hcju/proofwikips huggingface dataset:
Construction of Apotome is Unique
Let $D$ be the domain $\left\{{x \in \R_{>0} : x^2 \in \Q}\right\}$, the rationally expressible numbers. Let $a, b \in D$ be two rationally expressible numbers such that $a - b$ is an apotome. Then, there exists only one $x \in D$ such that $a - b + x$ and $a$ are commensurable in square only. {{:Euclid:Proposition/X/79}}
Lower Bound of Natural Logarithm
:$\forall x \in \R_{>0}: 1 - \dfrac 1 x \le \ln x$ where $\ln x$ denotes the natural logarithm of $x$.
Bounds of Natural Logarithm
Let $\ln y$ be the natural logarithm of $y$ where $y \in \R_{>0}$. Then $\ln$ satisfies the compound inequality: :$\displaystyle 1 - \frac 1 y \le \ln y \le y - 1$
Construction of Solid Angle from Three Plane Angles any Two of which are Greater than Other Angle/Lemma
{{:Euclid:Proposition/XI/23/Lemma}}
Tetrahedra are Equal iff Bases are Reciprocally Proportional to Heights
{{:Euclid:Proposition/XII/9}}
Construction of Polyhedron in Outer of Concentric Spheres/Porism
{{:Euclid:Proposition/XII/17/Porism}}
Definition:Elevated/Point
Let $P$ be a plane. An '''elevated point (relative to $P$)''' is a point in space which does not lie on $P$.
Definition:Elevated/Line
Let $P$ be a plane. An '''elevated line (relative to $P$)''' is a line in space which does not lie entirely in $P$.
Common Ratio in Rational Geometric Sequence is Rational
Let $\sequence {a_k}$ be a geometric sequence whose terms are rational. Then the common ratio of $\sequence {a_k}$ is rational.
Common Ratio in Integer Geometric Sequence is Rational
Let $\sequence {a_k}$ be a geometric sequence whose terms are all integers. Then the common ratio of $\sequence {a_k}$ is rational.
Form of Geometric Sequence of Integers
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n + 1$ consisting of integers only. Then the $j$th term of $P$ is given by: :$a_j = k p^{n - j} q^j$ where: : the common ratio of $P$ expressed in canonical form is $\dfrac q p$ : $k$ is an integer.
Geometric Sequence of Integers with Integer Common Ratio
Let $P = \sequence {a_j}_{1 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting entirely of integers. Let $r$ be the common ratio of $P$. Then $r$ is an integer : :$\forall i, j \in \set {1, 2, \ldots, n}, i \le j: a_i \divides a_j$ That is, terms of $P$ divide later terms of $P$  $r$ is an integer.
Integers are Coprime iff Powers are Coprime
Let $a, b \in \Z$ be integers. Then: :$a \perp b \iff \forall n \in \N: a^n \perp b^n$ That is, two integers are coprime  all their positive integer powers are coprime.
Integer Addition is Cancellable
The operation of addition on the set of integers $\Z$ is cancellable: :$\forall x, y, z \in \Z: x + z = y + z \implies x = y$
Complement Relative to Subset is Subset of Complement Relative to Superset
Let $A, B, C$ be sets such that $A \subseteq B \subseteq C$. Then: :$\relcomp B A \subseteq \relcomp C A$
Complement of Relative Complement is Union with Complement
Let $A, B, C$ be sets such that $A \subseteq B \subseteq C$. Then: :$\relcomp C {\relcomp B A} = A \cup \relcomp C B$
Set Union Preserves Subsets/Families of Sets
Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$. Let: :$\forall \beta \in I: A_\beta \subseteq B_\beta$ Then: :$\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha \subseteq \bigcup_{\alpha \mathop \in I} B_\alpha$
Set Union Preserves Subsets/General Result
Let $\mathbb S, \mathbb T$ be sets of sets. Suppose that for each $S \in \mathbb S$ there exists a $T \in \mathbb T$ such that $S \subseteq T$. Then $\bigcup \mathbb S \subseteq \bigcup \mathbb T$.
