nyuuzyou/wikis huggingface dataset:
Continued Fraction Expansion of Irrational Square Root/Examples/5

== Examples of Continued Fraction Expansion of Irrational Square Root ==
The continued fraction expansion of the square root of $5$ is given by:
:$\sqrt 5 = \sqbrk {2, \sequence 4}$
=== Convergents ===
== Proof ==
Let $\sqrt 5 = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$
From Partial Denominators of Continued Fraction Expansion of Irrational Square Root, the partial denominators of this continued fraction expansion can be calculated as:
:$a_r = \floor {\dfrac {\floor {\sqrt 5} + P_r} {Q_r} }$
where:
:$P_r = \begin {cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end {cases}$
:$Q_r = \begin {cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end {cases}$
|}
and the pattern is established:
:$\sequence 4$
== Sources ==
* 
* 
Category:Examples of Continued Fractions
Category:5
Continued Fraction Expansion of Irrational Square Root/Examples/5/Convergents

== Convergents to Continued Fraction Expansion of $\sqrt 5$ ==
The sequence of convergents to the continued fraction expansion of the square root of $5$ begins:
:$\dfrac 2 1, \dfrac 9 4, \dfrac {38} {17}, \dfrac {161} {72}, \dfrac {682} {305}, \dfrac {2889} {1292}, \dfrac {12238} {5473}, \dfrac {51841} {23184}, \ldots$
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then the $n$th convergent is $\dfrac {p_n} {q_n}$.
By definition:
:$p_k = \begin {cases} a_0 & : k = 0 \\
a_0 a_1 + 1 & : k = 1 \\
a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$
:$q_k = \begin {cases} 1 & : k = 0 \\
a_1 & : k = 1 \\
a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$
From Continued Fraction Expansion of $\sqrt 5$:
:$\sqrt 5 = \sqbrk {2, \sequence 4}$
Thus the convergents are assembled:
|}
== Sources ==
* 
Category:Examples of Continued Fractions
Category:5
Continued Fraction Expansion of Irrational Square Root/Examples/61

== Examples of Continued Fraction Expansion of Irrational Square Root ==
The continued fraction expansion of the square root of $61$ is given by:
:$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$
=== Convergents ===
== Proof ==
Let $\sqrt {61} = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$
From Partial Denominators of Continued Fraction Expansion of Irrational Square Root, the partial denominators of this continued fraction can be calculated as:
:$a_r = \floor {\dfrac {\floor {\sqrt {61} } + P_r} {Q_r} }$
where:
:$P_r = \begin {cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end {cases}$
:$Q_r = \begin {cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end {cases}$
|}
and the cycle is complete:
:$\sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}$
Category:Examples of Continued Fractions
Category:61
Continued Fraction Expansion of Irrational Square Root/Examples/61/Convergents

== Convergents to Continued Fraction Expansion of $\sqrt {61}$ ==
The sequence of convergents to the continued fraction expansion of the square root of $61$ begins:
:$\dfrac 7 1, \dfrac {8} 1, \dfrac {39} 5, \dfrac {125} {16}, \dfrac {164} {21}, \dfrac {453} {58}, \dfrac {1070} {137}, \dfrac {1523} {195}, \dfrac {5639} {722}, \dfrac {24079} {3083}, \ldots$
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \ge \mathop 0}$ and $\sequence {q_n}_{n \ge \mathop 0}$ be its numerators and denominators.
Then the $n$th convergent is $p_n / q_n$.
By definition:
:$p_k = \begin {cases} a_0 & : k = 0 \\
a_0 a_1 + 1 & : k = 1 \\
a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$
:$q_k = \begin {cases} 1 & : k = 0 \\
a_1 & : k = 1 \\
a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$
From Continued Fraction Expansion of $\sqrt {61}$:
:$\sqrt {61} = \sqbrk {7, \sequence {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14} }$
Thus the convergents are assembled:
|}
Category:Examples of Continued Fractions
Category:61
Continued Fraction Expansion of Irrational Square Root/Examples/8

