nyuuzyou/wikis huggingface dataset:
Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each

== Theorem ==
Consider the sequence of Wythoff pairs arranged in sequential order:
:$\tuple {0, 0}, \tuple {1, 2}, \tuple {3, 5}, \tuple {4, 7}, \tuple {6, 10}, \tuple {8, 13}, \ldots$
Apart from the first Wythoff pair $\tuple {0, 0}$, every positive integer appears in this sequence exactly once.
== Proof ==
By definition, the $n$th Wythoff pair is $\tuple {\floor {n \phi}, \floor {n \phi^2} }$.
Thus the coordinates of the sequence of Wythoff pairs are the terms of the lower and upper Wythoff sequences.
By definition:
:the lower Wythoff sequence is the Beatty sequence $\BB_x$ on the golden section $\phi$.
:the upper Wythoff sequence is the complementary Beatty sequence on the golden section $\phi$.
Also by definition, the complementary Beatty sequence on $x$ is the integer sequence formed by the integers which are missing from $\BB_x$.
Hence the result.
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Beatty Sequences
Sequence on Product Space Converges to Point iff Projections Converge to Projections of Point

== Theorem ==
Let $I$ be an arbitrary index set.
For all $i \in I$, let $T_i = \struct {X_i, \tau_i}$ be topological spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $\tau$ be the product topology on $X$.
Denote by $\pr_i : X \to X_i$ the projection from $X$ onto $X_i$.
Let $\sequence {x_n}$ be a sequence on $X$ and let $x \in X$.
Then $\sequence {x_n}$ converges to $x$ :
:for all $i \in I$ the sequence $\sequence {\map {\pr_i} {x_n} }$ converges to $\map {\pr_i} x$.
== Proof ==
=== Necessary Condition ===
Let $x_n \to x$.
Let $i \in I$.
From Projection from Product Topology is Continuous it follows that $\pr_i$ is continuous.
By Continuous Mapping is Sequentially Continuous, $\pr_i$ is also sequentially continuous.
Hence $\map {\pr_i} {x_n} \to \map {\pr_i} x$.
=== Sufficient Condition ===
Let $\map {\pr_i} {x_n} \to \map {\pr_i} x$ for all $i \in I$.
Let $U \in \tau$ be an open neighborhood of $x$.
Let $\BB$ be the collection of sets of the form $\ds \prod_{i \mathop \in I} U_i$, where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
By Natural Basis of Product Topology and Synthetic Basis and Analytic Basis are Compatible it follows that $\BB$ is an analytic basis for $\tau$.
As $x \in U$ it follows that there exists $\ds \prod_{i \mathop \in I} U_i \in \BB$ such that:
:$\ds x \in \prod_{i \mathop \in I} U_i \subseteq U$ 
Let $F := \set {i \in I : U_i \neq X_i}$.
By definition of $\BB$, $F$ is finite.
By our hypothesis $\map {\pr_i} {x_n} \to \map {\pr_i} x$ it follows that:
:$\forall i \in F: \exists M_i \in \N : \forall n \ge M_i: \map {\pr_i} {x_n} \in U_{i}$
Furthermore, for $i \in I \setminus F$ we have $\map {\pr_i} {x_n} \in X_{i} = U_{i}$.
Thus for all $n \ge M := \max \set {M_i : i \in F}$ it holds that:
:$ x_n = \tuple {\map {\pr_i} {x_n} }_{i \mathop \in I} \in \prod_{i \mathop \in I} U_i \subseteq U$
Hence the result.
== Also see ==
* Filter on Product Space Converges to Point iff Projections Converge to Projections of Point
== Source ==
* : $7$ Complete Metric Spaces and Function Spaces: $\S 43$ Complete Metric Spaces: Lemma $43.3$
Category:Topology
Category:Convergence
Category:Sequences
Category:Projections
Sequences of 3 Consecutive Triangular Numbers which are Sphenic