Intersection of Family is Subset of Intersection of Subset of Family
Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$. Let $J \subseteq I$. Then: :$\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in J} A_\alpha$ where $\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}_{\alpha \mathop \in I}$.
Union of Subset of Family is Subset of Union of Family
Let $I$ be an indexing set. Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$. Let $J \subseteq I$ Then: :$\displaystyle \bigcup_{\alpha \mathop \in J} A_\alpha \subseteq \bigcup_{\alpha \mathop \in I} A_\alpha$ where $\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
Intersection of Open Intervals of Positive Reals is Empty
Let $\R_{> 0}$ be the set of strictly positive real numbers. For all $x \in \R_{> 0}$, let $A_x$ be the open real interval $\openint 0 x$. Then: :$\displaystyle \bigcap_{x \mathop \in \R_{> 0} } A_x = \O$
Union of Closed Intervals of Positive Reals is Set of Positive Reals
Let $\R_{> 0}$ be the set of strictly positive real numbers. For all $x \in \R_{> 0}$, let $B_x$ be the closed real interval $\closedint 0 x$. Then: :$\displaystyle \bigcup_{x \mathop \in \R_{> 0} } B_x = \R_{\ge 0}$
Subset of Cartesian Product not necessarily Cartesian Product of Subsets
Let $A$ and $B$ be sets. Let $A$ and $B$ both have at least two distinct elements. Then there exists $W \subseteq A \times B$ such that $W$ is not the cartesian product of a subset of $A$ and a subset of $B$.
Subset equals Preimage of Image implies Injection
Let $f: S \to T$ be a mapping. Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$. Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$. Let: :$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ Then $f$ is an injection.
Preimage of Image of Subset under Injection equals Subset
Let $f: S \to T$ be an injection. Then: :$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ where: :$f \sqbrk A$ denotes the image of $A$ under $f$ :$f^{-1}$ denotes the inverse of $f$ :$f^{-1} \circ f$ denotes composition of $f^{-1}$ and $f$.
Subset equals Image of Preimage implies Surjection
Let $f: S \to T$ be a mapping. Let: :$\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$ where $f \sqbrk B$ denotes the image of $B$ under $f$. Then $f$ is a surjection.
Image of Preimage of Subset under Surjection equals Subset
Let $f: S \to T$ be a surjection. Then: :$\forall B \subseteq T: B = \left({f \circ f^{-1} }\right) \sqbrk B$ where: :$f \sqbrk B$ denotes the image of $B$ under $f$ :$f^{-1}$ denotes the inverse of $f$ :$f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
Intersection of Image with Subset of Codomain
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $A \subseteq S$ and $B \subseteq T$. Then: :$f \sqbrk {A \cap f^{-1} \sqbrk B} = f \sqbrk A \cap B$
Projection is Injection iff Factor is Singleton
Let $S_1, S_2, \ldots, S_n$ be non-empty sets. Let $\displaystyle S = \prod_{i \mathop = 1}^n S_i$ be the cartesian product of $S_1, S_2, \ldots, S_n$. Let $\pr_j: S \to S_j$ be the $j$th projection on $S$. Then $\pr_j$ is an injection  $S_k$ is a singleton for all $k \in \set {1, 2, \dotsc, n}$ where $k \ne j$.