== Example of Continued Fraction Expansion of Irrational Square Root ==
The continued fraction expansion of the square root of $8$ is given by:
:$\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$
=== Convergents ===
== Proof ==
Let $\sqrt 8 = \sqbrk {a_0, a_1, a_2, a_3, \ldots}$
From Partial Denominators of Continued Fraction Expansion of Irrational Square Root, the partial denominators of this continued fraction expansion can be calculated as:
:$a_r = \floor {\dfrac {\floor {\sqrt 8} + P_r} {Q_r} }$
where:
:$P_r = \begin {cases} 0 & : r = 0 \\
a_{r - 1} Q_{r - 1} - P_{r - 1} & : r > 0 \\
\end {cases}$
:$Q_r = \begin {cases} 1 & : r = 0 \\
\dfrac {n - {P_r}^2} {Q_{r - 1} } & : r > 0 \\
\end {cases}$
|}
and the cycle is complete:
:$\sequence {1, 4}$
Category:Examples of Continued Fractions
Category:8
Continued Fraction Expansion of Irrational Square Root/Examples/8/Convergents

== Convergents to Continued Fraction Expansion of $\sqrt 8$ ==
The sequence of convergents to the continued fraction expansion of the square root of $8$ begins:
:$\dfrac 2 1, \dfrac 3 1, \dfrac {14} 5, \dfrac {17} 6, \dfrac {82} {29}, \dfrac {99} {35}, \dfrac {478} {169}, \dfrac {577} {204}, \dfrac {2786} {985}, \dfrac {3363} {1189}, \ldots$
== Proof ==
Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then the $n$th convergent is $\dfrac {p_n} {q_n}$.
By definition:
:$p_k = \begin {cases} a_0 & : k = 0 \\
a_0 a_1 + 1 & : k = 1 \\
a_k p_{k - 1} + p_{k - 2} & : k > 1 \end {cases}$
:$q_k = \begin {cases} 1 & : k = 0 \\
a_1 & : k = 1 \\
a_k q_{k - 1} + q_{k - 2} & : k > 1 \end {cases}$
From Continued Fraction Expansion of $\sqrt 8$:
:$\sqrt 8 = \sqbrk {2, \sequence {1, 4} }$
Thus the convergents are assembled:
|}
Category:Examples of Continued Fractions
Category:8
Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself

== Theorem ==
Let $\sequence {a_n}_{n \mathop \ge 0}$ be a simple infinite continued fractions in $\R$.
Then $\sequence {a_n}_{n \mathop \ge 0}$ converges to an irrational number, whose continued fraction expansion is $\sequence {a_n}_{n \mathop \ge 0}$.
== Proof ==
By Simple Infinite Continued Fraction Converges to Irrational Number, the value of $\sequence {a_n}_{n \mathop \ge 0}$ exists and is irrational.
Let $\sequence {b_n}_{n \mathop \ge 0}$ be its continued fraction expansion.
By Continued Fraction Expansion of Irrational Number Converges to Number Itself, $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same value.
The result will be achieved by the Second Principle of Mathematical Induction.
First we note that if $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ then $a_0 = b_0$ since both are equal to the integer part of the common value.
This is our basis for the induction.
Now suppose that for some $k \ge 1$, we have:
:$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$.
Then all need to do is show that $a_{k + 1} = b_{k + 1}$.
Now:
:$\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {a_0, a_1, \ldots, a_k, \sqbrk {a_{k + 1}, a_{k + 2}, \ldots} }$
and similarly:
:$\sqbrk {b_0, b_1, b_2, \ldots} = \sqbrk {b_0, b_1, \ldots, b_k, \sqbrk {b_{k + 1}, b_{k + 2}, \ldots} }$
As these have the same value and have the same first $k$ partial denominators, it follows that:
:$\sqbrk {a_{k + 1}, a_{k + 2}, \ldots} = \sqbrk {b_{k + 1}, b_{k + 2}, \ldots}$
But now $a_{k + 1} = b_{k + 1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.
Hence the result.
== Also see ==
* Simple Infinite Continued Fraction is Uniquely Determined by Limit
Category:Simple Continued Fractions
Continued Fraction Expansion of Pi

