nyuuzyou/wikis huggingface dataset:
Constant Function is Continuous/Metric Space

== Theorem ==
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
== Proof 1 ==
== Proof 2 ==
Category:Constant Mappings
Category:Continuous Mappings on Metric Spaces
Constant Function is Continuous/Metric Space/Proof 1

== Theorem ==
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
== Proof ==
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
From Constant Function is Uniformly Continuous, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
The result follows from Uniformly Continuous Function is Continuous.
Category:Constant Mappings
Category:Continuous Mappings on Metric Spaces
Constant Function is Continuous/Metric Space/Proof 2

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: \map {f_c} a = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
== Proof ==
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$.
Let $\epsilon \in \R_{>0}$.
Let $x \in A_1$.
Pick any $\delta \in \R_{>0}$.
Let $y \in A_1$ such that $\map {d_1} {x, y} < \delta$.
Now we have:
:$\map {f_c} x = c = \map {f_c} y$
Hence:
:$\map {d_2} {\map {f_c} x, \map {f_c} y} = \map {d_2} {c, c}$
By definition of a metric:
:$\map {d_2} {c, c} = 0 < \epsilon$
Thus we have that:
:$\forall \epsilon > 0: \forall x \in A_1: \exists \delta \in \R: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map {f_c} x, \map {f_c} y} < \epsilon$
Hence by definition of continuity on a metric space, $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Theorem $3.3$
Category:Constant Mappings
Category:Continuous Mappings on Metric Spaces
Constant Function is Continuous/Real Function

== Theorem ==
Let $f_c: \R \to \R$ be the constant mapping:
:$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$
Then $f_c$ is continuous on $\R$.
== Proof ==
Follows directly from:
:Constant Real Function is Uniformly Continuous
:Uniformly Continuous Real Function is Continuous.
== Sources ==
* 
Category:Constant Mappings
Category:Continuous Real Functions
Constant Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $f : X \to \overline \R$ be a constant extended real-valued function.
That is, there exists $c \in \overline \R$ such that: 
:$\map f x = c$ for all $x \in X$. 
Then $f$ is $\Sigma$-measurable.
== Proof 1 ==
 
== Proof 2 ==
Category:Constant Function is Measurable
Category:Measurable Functions
Constant Function is Measurable/Proof 1

== Theorem ==
== Proof ==
By the definition of a $\Sigma$-measurable function, we aim to show that: 
:$\set {x \in X : \map f x \le r}$ is $\Sigma$-measurable for each $r \in \R$.
First suppose that $\size c < \infty$. 
Note that there are no $x \in X$ such that $\map f x < c$.
So for $r < c$, we have: 
:$\set {x \in X : \map f x \le r} = \O$
From $\sigma$-Algebra Contains Empty Set, we then have: 
:$\set {x \in X : \map f x \le r} \in \Sigma$ if $r < c$. 
Since for all $x \in X$ we have $\map f x = c$, we also have $\map f x \le c$ for all $x \in X$.
So, if $r \ge c$:
:$\set {x \in X : \map f x \le r} = X$
From the definition of a $\sigma$-algebra, we then have: 
:$\set {x \in X : \map f x \le r} \in \Sigma$ if $r \ge c$. 
So:
:$\set {x \in X : \map f x \le r} \in \Sigma$ for all $r \in \R$
which was the demand.
Now suppose that $c = \infty$. 
Then we have: 
:$\map f x \ge r$
for all $x \in X$ and $r \in \R$. 
So:
:$\set {x \in X : \map f x \le r} = \O$
for each $r \in \R$. 
From $\sigma$-Algebra Contains Empty Set we have $\O \in \Sigma$, so:
:$\set {x \in X : \map f x \le r} \in \Sigma$ for all $r \in \R$.
Now suppose that $c = -\infty$. 
Then we have: 
:$\map f x \le r$
for all $x \in X$ and $r \in \R$.
So:
:$\set {x \in X : \map f x \le r} = X$
for each $r \in \R$. 
From the definition of a $\sigma$-algebra, we have that $X \in \Sigma$, so:
:$\set {x \in X : \map f x \le r} \in \Sigma$ for all $r \in \R$.
Category:Constant Function is Measurable
Constant Function is Measurable/Proof 2