== Theorem ==
Let $n \in \N$ be a natural number such that:
:$T_n$, $T_{n + 1}$ and $T_{n + 2}$ are all sphenic numbers
where $T_n$ denotes the $n$th triangular number.
The sequence of such $n$ begins:
:$406$, $861$, $39 \, 621$, $2 \, 166 \, 321$, $3 \, 924 \, 201$, $11 \, 146 \, 281$, $14 \, 804 \, 961$, $19 \, 198 \, 306$, $73 \, 951 \, 041$, $83 \, 417 \, 986$, $97 \, 951 \, 006$, $209 \, 643 \, 526$, $310 \, 415 \, 986$, $522 \, 339 \, 681$, $526 \, 225 \, 461$, $583 \, 333 \, 246$, $611 \, 153 \, 241$, $801 \, 460 \, 666$, $1 \, 601 \, 581 \, 906$, $2 \, 520 \, 251 \, 506$, $2 \, 690 \, 954 \, 841$, $4 \, 455 \, 349 \, 606$, $6 \, 681 \, 853 \, 401$, $9 \, 895 \, 642 \, 221$, $13 \, 878 \, 029 \, 901$
== Also see ==
* There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers‎
Category:Sphenic Numbers
Category:Triangular Numbers
Sequences of 4 Consecutive Integers with Falling Divisor Sum

== Theorem ==
The following ordered quadruple of consecutive integers have divisor sums which are strictly decreasing:
:$44, 45, 46, 47$
:$104, 105, 106, 107$
== Proof ==
== Also see ==
* Sequences of 4 Consecutive Integers with Rising Divisor Sum
== Sources ==
* : $44$
Category:Divisor Sum Function
Sequences of 4 Consecutive Integers with Rising Divisor Sum

== Theorem ==
The following ordered quadruples of consecutive integers have divisor sum values which are strictly increasing:
:$61, 62, 63, 64$
:$73, 74, 75, 76$
== Proof ==
== Also see ==
* Sequences of 4 Consecutive Integers with Falling Divisor Sum
== Sources ==
* : $61$
Category:Divisor Sum Function
Sequences of Projections in 2-Sequence Space Converges in Strong Operator Topology

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $p$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
For $n \in \N$ let $P_n \in \map {CL} {\ell^2}$ be the projection operator over $\ell^2$.
Let $\sequence {P_n}_{n \mathop \in \N}$ be a sequence.
Let $I \in \map {CL} {\ell^2}$ be the identity operator.
Then $\sequence {P_n}_{n \mathop \in \N}$ converges to $I$ in the strong operator topology.
== Proof ==
Let $\mathbf a = \tuple {a_1, a_2, \ldots} \in \ell^2$.
Then:
Hence:
:$\ds \lim_{n \mathop \to \infty } \norm {I \mathbf a - P_n \mathbf a}_2 = 0$
By definition, $\sequence {P_n}_{n \mathop \in \N}$ converges to $I$ in the strong operator topology.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$
Category:Convergence
Sequences of Projections in 2-Sequence Space do not Converge in Uniform Operator Topology

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $p$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
For $n \in \N$ let $P_n \in \map {CL} {\ell^2}$ be the projection operator over $\ell^2$.
Let $\sequence {P_n}_{n \mathop \in \N}$ be a sequence.
Let $I \in \map {CL} {\ell^2}$ be the identity operator.
Then $\sequence {P_n}_{n \mathop \in \N}$ does not converge to $I$ in the uniform operator topology.
== Proof ==
 $\sequence {P_n}_{n \mathop \in \N}$ converges to $I$ in the uniform operator topology.
By definition:
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {I - P_n} < \epsilon$
Suppose $\mathbf e_{N + 1} \in \ell^2$ is such that:
:$\mathbf e_{N + 1} = \tuple {\underbrace {0, \ldots, 0}_N, 1, 0 \ldots}$
Then:
This has to hold for all $\epsilon > 0$.
However, for $\epsilon < 1$ this does not hold.
Hence, we have a contradiction.
Therefore, $\sequence {P_n}_{n \mathop \in \N}$ does not converge to $I$ in the uniform operator topology.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$
Category:Convergence
Sequences of Three Consecutive Strictly Increasing Euler Phi Values