== Theorem ==
The constant $\pi$ (pi) has the continued fraction expansion:
:$\pi = \sqbrk {3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, \ldots}$
=== Convergents ===
== Proof ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Pi
Category:Examples of Continued Fractions
Continued Fraction Expansion of Pi/Convergents

== Theorem ==
The convergents of the continued fraction expansion to $\pi$ (pi) are:
:$3, \dfrac {22} 7, \dfrac {333} {106}, \dfrac {355} {113}, \dfrac {103993} {33102}, \dfrac {104348} {33215}$
These best rational approximations are accurate to $0, 2, 4, 6, 9, 9, 9, 10, 11, 11, 12, 13, \ldots$ decimals.
=== Zu Chongzhi Fraction ===
== Proof ==
== Historical Note ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
Category:Pi
Category:Examples of Continued Fractions
Continued Fraction Expansion of Pi/Convergents/Historical Note

== Historical Note on $\pi$ is Irrational ==
The convergents of the continued fraction expansion to $\pi$ (pi) were calculated from $\dfrac {103 \, 993} {33 \, 102}$ up to $\dfrac {1 \, 019 \, 514 \, 486 \, 099 \, 146} {324 \, 521 \, 540 \, 032 \, 945}$ by .
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Historical Notes
Continued Fraction Expansion via Gauss Map

== Theorem ==
Let $T : \closedint 0 1 \to \closedint 0 1$ be the Gauss map.
Let $x \in \closedint 0 1 \setminus \Q$.
Then $x$ has the simple infinite continued fraction:
where:
:$\map {a_n} x := \floor {\dfrac 1 {\map {T^{n - 1} } x} }$
:$\floor \cdot$ denotes the floor.
== Proof ==
For $x \in \closedint 0 1 \setminus \Q$, we have:
:$\map {a_1} x = \floor {\dfrac 1 x}$
and:
:$\forall n \in \N_{>0} : \map {a_n} x = \map {a_1} { T^{n-1} x }$
Category:Continued Fractions
Continued Fraction Identities/First/Infinite

== Theorem ==
Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a simple infinite continued fraction.
Then:
:$\left[{a_1, a_2, a_3, \ldots}\right] = a_1 + \dfrac 1 {\left[{a_2, a_3, \ldots}\right]}$
== Proof ==
Category:Continued Fractions
Continued Fraction for Exponential Function

== Theorem ==
:$e^x = \cfrac 1 {1 - \cfrac x {1 + x - \cfrac x {2 + x - \cfrac {2 x } {3 + x - \cfrac {3 x} {4 + x - \cfrac {\ddots} {\ddots } } } } } }$
=== Corollary ===
== Proof ==
From Euler's Continued Fraction Formula, we have:
:$a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots } } } } }$
Therefore:
== Also see ==
* Continued Fraction for Real Arctangent Function
* Continued Fraction for Logarithm of 1 + x
* Continued Fraction for Real Arcsine Function
== Sources ==
* 
Category:Continued Fractions
Category:Examples of Euler's Continued Fraction Formula
Category:Exponential Function
Continued Fraction for Exponential Function/Corollary

== Corollary to Continued Fraction for Exponential Function == 
:$e = \cfrac 1 {1 - \cfrac 1 {2 - \cfrac 1 {3 - \cfrac 2 {4 - \cfrac {\ddots} {\ddots } } } } }$
== Proof ==
From Continued Fraction for Exponential Function, we have:
:$e^x = \cfrac 1 {1 - \cfrac x {1 + x - \cfrac x {2 + x - \cfrac {2 x } {3 + x - \cfrac {\ddots} {\ddots } } } } }$
Therefore:
Category:Continued Fractions
Category:Continued Fraction Expansion of Euler's Number
Category:Euler's Number
Category:Examples of Continued Fractions
Continued Fraction for Logarithm of 1 + x