Preimage of Element under Projection
Let $A$ and $B$ be sets. Let $A \times B$ be the cartesian product of $A$ and $B$. Let $\pr_1: A \times B \to A$ be the first projection of $A \times B$. Let $a \in A$. Then: :$\pr_1^{-1} \sqbrk {\set a} = \set {\tuple {a, b}: b \in B}$ that is: :$\pr_1^{-1} \sqbrk {\set a} = \set a \times B$
Divisors of Factorial
Let $n \in \N_{>0}$. Then all natural numbers less than or equal to $n$ are divisors of $n!$: :$\forall k \in \left\{{1, 2, \ldots, n}\right\}: n! \equiv 0 \pmod k$
Positive Difference Relation on Reals is Transitive
Let $P \subseteq \R$ be a subset of the real numbers such that: :$(1): \quad 1 \in P$ :$(2): \quad a, b \in P \implies a + b \in P$ :$(3): \quad$ For all $x \in \R$, exactly one of these is true: :::$x \in P$ :::$x = 0$ :::$-x \in P$ Let $Q \subseteq \R \times \R$ be the relation on $\R$ defined as: :$Q = \set {\tuple {a, b} \in \R:  a - b \in P}$ Then $Q$ is a transitive relation.
Composition of Repeated Compositions of Injections
Let $S$ be a set. Let $f: S \to S$ be an injection. Let the sequence of mappings: :$f^0, f^1, f^2, \ldots, f^n, \ldots$ be defined as: :$\forall n \in \N: f^n \left({x}\right) = \begin{cases} x & : n = 0 \\ f \left({x}\right) & : n = 1 \\ f \left({f^{n-1} \left({x}\right)}\right) & : n > 1 \end{cases}$ Then: :$\forall m, n \in \Z_{\ge 0}: f^n \circ f^m = f^{m + n}$ where $f^n \circ f^m$ denotes composition of mappings.
Derivative Function on Set of Functions induces Equivalence Relation
Let $X$ be the set of real functions $f: \R \to \R$ which possess continuous derivatives. Let $\mathcal R \subseteq X \times X$ be the relation on $X$ defined as: :$\mathcal R = \set {\tuple {f, g} \in X \times X: D f = D g}$ where $D f$ denotes the first derivative of $f$. Then $\mathcal R$ is an equivalence relation.
Factorial Divides Product of Successive Numbers
Let $m, n \in \N_{\ge 1}$ be natural numbers Let $m^{\overline n}$ be $m$ to the power of $n$ rising. Then: :$m^{\overline n} \equiv 0 \bmod n!$ That is, the factorial of $n$ divides the product of $n$ successive numbers.
Binomial Coefficient is Integer
Let $\dbinom n k$ be a binomial coefficient. Then $\dbinom n k$ is an integer.
Set of Mappings which map to Same Element induces Equivalence Relation
Let $X$ and $Y$ be sets. Let $E$ be the set of all mappings from $X$ to $Y$. Let $b \in X$. Let $\mathcal R \subseteq E \times E$ be the relation on $E$ defined as: :$\mathcal R := \set {\tuple {f, g} \in \mathcal R: \map f b = \map g b}$ Then $\mathcal R$ is an equivalence relation.
Equivalence of Definitions of Inverse Mapping
Let $S$ and $T$ be sets. {{TFAE|def = Inverse Mapping}}
Inverse Mapping is Bijection
Let $S$ and $T$ be sets. Let $f: S \to T$ and $g: T \to S$ be inverse mappings of each other. Then $f$ and $g$ are bijections.
Mapping is Injection and Surjection iff Inverse is Mapping/Proof 2
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Then: : $f: S \to T$ can be defined as a bijection in the sense that: ::$(1): \quad f$ is an injection ::$(2): \quad f$ is a surjection. : :the inverse $f^{-1}$ of $f$ is such that: ::for each $y \in T$, the preimage $\map {f^{-1} } y$ has exactly one element. :That is, such that $f^{-1} \subseteq T \times S$ is itself a mapping.
Mapping is Extension iff Composite with Inclusion
Let $S$ and $T$ be sets. Let $A \subseteq S$. Let $f: S \to T$ and $g: A \to T$ be mappings. Then $f$ is an extension of $g$ : :$f = g \circ i_A$ where $i_A$ is the inclusion mapping on $A$. This can be illustrated using a commutative diagram as follows: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  A \ar[r]^*{i_A}      \ar@{-->}[rd]_*{f = g \circ i_A} &  S \ar[d]^*{g} \\ &  T } \end {xy}$
Cardinality of Set of Restrictions of Mapping
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let the cardinality of $S$ be $n$. Let $F$ be the set of restrictions of $f$ to a subset of $S$. Then there are $2^n$ elements of $F$.