== Theorem ==
The following sequences of $3$ consecutive positive integers have the property that their Euler $\phi$ values are strictly increasing:
:$\tuple {105, 106, 107}, \tuple {165, 166, 167}, \tuple {315, 316, 317}, \tuple {525, 526, 527}, \dots$
== Proof ==
== Sources ==
* : $523$
Category:Euler Phi Function
Sequential Characterization of Closed Linear Transformation

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces.
Let $\map D T$ be a linear subspace of $X$.
Let $T : \map D T \to Y$ be a linear transformation.
Then $T$ is closed :
:for each convergent sequence $\sequence {x_n}_{n \in \N}$ in $\struct {X, \norm {\, \cdot \,}_X}$ valued in $\map D T$ such that $\sequence {T x_n}_{n \in \N}$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$, we have:
::$\ds \lim_{n \mathop \to \infty} x_n \in \map D T$
:and:
::$\ds \lim_{n \mathop \to \infty} T x_n = \map T {\lim_{n \mathop \to \infty} x_n}$
== Proof ==
Let $\struct {X \times Y, \norm \cdot_{X \times Y} }$ be the direct product $X \times Y$ equipped with the direct product norm.
From Direct Product of Banach Spaces is Banach Space, $\struct {X \times Y, \norm \cdot_{X \times Y} }$ is a Banach space. 
Let:
:$\map G T = \set {\tuple {x, T x} \in X \times Y : x \in \map D T}$ 
be the graph of $T$.
=== Necessary Condition ===
Suppose that $T$ is closed.
Then $\map G T$ is closed in $X \times Y$. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $\struct {X, \norm {\, \cdot \,}_X}$ with limit $x$ such that $\sequence {T x_n}_{n \in \N}$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$ to a limit $y$.
From Convergence in Direct Product Norm, we have:
:$\tuple {x_n, T x_n} \to \tuple {x, y}$ in $\struct {X \times Y, \norm \cdot_{X \times Y} }$.
Since $\map G T$ is closed in $X \times Y$ and $\tuple {x_n, T x_n} \in \map G T$ for each $n \in \N$, we have:
:$\tuple {x, y} \in \map G T$
Hence $x \in \map D T$ and $y = T x$. 
=== Sufficient Condition ===
Suppose that:
:for each convergent sequence $\sequence {x_n}_{n \in \N}$ in $\struct {X, \norm {\, \cdot \,}_X}$ valued in $\map D T$ such that $\sequence {T x_n}_{n \in \N}$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$, we have:
::$\ds \lim_{n \mathop \to \infty} x_n \in \map D T$
:and:
::$\ds \lim_{n \mathop \to \infty} T x_n = \map T {\lim_{n \mathop \to \infty} x_n}$
Let $\sequence {\tuple {x_n, T x_n} }_{n \in \N}$ be a sequence in $\struct {X \times Y, \norm \cdot_{X \times Y} }$ valued in $\map G T$ converging to $\tuple {x, y}$.
From Convergence in Direct Product Norm, we have:
:$x_n \to x$
and:
:$T x_n \to y$ 
for some $x \in X$ and $y \in Y$.
By hypothesis, we have $x \in \map D T$ and $y = T x$. 
Hence from Convergence in Direct Product Norm, we have:
:$\ds \lim_{n \mathop \to \infty} \tuple {x_n, T x_n} = \tuple {x, T x} \in \map G T$
Hence $\map G T$ is closed.
Category:Closed Linear Transformations
Sequential Characterization of Limit at Minus Infinity of Real Function/Corollary