== Theorem ==
== Proof ==
First, suppose that $\size c < \infty$. 
From Characteristic Function of Universe, we can write: 
:$\map f x = c \map {\chi_X} x$
for each $x \in X$.
From the definition of a $\sigma$-algebra, we have: 
:$X \in \Sigma$
So: 
:$f$ is a simple function.
Then, from Simple Function is Measurable, we have: 
:$f$ is $\Sigma$-measurable. 
Now suppose that $c = \infty$. 
For each $n \in \N$, define $f_n : X \to \R$ by: 
:$\map {f_n} x = n$
for each $x \in X$. 
Then: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$ 
for each $x \in X$.
We have shown that each $f_n$ is $\Sigma$-measurable for each $n \in \N$, so this is the limit of $\Sigma$-measurable functions.
So, by Pointwise Limit of Measurable Functions is Measurable:
:$f$ is $\Sigma$-measurable in this case. 
Now suppose that $c = -\infty$.
For each $n \in \N$, define $g_n : X \to \R$ by: 
:$\map {g_n} x = -n$
for each $x \in X$. 
Then: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {g_n} x$ 
for each $x \in X$.
We have shown that each $g_n$ is $\Sigma$-measurable for each $n \in \N$, so this is the limit of $\Sigma$-measurable functions.
So, by Pointwise Limit of Measurable Functions is Measurable:
:$f$ is $\Sigma$-measurable in this case.
Category:Constant Function is Measurable
Constant Function is Primitive Recursive

== Theorem ==
The constant function $f_c: \N \to \N$, defined as:
:$\map {f_c} n = c$ where $c \in \N$
is primitive recursive‎.
=== General Case ===
== Proof ==
The proof proceeds by the Principle of Mathematical Induction.
=== Base Case ===
First we note that $f_0: \N \to \N$ is the zero function, which is a basic primitive recursive function.
This is our base case.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$f_k: \N \to \N$ is primitive recursive‎ for some given $k \in \N$.
Then we need to show:
:$f_{k + 1}: \N \to \N$ is primitive recursive‎.
=== Induction Step ===
This is our induction step:
:$\map {f_{k + 1} } n = k + 1 = \map \Succ k = \map \Succ {\map {f_k} n}$
Now $\map {f_k} n$ is primitive recursive‎ from our induction hypothesis.
Thus $\map {f_{k + 1} } n$ is obtained from the basic primitive recursive function $\Succ$ and $\map {f_k} n$ by substitution.
The result follows by the Principle of Mathematical Induction.
Category:Primitive Recursive Functions
Category:Constant Mappings
Constant Function is Primitive Recursive/General Case

== Theorem ==
The constant function of $k$ variables: $f_c^k: \N^k \to \N$, defined as:
:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = c$ where $c \in \N$
is primitive recursive‎.
== Proof ==
For $k \ge 1$, let $f_c^k$ be the constant function of $k$ variables with value $c$.
We know from Constant Function is Primitive Recursive that $f_c^1$ is primitive recursive‎.
Now:
:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = \map {f_c^1} {n_1} = \map {f_c^1} {\map {\pr_1^k} {n_1, n_2, \ldots, n_k} }$
where $\pr_1^k$ is a projection function which is basic primitive recursive.
So $f_c^k$ is obtained from the primitive recursive‎ function $f_c^1$ and the basic primitive recursive function $\pr_1^k$ by substitution.
Hence by definition, $f_c^k$ is primitive recursive‎.
Category:Primitive Recursive Functions
Category:Constant Mappings
Constant Function is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T_\ast$ be an extended natural number.
Define $T : \Omega \to \Z_{\ge 0} \cup \set \infty$ by: 
:$\map T \omega = T_\ast$ 
for each $\omega \in \Omega$.
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
First, if $T_\ast = \infty$, we have: 
:$\set {\omega \in \Omega : \map T \omega \le t} = \O$
for all $t \in \Z_{\ge 0}$.
Since each $\FF_t$ is a $\sigma$-algebra we therefore have: 
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$ in the case $T_\ast = \infty$. 
Now let $T_\ast < \infty$.
For $t < T_\ast$, we have: 
:$\set {\omega \in \Omega : \map T \omega \le t} = \O$
and so:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$ again. 
For $t \ge T_\ast$, we have:
:$\set {\omega \in \Omega : \map T \omega \le t} = \Omega$
Since each $\FF_t$ is a $\sigma$-algebra we therefore have: 
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$.
So $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times
Constant Function is Uniformly Continuous/Metric Space