Projection is Surjection/Family of Sets
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets. Let $\displaystyle \prod_{\alpha \mathop \in I} S_\alpha$ be the Cartesian product of $\family {S_\alpha}_{\alpha \mathop \in I}$. Let each of $S_\alpha$ be non-empty. For each $\beta \in I$, let $\displaystyle \pr_\beta: \prod_{\alpha \mathop \in I} S_\alpha \to S_\beta$ be the $\beta$th projection on $\displaystyle S = \prod_{\alpha \mathop \in I} S_\alpha$. Then $\pr_\beta$ is a surjection.
Chebyshev Distance on Real Vector Space is Metric
The Chebyshev distance on $\R^n$: : $\displaystyle \forall x, y \in \R^n: d_\infty \left({x, y}\right):= \max_{i \mathop = 1}^n {\left\vert{x_i - y_i}\right\vert}$ is a metric.
Chebyshev Distance is Metric
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces. Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$. Let $d_\infty: \mathcal A \times \mathcal A \to \R$ be the Chebyshev distance on $\mathcal A$: :$\displaystyle \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$ where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \mathcal A$. Then $d_\infty$ is a metric,
Set of Successive Numbers contains Unique Multiple
Let $m \in \Z_{\ge 1}$. Then $\set {m, m + 1, \ldots, m + n - 1}$ contains a unique integer that is a multiple of $n$. That is, in any set containing $n$ successive integers, $n$ divides exactly one of those integers.
Taxicab Metric is Topologically Equivalent to Chebyshev Distance on Real Vector Space
For $n \in \N$, let $\R^n$ be a real vector space. Let $d_1$ be the taxicab metric on $\R^n$. Let $d_\infty$ be the Chebyshev distance on $\R^n$. Then :$\forall x, y \in \R^n: d_\infty \left({x, y}\right) \le d_1 \left({x, y}\right) \le n \cdot d_\infty \left({x, y}\right)$ It follows that $d_1$ and $d_\infty$ are Lipschitz equivalent.
Euclidean Metric is Metric
Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be metric spaces. Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$. The Euclidean metric on $\mathcal A$ is a metric.
Euclidean Space is Normed Space
The Euclidean norm is a norm on the Euclidean space $\R^n$.
Number of Multiples less than Given Number
Let $m, n \in \N_{\ge 1}$. The number of multiples of $m$ not greater than $n$ is given by: :$q = \floor {\dfrac n m}$ where $\floor {\cdot}$ denotes the floor function
Restriction of Non-Continuous Mapping on Metric Space to Subspace may be Continuous
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be a metric spaces. Let $f: A_1 \to A_2$ be a mapping. Let $Y \subseteq A_1$. Let $f {\restriction_Y}: Y \to A_2$ be the restriction of $f$ to $Y$. Let $f {\restriction_Y}$ be $\left({d_Y, d_2}\right)$-continuous. Then it is not necessarily the case that $f$ is also $\left({d_1, d_2}\right)$-continuous.
Supremum Metric on Bounded Real Functions on Closed Interval is Metric
Let $\closedint a b \subseteq \R$ be a closed real interval. Let $A$ be the set of all bounded real functions $f: \closedint a b \to \R$. Let $d: A \times A \to \R$ be the supremum metric on $A$. Then $d$ is a metric.