== Corollary ==
Let $f : \R \to \R$ be a real function. 
Let $L$ be a real number. 
Then: 
:$\ds \lim_{x \to -\infty} \map f x = L$ 
:
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$
where:
:$\ds \lim_{x \mathop \to -\infty} \map f x$ denotes the limit at $-\infty$ of $f$.
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$\ds \lim_{x \to -\infty} \map f x = L$ 
Then, from Sequential Characterization of Limit at Minus Infinity of Real Function, we have: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$. 
So, in particular: 
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$. 
=== Sufficient Condition ===
Suppose that: 
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$. 
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
From Peak Point Lemma, there exists a monotone subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$ of $\sequence {x_{n_j} }_{j \mathop \in \N}$.
From Subsequence of Real Sequence Diverging to Negative Infinity Diverges to Negative Infinity, we have: 
:$x_{n_{j_k} } \to -\infty$
So, from the hypothesis, we have: 
:$\map f {x_{n_{j_k} } } \to L$
So: 
:any subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$ has a subsequence converging to $L$.
So, from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit, we have: 
:$\map f {x_n} \to L$
Category:Sequential Characterization of Limit at Minus Infinity of Real Function
Sequential Characterization of Limit at Positive Infinity of Real Function/Corollary

== Corollary ==
Let $f : \R \to \R$ be a real function. 
Let $L$ be a real number. 
Then: 
:$\ds \lim_{x \to \infty} \map f x = L$ 
:
:for all increasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$
where:
:$\ds \lim_{x \mathop \to \infty} \map f x$ denotes the limit at $+\infty$ of $f$.
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$\ds \lim_{x \to \infty} \map f x = L$ 
Then, from Sequential Characterisation of Limit at Positive Infinity of Real Function, we have: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$. 
So, in particular: 
:for all increasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$. 
=== Sufficient Condition ===
Suppose that: 
:for all increasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to \infty$. 
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
From Peak Point Lemma, there exists a monotone subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$ of $\sequence {x_{n_j} }_{j \mathop \in \N}$.
From Subsequence of Real Sequence Diverging to Positive Infinity Diverges to Positive Infinity, we have: 
:$x_{n_{j_k} } \to \infty$
So, from the hypothesis, we have: 
:$\map f {x_{n_{j_k} } } \to L$
So: 
:any subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$ has a subsequence converging to $L$.
So, from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit, we have: 
:$\map f {x_n} \to L$
Category:Sequential Characterization of Limit at Positive Infinity of Real Function
Sequential Characterization of von Neumann-Boundedness in Topological Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $E \subseteq X$.
Then $E$ is von Neumann-bounded :
:for each sequence $\sequence {x_n}_{n \in \N}$ in $E$ and each sequence $\sequence {\alpha_n}_{n \in \N}$ in $\GF$ such that $\alpha_n \to 0$ as $n \to \infty$, we have $\alpha_n x_n \to {\mathbf 0}_X$ as $n \to \infty$. 
== Proof ==
=== Necessary Condition ===
Let $E \subseteq X$ be von Neumann-bounded.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists a balanced open neighborhood $V$ of ${\mathbf 0}_X$ such that $V \subseteq X$. 
Since $E$ is von Neumann-bounded, there exists $t > 0$ such that: 
:$E \subseteq t V$ 
so that: 
:$\ds \frac 1 t E \subseteq V$ 
Then we have: 
:$\ds \frac {x_n} t \in V$ 
Since $\alpha_n \to 0$, there exists $N \in \N$ such that: 
:$\ds \cmod {\alpha_n} < \frac 1 t$ for $n \ge N$.
That is: 
:$\ds \cmod {\alpha_n} t = \cmod {\alpha_n t} < 1$ for $n \ge N$.
Since $V$ is balanced, we have: 
:$\alpha_n t V \subseteq V$ 
so that: 
:$\alpha_n x_n \in V$ for $n \ge N$
since $x_n/t \in V$.
So, we have: 
:$\alpha_n x_n \in V$ for $n \ge N$.
So $\alpha_n x_n \to \mathbf 0_X$ as $n \to \infty$. 
=== Sufficient Condition ===
Let $E \subseteq X$ be such that:
:for all sequences $\sequence {x_n}_{n \in \N}$ is a sequence in $E$ and $\sequence {\alpha_n}_{n \in \N}$ is a sequence in $\GF$ such that $\alpha_n \to 0$, we have $\alpha_n x_n \to {\mathbf 0}_X$.
Suppose that $E$ is not von Neumann-bounded.
Then for some open neighbourhood $V$ of ${\mathbf 0}_X$ and all $s > 0$, there exists $t > s$ such that $E \not \subseteq t V$.
That is, there exists a sequence $\sequence {r_n}_{n \in \N}$ of positive real numbers such that $r_n \to \infty$ and $E \not \subseteq r_n V$ for each $n \in \N$.
Then $r_n^{-1} \to 0$ as $n \to \infty$. 
For each $n \in \N$, pick $x_n \in E \setminus r_n V$.
Then we have $r_n^{-1} x_n \not \in V$ for each $n \in \N$. 
So $\sequence {r_n^{-1} x_n}_{n \in \N}$ cannot converge to $\mathbf 0_X$.
This is a contradiction, so $E$ is von Neumann-bounded.
== Sources ==
* : $1.30$: Theorem
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Sequential Compactness is Preserved under Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is sequentially compact, then $T_B$ is also sequentially compact.
== Proof ==
Let $T_A$ be a sequentially compact space.
Take an infinite sequence $\sequence {x_n} \subseteq S_B$.
From the surjectivity of $\phi$, there exists another infinite sequence $\sequence {y_n} \subseteq S_A$ such that $\map \phi {y_n} = x_n$.
By the definition of sequential compactness, there exists a subsequence of $\sequence {y_n}$
Let this subsequence be named $\sequence {y_{n_k} }$.
Let $\sequence {y_{n_k} }$ converge to $y \in S_A$ with respect to $T_A$.
From the continuity of $\phi$, it is concluded that $\map \phi {y_{n_k} } = x_{n_k}$ converges to $\map \phi y \in S_B$.
Thus, $\sequence {x_n}$ has a subsequence that converges.
By definition, $T_B$ is sequentially compact.
== Also see ==
* Compactness Properties Preserved under Continuous Surjection
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Sequentially Compact Spaces
Category:Continuous Mappings
Category:Surjections
Sequential Continuity is Equivalent to Continuity in Metric Space