== Theorem ==
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
== Proof ==
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
Let $\epsilon \in \R: \epsilon > 0$.
Pick any $\delta \in \R$ such that $\delta > 0$.
Let $x, y \in A_1$ such that $d_1 \left({x, y}\right) < \delta$.
Now we have:
:$f_c \left({x}\right) = c = f_c \left({y}\right)$
Hence:
:$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$
By definition of a metric:
:$d_2 \left({c, c}\right) = 0 < \epsilon$.
Thus we have that:
:$\forall \epsilon > 0: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$
Hence by definition of uniform continuity on a metric space, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
Category:Constant Mappings
Category:Metric Spaces
Category:Uniformly Continuous Mappings
Constant Function is Uniformly Continuous/Real Function

== Theorem ==
Let $f_c: \R \to \R$ be the constant mapping:
:$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$
Then $f_c$ is uniformly continuous on $\R$.
== Proof ==
Follows directly from:
:Constant Function is Uniformly Continuous: Metric Space
:Real Number Line is Metric Space.
== Sources ==
* : $8.2$: Definition and examples: Remarks $8.2.4 \ \text{(a)}$
Category:Constant Mappings
Category:Real Analysis
Category:Uniformly Continuous Real Functions
Constant Function is of Exponential Order Zero

== Theorem ==
Let $f_C: \R \to \GF: t \mapsto C$ be a constant function, where $\GF \in \set {\R, \C}$.
Then $f_C$ is of exponential order $0$.
== Proof ==
The result follows from the definition of exponential order, with $M = 1$, $K = \size C + 1$, and $a = 0$.
Category:Exponential Order
Constant Function of Computable Real Number is Sequentially Computable

== Theorem ==
Let $c \in \R$ be a computable real number.
Then, $f : \R^n \to \R$, defined as:
:$\map f \bsx = c$
is sequentially computable.
== Proof ==
For any real sequence $\sequence {x_n}$, we have:
:$\sequence {\map f {x_n}} = \sequence c$
Thus, the result follows from Constant Sequence of Computable Real Number is Computable.
Category:Computability Theory
Constant Loop is Loop

== Theorem ==
Let $\struct {T, \tau}$ be a topological space.
Let $p \in T$.
Let $c_p : \closedint 0 1 \to T$ be the constant mapping defined by:
:$\forall t \in \closedint 0 1 : \map {c_p} t = p$
Then $c_p$ is a loop in $T$.
== Proof ==
From Constant Mapping is Continuous, it follows that $c_p$ is continuous.
By definition of path, it follows that $c_p$ is a path in $T$.
We have:
:$\map {c_p} 0 = \map {c_p} 1 = p$
Hence, $c_p$ is a loop in $T$.
Category:Loops (Topology)
Constant Mapping/Examples/Constant Mappings on Set of 3

== Examples of Constant Mappings ==
Let $X = \set {a, b, c}$.
Let $S = \set {f_a, f_b, f_c}$ be the constant mappings from $X$ to $X$.
The Cayley table for composition on $S$ is as follows:
:$\begin{array}{c|cccc}
\circ & f_a & f_b & f_c \\
\hline
f_a & f_a & f_a & f_a \\
f_b & f_b & f_b & f_b \\
f_c & f_c & f_c & f_c \\
\end{array}$
As can be seen, there is no identity element, so $\struct {S, \circ}$ is not a group.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Exercise $3$
Category:Examples of Constant Mappings
Constant Mapping is Continuous

== Theorem ==
Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $b \in B$ be any point in $B$.
Let $f_b: A \to B$ be the constant mapping defined by:
:$\forall x \in A: \map {f_b} x = b$
Then $f_b$ is continuous.
== Proof ==
We have by definition that:
:$\forall x \in A: \map {f_b} x = b$
So:
$\map {f^{-1} } b = A$
For $c \in B: c \ne B$ we have that:
$\map {f^{-1} } c = \O$
Let $U \in \tau_B$ such that $b \in U$.
Then $f^{-1} \sqbrk U = A$
Let $V \in \tau_B$ such that $b \notin V$.
Then $f^{-1} \sqbrk V = \O$
From the definition of topology, $A$ is open in $T$.
From Empty Set is Element of Topology, $\O$ is also open in $T$.
So $f_b$ fulfils the conditions for it to be continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Proposition $3.1.10 \ \text{(b)}$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 5$
Category:Continuous Mappings
Category:Constant Mappings
Constant Mapping is Non-Commutative