== Theorem ==
For all $x \in \R$ such that $-1 < x \le 1$:
:$\map \ln {1 + x} = \cfrac x {1 + \cfrac x {2 - x + \cfrac {2^2 x} {3 - 2 x + \cfrac {3^2 x } {4 - 3 x + \cfrac {\ddots} {\ddots } } } } }$
== Proof ==
From Euler's Continued Fraction Formula, we have:
:$a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} \ddots} } } }$
Therefore:
== Also see ==
* Continued Fraction for Real Arctangent Function
* Continued Fraction for Exponential Function
* Continued Fraction for Real Arcsine Function
== Sources ==
* : Pages $\text {32 - 81}$
Category:Continued Fractions
Category:Examples of Euler's Continued Fraction Formula
Category:Natural Logarithms
Continued Fraction for Real Arcsine Function

== Theorem ==
Let $-1 \le x \le 1$
Then:
:$\arcsin x = \cfrac x {1 - \cfrac {x^2} {2 \times 3 + x^2 - \cfrac {2 \times 3 \times \paren {3 x}^2} {4 \times 5 + \paren {3 x}^2 - \cfrac {4 \times 5 \times \paren {5 x}^2} {6 \times 7 + \paren {5 x}^2 - \cfrac {6 \times 7 \times \paren {7 x}^2} {\ddots } } } } }$
== Proof ==
From Euler's Continued Fraction Formula, we have:
:$a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots } } } } }$
Therefore:
== Also see ==
* Continued Fraction for Exponential Function
* Continued Fraction for Logarithm of 1 + x
* Continued Fraction for Real Arctangent Function
== Sources ==
* : Pages $\text {32 - 81}$
Category:Arcsine Function
Category:Continued Fractions
Category:Examples of Euler's Continued Fraction Formula
Continued Fraction for Real Arctangent Function

== Theorem ==
Let $-1 \le x \le 1$
Then:
:$\arctan x = \cfrac x {1 + \cfrac {x^2} {3 - x^2 + \cfrac {\paren {3 x}^2} {5 - 3 x^2 + \cfrac {\paren {5 x}^2} {7 - 5 x^2 + \cfrac {\paren {7 x}^2} {\ddots } } } } }$
== Proof ==
From Euler's Continued Fraction Formula, we have:
:$a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots } } } } }$
Therefore:
== Also see ==
* Brouncker's Formula
* Continued Fraction for Exponential Function
* Continued Fraction for Logarithm of 1 + x
* Continued Fraction for Real Arcsine Function
== Sources ==
* : Pages $\text {32 - 81}$
Category:Arctangent Function
Category:Continued Fractions
Category:Examples of Euler's Continued Fraction Formula
Continued Fraction for Real Hyperbolic Tangent Function

== Theorem ==
Let $x \in \R$.
Then:
:$\tanh x = \cfrac x {1 + \cfrac {x^2} {3 + \cfrac {x^2} {5 + \cfrac {x^2} {7 + \cfrac {x^2} {9 + \cfrac {x^2} \ddots} } } } }$
== Proof ==
From Hypergeometric Continued Fraction Formulation 1, we have:
:$\dfrac {\map { {}_0 \operatorname F_1} {a + 1; x} } {\map { {}_0 \operatorname F_1} {a; x} } = \cfrac a {a + \cfrac x {\paren {a + 1} + \cfrac x {\paren {a + 2} + \cfrac x {\paren {a + 3} + \cfrac x {\ddots + \cfrac \ddots \ddots} } } } }$
Therefore:
== Also see ==
* Continued Fraction for Exponential Function
* Continued Fraction for Logarithm of 1 + x
* Continued Fraction for Real Arctangent Function
* Continued Fraction for Tangent Function
Category:Hyperbolic Tangent Function
Category:Continued Fractions
Category:Hypergeometric Functions
Continued Fraction for Tangent Function