== Theorem ==
Let $\struct {X, d}$ and $\struct {Y, e}$ be metric spaces.
Let $f: X \to Y$ be a mapping.
Let $x \in X$.
Then $f$ is continuous at $x$ $f$ is sequentially continuous at $x$.
=== Corollary ===
== Proof ==
We have that a Continuous Mapping is Sequentially Continuous.
To prove the converse, by the Rule of Transposition we may prove the contrapositive:
:If $f$ is not continuous at $x$, then $f$ is not sequentially continuous at $x$.
We suppose therefore that there exists $\epsilon_0 > 0$ such that for all $\delta > 0$ there exists $y \in X$ such that $\map d {x, y} < \delta$ and $\map e {\map f x, \map f y} \ge \epsilon_0$.
For $n \ge 1$, define $\delta_n = \dfrac 1 n$.
For $n \ge 1$, we may choose $y_n \in X$ such that $\map d {x, y_n} < \delta_n$ and $\map e {\map f x, \map f {y_n} } \ge \epsilon_0$.
Therefore, by definition the sequence $\sequence {y_n}_{n \mathop \ge 1}$ converges to $x$.
However, by definition the sequence $\sequence {\map f {y_n} }_{n \mathop \ge 1}$ does not converge to $\map f x$.
That is, $f$ is not sequentially continuous at $x$.
Category:Metric Spaces
Category:Sequential Continuity
Category:Continuous Mappings on Metric Spaces
Sequential Continuity is Equivalent to Continuity in Metric Space/Corollary