== Theorem ==
Let $S$ be a set whose cardinality is greater than one.
Let $f: S \to S$ and $g: S \to S$ be constant mappings on $S$.
Then:
:$f \circ g \ne g \circ f$
where $\circ$ denotes composition of mappings.
== Proof ==
First note that if $S$ is a singleton, then there exists only one constant mapping on $S$.
In such a circumstance, $f = g$ and so $f \circ g \ne g \circ f$.
So, let $\card S > 1$.
Then there exist at least $2$ distinct elements $a$ and $b$ of $S$.
Thus, let $f$ and $g$ be defined as:
:$\forall x \in S: \map f x = a$
:$\forall x \in S: \map g x = b$
for some $a, b \in S$ such that $a \ne b$.
We have that:
So $f \circ g \ne g \circ f$ as required.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $8$
Category:Constant Mappings
Constant Mapping to Identity is Homomorphism/Groups

== Theorem ==
Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1$ and $e_2$ respectively.
Let $\phi_e: \struct {G_1, \circ_1} \to \struct {G_2, \circ_2}$ be the constant mapping defined as:
:$\forall x \in G_1: \map {\phi_e} x = e_2$
Then $\phi_e$ is a group homomorphism whose image is $\set {e_2}$ and whose kernel is $G_1$.
== Proof ==
Let $x, y \in G_1$.
Then:
So $\phi_e$ is a group homomorphism.
The results about image and kernel follow directly by definition.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $1$
Category:Constant Mapping to Identity is Homomorphism
Category:Group Homomorphisms
Constant Mapping to Identity is Homomorphism/Rings

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings with zeroes $0_1$ and $0_2$ respectively.
Let $\zeta$ be the zero homomorphism from $R_1$ to $R_2$, that is:
:$\forall x \in R_1: \map \zeta x = 0_2$
Then $\zeta$ is a ring homomorphism whose image is $\set {0_2}$ and whose kernel is $R_1$.
== Proof ==
The additive groups of $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ respectively.
Their identities are $0_1$ and $0_2$ respectively.
Thus from the Constant Mapping to Group Identity is Homomorphism we have that $\zeta: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a group homomorphism:
:$\map \zeta {x +_1 y} = \map \zeta x +_2 \map \zeta y$
Then we have:
The results about image and kernel follow directly by definition.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Example $1$
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $42$
Category:Constant Mapping to Identity is Homomorphism
Category:Ring Homomorphisms
Constant Multiple of Replicative Function is Replicative

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $f$ be a replicative function.
Let $c \in \R$ be a constant.
Let $g: \R \to \R$ be the real function defined as:
:$\map g x = c \map f x$
Then $g$ is also a replicative function.
== Proof ==
Hence the result by definition of replicative function.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{h)}$
Category:Replicative Functions
Constant Net is Convergent

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x \in X$. 
Define a net $\family {x_\lambda}_{\lambda \in \Lambda}$ by: 
:$x_\lambda = x$ for each $\lambda \in \Lambda$. 
Then $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$. 
== Proof ==
Let $U$ be an open neighborhood of $x$.
Let $\lambda_0 \in \Lambda$. 
Then we have $x_\lambda \in U$ for all $\lambda \in \Lambda$, and in particular all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$. 
So $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.
Category:Nets (Set Theory)
Constant Operation is Associative

== Theorem ==
Let $S$ be a set.
Let $x \sqbrk c y = c$ be a constant operation on $S$.
Then $\sqbrk c$ is an associative operation:
:$\forall x, y, z \in S: \paren {x \sqbrk c y} \sqbrk c z = x \sqbrk c \paren {y \sqbrk c z}$
== Proof ==
Category:Constant Operation
Category:Examples of Associative Operations
Constant Operation is Commutative

== Theorem ==
Let $S$ be a set.
Let $x \sqbrk c y = c$ be a constant operation on $S$.
Then $\sqbrk c$ is a commutative operation:
:$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$
== Proof ==
Hence the result by definition of commutativity.
Category:Constant Operation
Category:Examples of Commutative Operations
Constant Real-Valued Function is Bounded

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $c \in \R$.
Let $c_{\R^S} : S \to R$ be the constant mapping defined by:
:$\forall s \in S : \map {c_{\R^S} } s = c$
Then $c_{\R^S}$ is a bounded real-valued function.
== Proof ==
We have:
It follows that $c_{\R^S}$ is a bounded real-valued function by definition.
Category:Constant Mappings
Category:Bounded Real-Valued Functions
Constant Real Function is Absolutely Continuous