Supremum Metric on Continuous Real Functions is Subspace of Bounded
Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a closed real interval. Let $\mathscr C \left[{a \,.\,.\, b}\right]$ be the set of all continuous functions $f: \left[{a \,.\,.\, b}\right] \to \R$. Let $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$ be the set of all bounded real functions $f: \left[{a \,.\,.\, b}\right] \to \R$. Let $d$ be the supremum metric on $\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right)$. Then $\left({\mathscr C \left[{a \,.\,.\, b}\right], d_{\mathscr C} }\right)$ is a subspace of $\left({\mathscr B \left({\left[{a \,.\,.\, b}\right], \R}\right), d}\right)$.
Zero Definite Integral of Nowhere Negative Function implies Zero Function
Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a closed real interval. Let $h: \left[{a \,.\,.\, b}\right] \to \R$ be a continuous real function such that: :$\forall x \in \left[{a \,.\,.\, b}\right]: h \left({x}\right) \ge 0$ Let: :$\displaystyle \int_a^b h \left({x}\right) \rd x = 0$ Then: :$\forall x \in \left[{a \,.\,.\, b}\right]: h \left({x}\right) = 0$
L2 Metric on Closed Real Interval is Metric
Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$. Let $d: S \times S \to \R$ be the $L^2$ metric on $\closedint a b$: :$\displaystyle \forall f, g \in S: \map {d_2} {f, g} := \paren {\int_a^b \paren {\map f t - \map g t}^2 \rd t}^{\frac 1 2}$ Then $d_2$ is a metric.
Element in Bounded Metric Space has Bound
Let $M = \left({X, d}\right)$ be a metric space. Let $M' = \left({Y, d_Y}\right)$ be a subspace of $M$. Let $M'$ be bounded in $M$. Then: :$\forall a' \in X: \exists K' \in \R: \forall x \in Y: d \left({x, a}\right) \le K'$ That is, if there is one element of $X$ which satisfies the condition for $Y$ to be bounded in $M$, they ''all'' do.
Convergence of Square of Linear Combination of Sequences whose Squares Converge
Let $\sequence {x_i}$ and $\sequence {y_i}$ be real sequences such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ and $\ds \sum_{i \mathop \ge 0} y_i^2$ are convergent. Let $\lambda, \mu \in \R$ be real numbers. Then $\ds \sum_{i \mathop \ge 0} \paren {\lambda x_i + \mu y_i}^2$ is convergent.
Hilbert Sequence Space is Metric Space
Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $\ell^2 = \left({A, d_2}\right)$ be the Hilbert sequence space on $\R$. Then $\ell^2$ is a metric space.
Supremum Metric on Bounded Real-Valued Functions is Metric
Let $X$ be a set. Let $A$ be the set of all bounded real-valued functions $f: X \to \R$. Let $d: A \times A \to \R$ be the supremum metric on $A$. Then $d$ is a metric.
Supremum Metric on Bounded Real Sequences is Metric
Let $A$ be the set of all bounded real sequences. Let $d: A \times A \to \R$ be the supremum metric on $A$. Then $d$ is a metric.
Supremum Metric is Metric
Let $S$ be a set. Let $M = \left({A', d'}\right)$ be a metric space. Let $A$ be the set of all bounded mappings $f: S \to M$. Let $d: A \times A \to \R$ be the supremum metric on $A$. Then $d$ is a metric.
Supremum Metric on Bounded Continuous Mappings is Metric
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces. Let $A$ be the set of all continuous mappings $f: M_1 \to M_2$ which are also bounded. Let $d: A \times A \to \R$ be the supremum metric on $A$. Then $d$ is a metric.
Harmonic Number is not Integer
Let $H_n$ be the $n$th harmonic number. Then $H_n$ is not an integer for $n \ge 2$. That is, the only harmonic numbers that are integers are $H_0$ and $H_1$.
Supremum Metric on Differentiability Class is Metric
Let $\left[{a \,.\,.\, b}\right] \subseteq \R$ be a closed real interval. Let $r \in \N$ be a natural number. Let $A := \mathscr D^r \left[{a \,.\,.\, b}\right]$ be the set of all continuous functions $f: \left[{a \,.\,.\, b}\right] \to \R$ which are of differentiability class $r$. Let $d: A \times A \to \R$ be the supremum metric on $A$. Then $d$ is a metric.