== Corollary to Sequential Continuity is Equivalent to Continuity in Metric Space ==
Let $\struct {X, d}$ and $\struct {Y, e}$ be metric spaces.
Let $f: X \to Y$ be a mapping.
$f$ is continuous on $X$ $f$ is sequentially continuous on $X$.
== Proof ==
This follows immediately from the definitions:
By definition, a mapping is sequentially continuous everywhere in $X$ it is sequentially continuous at each point.
Also by definition, a mapping is continuous everywhere in $X$ it is continuous at each point.
The result follows from Sequential Continuity is Equivalent to Continuity in Metric Space.
Category:Metric Spaces
Category:Sequential Continuity
Category:Continuous Mappings on Metric Spaces
Sequential Continuity is Equivalent to Continuity in the Reals

== Theorem ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f : A \to \R$ be a real function.
Then $f$ is continuous at $c$ :
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Sources ==
* : $\S 5.1$: Theorem $2$
Category:Real Analysis
Category:Continuity
Category:Sequential Continuity
Category:Sequential Continuity is Equivalent to Continuity in the Reals
Sequential Continuity is Equivalent to Continuity in the Reals/Corollary

== Theorem ==
Let $I$ be a real interval. 
Let $x \in I$. 
Let $f : I \to \R$ be a real function.
Then $f$ is continuous at $x$ :
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have: 
::$\map f {x_n} \to \map f x$
== Proof ==
=== Necessary Condition ===
Suppose $f$ is continuous at $x$, then: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have:
::$\map f {x_n} \to \map f x$
from Sequential Continuity is Equivalent to Continuity in the Reals.
So in particular: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have: 
::$\map f {x_n} \to \map f x$
=== Sufficient Condition ===
Suppose that: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have: 
::$\map f {x_n} \to \map f x$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence converging to $x$.
Let $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ be a subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$. 
We aim to show that $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ has a subsequence converging to $\map f x$. 
We will then obtain $\map f {x_n} \to \map f x$ from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
From the Peak Point Lemma, $\sequence {x_{n_j} }_{j \mathop \in \N}$ has a monotonic subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$.
So from hypothesis: 
:$\map f {x_{n_{j_k} } } \to \map f x$ 
So:
:$\sequence {\map f {x_{n_{j_k} } } }_{j \mathop \in \N}$ is a subsequence of $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ converging to $\map f x$.
Since $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ was an arbitrary subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$, we have: 
:$\map f {x_n} \to \map f x$ 
from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have: 
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have $\map f {x_n} \to \map f x$.
So, from Sequential Continuity is Equivalent to Continuity in the Reals:
:$f$ is continuous at $x$.
Category:Sequential Continuity is Equivalent to Continuity in the Reals
Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition

== Theorem ==
Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f : A \to \R$ be a real function.
Then if $f$ is continuous at $c$:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
== Proof ==
Let $c \in \R$. 
Let $\sequence {x_n}$ be a sequence in $A$ that converges to $c$.
Let $\varepsilon \in \R_{> 0}$. 
Since $f$ is continuous at $c$, there exists $\delta > 0$ such that: 
:for all $x \in A$ with $\size {x - c} < \delta$, we have $\size {\map f x - \map f c} < \varepsilon$.
Additionally, since $\sequence {x_n}$ converges to $c$, there exists $N \in \N$ such that: 
:for all $n > N$ we have $\size {x_n - c} < \delta$.
Therefore, since $x_n \in A$ for all $n \in \N$: 
:for all $n > N$, we have $\size {\map f {x_n} - \map f c} < \varepsilon$.
Since $\varepsilon$ was arbitrary, we have: 
:$\sequence {\map f {x_n} }$ converges to $\map f c$.
Category:Sequential Continuity is Equivalent to Continuity in the Reals