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an constant real function.
Then $f$ is absolutely continuous.
== Proof ==
Let $\delta, \epsilon$ be positive real numbers.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of disjoint closed real intervals with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $f$ is constant, for all $i \in \set {1, 2, \ldots, n}$ we have:
:$\size {\map f {b_i} - \map f {a_i} } = 0$
so:
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } = 0 < \epsilon$
Since $\epsilon$ was arbitrary: 
:$f$ is absolutely continuous.
Category:Constant Mappings
Category:Absolutely Continuous Real Functions
Constant Sequence Converges to Constant in Normed Division Ring

== Theorem == 
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\lambda \in R$.
Then:
:the constant sequence $\tuple {\lambda, \lambda, \lambda, \dots}$ is convergent and $\ds \lim_{n \mathop \to \infty} \lambda = \lambda$
== Proof ==
Let $\sequence {x_n}$ be the constant sequence: 
:$\forall n \in \N: x_n = \lambda$
Given $\epsilon \in \R_{>0}$:
:$\forall n \ge 1: \norm {x_n - \lambda} = \norm {\lambda - \lambda} = \norm 0 = 0 < \epsilon$
The result follows.
Category:Sequences
Category:Normed Division Rings
Constant Sequence in Normed Vector Space Converges

== Theorem ==
Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$. 
Let $x \in X$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence with $x_n = x$ for all $n \in \N$. 
Then: 
:$x_n \to x$
== Proof ==
We have: 
:$\norm {x_n - x} = 0$ 
for all $n \in \N$. 
So, for all $\epsilon > 0$, we have: 
:$\norm {x_n - x} < \epsilon$ for all $n \in \N$.
So:
:$x_n \to x$
Category:Convergent Sequences (Normed Vector Spaces)
Constant Sequence in Topological Space Converges

== Theorem ==
Let $X$ be a topological space. 
Let $x \in X$. 
Define a sequence $\sequence {x_n}_{n \in \N}$ by:
:$x_n = x$ for each $n \in \N$. 
Then $\sequence {x_n}_{n \in \N}$ converges to $x$. 
== Proof ==
Let $U$ be an open neighborhood of $x$.
Then we have $x_n = x \in U$ for all $n \in \N$. 
Hence $\sequence {x_n}_{n \in \N}$ converges to $x$.
Category:Convergence
Constant Sequence of Computable Real Number is Computable

== Theorem ==
Let $a \in \R$ be a computable real number.
Let $\sequence {x_n}$ be defined as:
:$x_n = a$
Then, $\sequence {x_n}$ is a computable real sequence.
== Proof ==
By definition of computable real number, there exists a total recursive function $f : \N \to \N$ such that:
:For every $n \in \N$, $\map f n$ codes an integer $k$ such that:
::$\dfrac {k - 1} {n + 1} < a < \dfrac {k + 1} {n + 1}$
Let $g : \N^2 \to \N$ be defined as:
:$\map g {m, n} = \map f n$
As $f$ is total recursive, it follows immediately that $g$ is also total recursive.
Let $m, n \in \N$ be arbitrary.
By definition of $g$:
:$\map g {m, n} = \map f n$
Thus, by assumption:
:$\map g {m, n}$ codes an integer $k$ such that:
::$\dfrac {k - 1} {n + 1} < a < \dfrac {k + 1} {n + 1}$
But, as $x_m = a$:
:$\map g {m, n}$ codes an integer $k$ such that:
::$\dfrac {k - 1} {n + 1} < x_m < \dfrac {k + 1} {n + 1}$
As $m$ and $n$ were arbitrary, $\sequence {x_n}$ is computable by definition.
Category:Computability Theory
Constant Sequence of Rational Number is Computable

== Theorem ==
Let $r \in \Q$ be a rational number.
Let $\sequence {x_n}$ be defined as:
:$x_n = r$
Then, $\sequence {x_n}$ is a computable rational sequence.
== Proof ==
By Existence of Canonical Form of Rational Number, let:
:$r = \dfrac p q$
where:
:$p \in \Z$
:$q \in \Z_{>0}$
Let $m$ be the code number for the integer $p$.
As $q \ge 1$, it follows that:
:$q - 1 \in \N$
We define $N, D : \N \to \N$ as:
:$\map N n = m$
:$\map D n = q - 1$
which are total recursive by:
* Constant Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
Then:
:$\dfrac {k_n} {\map D n + 1} = \dfrac p q = r = x_n$
where $\map N n = m$ codes the integer $k_n = p$
Therefore, $\sequence {x_n}$ is computable by definition.
Category:Computability Theory