== Theorem ==
Let $x \in \R$.
Then:
:$\tan x = \cfrac x {1 - \cfrac {x^2} {3 - \cfrac {x^2} {5 - \cfrac {x^2} {7 - \cfrac {x^2} {9 - \cfrac {x^2} \ddots} } } } }$
== Proof ==
From Hypergeometric Continued Fraction Formulation 1, we have:
:$\dfrac {\map { {}_0 \operatorname F_1} {a + 1; x} } {\map { {}_0 \operatorname F_1} {a; x} } = \cfrac a {a + \cfrac x {\paren {a + 1} + \cfrac x {\paren {a + 2} + \cfrac x {\paren {a + 3} + \cfrac x {\ddots + \cfrac \ddots \ddots} } } } }$
Therefore:
== Also see ==
* Continued Fraction for Exponential Function
* Continued Fraction for Real Hyperbolic Tangent Function
* Continued Fraction for Logarithm of 1 + x
* Continued Fraction for Real Arctangent Function
Category:Tangent Function
Category:Continued Fractions
Category:Hypergeometric Functions
Continued Square Root of 1, 2, 3, 4, .../Mistake

== Source Work ==
:
:The Dictionary
:: $3$
== Mistake ==
:The value of the infinite nested root, $\sqrt {1 + \sqrt {2 + \sqrt {3 + \sqrt {4 + \cdots} } } }$ [equals $3$].
It is suspected that the given expression is a misprint for:
:$\sqrt {1 + 2 \sqrt {1 + 3 \sqrt { 1 + \cdots} } }$
which famously equals $3$.
== Also see ==
* Ramanujan's Infinite Nested Roots
== Sources ==
* : $3$
Category:Mistakes/Curious and Interesting Numbers
Continuity Correction/Examples/Binomial Distribution

== Example of Continuity Correction ==
Let $X$ be a discrete random variable which obeys a binomial distribution.
Let us approximate $X$ by a normal distribution.
Then we apply a continuity correction of $0 \cdotp 5$ such that:
:$\map \Pr {X \le 15}$
is best approximated by:
:$\map \Pr {X \le 15 \cdotp 5}$ (normal)
== Sources ==
* 
* 
Category:Examples of Continuity Corrections
Continuity Defined by Closure

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous :
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$ in $T_1$.
That is, the image of the closure is a subset of the closure of the image.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Closure of Image under Continuous Mapping is not necessarily Image of Closure demonstrating that it is not necessarily the case that $f \sqbrk {H^-} = \paren {f \sqbrk H}^-$
* Closed Image of Closure of Set under Continuous Mapping equals Closure of Image demonstrating that if $f \sqbrk {H^-}$ s closed then it does follow that $f \sqbrk {H^-} = \paren {f \sqbrk H}^-$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Set Closures
Category:Continuous Mappings
Category:Continuity Defined by Closure
Continuity Defined by Closure/Proof 1

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $f$ be continuous.
Let $y \in f \sqbrk {\map \cl H}$.
Then:
:$\exists x \in \map \cl H: y = \map f x$
Let $U$ be an open set of $T_2$ such that $y \in U$.
Then by definition of continuous mapping:
:$f^{-1} \sqbrk U$ is an open set of $T_1$ such that:
::$x \in f^{-1} \sqbrk U$
Hence:
:$f^{-1} \sqbrk U \cap H \ne \O$
as $x \in \map \cl H$.
Hence:
and:
:$y \in \map \cl {f \sqbrk H}$
That is:
:$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
=== Sufficient Condition ===
Suppose that for all $H \subseteq S_1$:
:$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
Let $V \subseteq S_2$ be closed in $T_2$.
Then:
:$f \sqbrk {\map \cl {f^{-1} \sqbrk V} } \subseteq \map \cl V = V$
So by Set is Closed iff Equals Topological Closure:
:$\map \cl {f^{-1} \sqbrk V} \subseteq f^{-1} \sqbrk V$
and so $f^{-1} \sqbrk V$ is closed in $T_1$.
Hence from Continuity Defined from Closed Sets:
:$f$ is continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 25$
Category:Continuity Defined by Closure
Continuity Defined from Closed Sets

