = j_l} U_{i, l} \times X_{k + 1} \times \cdots \times X_n} \in \tau'$
By definition of topology, it follows that $U \in \tau'$, so $\tau \subseteq \tau'$.
== Sources ==
* : $\S 19$
Category:Box Topology
Category:Product Topology
Brachistochrone Problem

== Problem ==
Let a point $A$ be joined by a wire to a lower point $B$.
Let the wire be allowed to be bent into whatever shape is required.
Let a bead be released at $A$ to slide down without friction to $B$.
What is the shape of the wire so that the bead takes least time to descend from $A$ to $B$?
== Also see ==
* Brachistochrone is Cycloid
* Definition:Brachistochrone
* Definition:Tautochrone
* Definition:Cycloid
== Historical Note ==
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : $\S 3$: Appendix $\text B$: Newton
* : Chapter $\text {A}.18$: Newton ()
* : Chapter $\text {B}.21$: The Cycloid
* 
* 
Category:Classic Problems
Category:Cycloids
Brachistochrone Problem/Historical Note

== Historical Note on Brachistochrone Problem ==
The Brachistochrone Problem was raised by to the readers of in June $1696$.
 interpreted the problem as a direct challenge to his abilities, and (despite being out of practice) solved the problem in the evening before going to bed.
He published it anonymously, but could tell whose solution it was, and commented:
:I recognise the lion by his print.
 published the solution in the in May $1697$, along with solutions by and .
:With justice we admire because he first discovered that a heavy particle slides down to the bottom of a cycloid in the same time, no matter where it starts. But you will be petrified with astonishment when I say that this very same cycloid, the tautochrone of , is also the brachistochrone we are seeking.
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== Also see ==
* Tautochrone Problem/Historical Note
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : Chapter $\text{VIII}$: Nature or Nurture?
* : $\S 3$: Appendix $\text B$: Newton
* : Chapter $\text {A}.18$: Newton ()
* : Chapter $\text {A}.20$: The Bernoulli Brothers
* : Chapter $\text {B}.21$: The Cycloid
* 
* 
* 
* 
* 
Category:Historical Notes
Brachistochrone is Cycloid

== Theorem ==
The shape of the brachistochrone is a cycloid.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Cycloid has Tautochrone Property, in which it is shown that a cycloid is also the shape for which it takes the same time for the bead to reach the bottom from wherever it is released.
== Historical Note ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples
* 
* 
* 
Category:Cycloids
Category:Brachistochrone is Cycloid
Brachistochrone is Cycloid/Proof 1

== Theorem ==
== Proof ==
:
Recall from the Snell-Descartes Law:
:$\dfrac {\sin \alpha_1} {v_1} = \dfrac {\sin \alpha_2} {v_2}$
Here, we invoke a generalization of the Snell-Descartes Law.
This is justified, as we are attempting to demonstrate the curve that takes the smallest time.
Thus we have $\dfrac {\sin \alpha} v = k$, where $k$ is some constant.
By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there.
Thus, at the point $\tuple {x, y}$, we have:
We have:
Combining all the above equations, we get:
where $c$ is another (more convenient) constant.
This is the differential equation which defines the brachistochrone.
Now we solve it:
Now we introduce a change of variable:
Let $\sqrt {\dfrac y {c - y} } = \tan \phi$
Thus:
Also:
Thus:
As the curve goes through the origin, we have $x = y = 0$ when $\phi = 0$ and so $c_1 = 0$.
Now we can look again at our expression for $y$:
To simplify the constants, we can substitute $a = c / 2$ and $\theta = 2 \phi$, and thus we get:
which are the parametric equations of the cycloid.
== Historical Note ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis
* : Chapter $\text {B}.21$: The Cycloid
* : Chapter $\text {B}.22$: Bernoulli's Solution of the Brachistochrone Problem
Category:Brachistochrone is Cycloid