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping.
Then $f$ is continuous for all $V$ closed in $T_2$, $f^{-1} \sqbrk V$ is closed in $T_1$.
== Proof ==
First we show the following.
Let $W \in T_2$.
We note that $f^{-1} \sqbrk {T_2} = T_1$.
Hence, from Preimage of Set Difference under Mapping, we have:
:$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$
=== Necessary Condition ===
Suppose the condition on closed sets holds.
Let $U$ be open in $T_2$.
By Relative Complement of Relative Complement, $T_2 \setminus \paren { T_2 \setminus U } = U$.
Therefore $T_2 \setminus \paren { T_2 \setminus U }$ is open in $T_2$.
Then $T_2 \setminus U$ is closed in $T_2$.
By hypothesis, $f^{-1} \sqbrk {T_2 \setminus U} = T_1 \setminus f^{-1} \sqbrk U$ is closed in $T_1$.
So $f^{-1} \sqbrk U$ is open in $T_1$.
This is true for any $U \in T_2$.
Hence $f$ is continuous.
=== Sufficient Condition ===
Now let $f$ be continuous.
Let $V$ be closed in $T_2$.
Then $T_2 \setminus V$ is open in $T_2$.
As $f$ is continuous, $f^{-1} \sqbrk {T_2 \setminus V} = T_1 \setminus f^{-1} \sqbrk V$ is open in $T_1$.
So $f^{-1} \sqbrk V$ is closed in $T_1$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.8$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Closed Sets
Category:Continuous Mappings
Continuity Equation/Examples/Increase of Fluid Mass

== Examples of Continuity Equations ==
The continuity equation for mass equates the rate of increase of mass in a volume of fluid to the net rate of flow of mass into that volume:
:$\dfrac {\partial \rho} {\partial t} + \nabla \cdot \paren {\rho \mathbf v} = 0$
where:
:$\rho$ is the density of the fluid
:$\dfrac {\partial \rho} {\partial t}$ is the rate of change of density at an arbitrary point
:$\mathbf v$ denotes the velocity of the fluid at that point
:$\nabla$ is the del operator.
== Sources ==
* 
* 
Category:Examples of Continuity Equations
Category:Hydrodynamics
Continuity Test for Real-Valued Functions

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $f: S \to \R$ be a real-valued function.
Let $x \in S$.
Then $f$ is continuous at $x$ :
:$\forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
=== Corollary ===
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Category:Continuous Mappings
Category:Continuity Test for Real-Valued Functions
Continuity Test for Real-Valued Functions/Everywhere Continuous

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $f: S \to \R$ be a real-valued function.
Then $f$ is everywhere continuous :
:$\forall x \in S : \forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
== Proof ==
By definition, $f$ is everywhere continuous $f$ is continuous at every point $x \in S$.
From Continuity Test for Real-Valued Functions, $f$ is continuous at every point $x \in S$ :
:$\forall x \in S : \forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
Category:Continuity Test for Real-Valued Functions
Continuity Test for Real-Valued Functions/Necessary Condition

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $f: S \to \R$ be a real-valued function.
Let $x \in S$.
Let $f$ be continuous at $x$
Then:
:$\forall \epsilon \in \R_{>0} : \exists U \in \tau : x \in U : \map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
== Proof ==
Let $f$ be continuous at $x$.
Then by definition:
:For every neighborhood $N$ of $\map f x$ in $\R$, there exists a neighborhood $M$ of $x$ in $S$ such that $\map {f^\to} M \subseteq N$.
From:
:Open Ball in Real Number Line is Open Interval
:Open Ball of Metric Space is Open Set
:Set is Open iff Neighborhood of all its Points 
it follows that:
:For every $\epsilon \in \R_{>0}$, there exists a neighborhood $M$ of $x$ in $S$ such that $\map {f^\to} M \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
By definition of a neighborhood in $\struct {S, \tau}$:
:for all neighborhoods $M$ there exists $U \in \tau$ such that $z \in U \subseteq M$
Thus:
:For every $\epsilon \in \R_{>0}$, there exists $U \in \tau$ with $x \in U$ such that $\map {f^\to} U \subseteq \openint {\map f x - \epsilon} {\map f x + \epsilon}$
Category:Continuity Test for Real-Valued Functions