Subspace of Metric Space is Metric Space
Let $M = \left({A, d}\right)$ be a metric space. Let $H \subseteq A$. Let $d_H: H \times H \to \R$ be the restriction $d \restriction_{H \times H}$ of $d$ to $H$. Let $\left({H, d_H}\right)$ be a metric subspace of $\left({A, d}\right)$. Then $d_H$ is a metric on $H$.
Lp Metric on Closed Real Interval is Metric
Let $S$ be the set of all real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$. Let $p \in \R_{\ge 1}$. Let $d_p: S \times S \to \R$ be the $L^p$ metric on $\left[{a \,.\,.\, b}\right]$: :$\displaystyle \forall f, g \in S: d \left({f, g}\right) := \left({\int_a^b \left\vert{f \left({t}\right) - g \left({t}\right)}\right\vert^p \ \mathrm d t}\right)^{\frac 1 p}$ Then $d_p$ is a metric.
P-Sequence Space of Real Sequences is Metric Space
Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent. Let $d_p$ be the $p$-sequence metric on $\R$. Then $\ell^p := \left({A, d_p}\right)$ is a metric space.
Peano Structure is Unique
Let $\left({P, s, 0}\right)$ and $\left({P', s', 0'}\right)$ be Peano structures. Then there is a unique bijection $f: P \to P'$ such that: {{begin-eqn}} {{eqn|l = f \left({0}\right)      |r = 0' }} {{eqn|l = \forall n \in P: f \left({s \left({n}\right)}\right)      |r = s' \left({f \left({n}\right)}\right) }} {{end-eqn}}
Constant Function is Continuous/Metric Space/Proof 2
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 $d_2$.
Constant Function is Continuous/Metric Space/Proof 1
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 $d_2$.
Constant Function is Continuous/Metric Space
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 $d_2$.
Constant Function is Continuous/Real Function
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 $\R$.
Identity Mapping is Continuous/Metric Space
Let $M = \left({A, d}\right)$ be a metric space. The identity mapping $I_A: A \to A$ defined as: :$\forall x \in A: I_A \left({x}\right) = x$ is a continuous mapping.
Identity Mapping on Real Vector Space from Chebyshev to Euclidean Metric is Continuous
Let $\R^n$ be an $n$-dimensional real vector space. Let $d_2$ be the Euclidean metric on $\R^n$. Let $d_\infty$ be the Chebyshev distance on $\R^n$. Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself. Then the mapping: :$I: \struct {\R^n, d_\infty} \to \struct {\R^n, d_2}$ is $\tuple {d_\infty, d_2}$-continuous.
Identity Mapping on Real Vector Space from Euclidean to Chebyshev Distance is Continuous
Let $\R^n$ be an $n$-dimensional real vector space. Let $d_2$ be the Euclidean metric on $\R^n$. Let $d_\infty$ be the Chebyshev distance on $\R^n$. Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself. Then the mapping: :$I: \left({\R^n, d_2}\right) \to \left({\R^n, d_\infty}\right)$ is $\left({d_2, d_\infty}\right)$-continuous.
Composite of Continuous Mappings at Point between Metric Spaces is Continuous at Point
Let $M_1 = \left({X_1, d_1}\right), M_2 = \left({X_2, d_2}\right), M_3 = \left({X_3, d_3}\right)$ be metric spaces. Let $f: M_1 \to M_2$ be continuous at $a \in X_1$. Let $g: M_2 \to M_3$ be continuous at $f \left({a}\right) \in X_2$. Then their composite $g \circ f: M_1 \to M_3$ is continuous at $a \in X_1$.
Composite of Continuous Mappings between Metric Spaces is Continuous
Let $M_1 = \left({X_1, d_1}\right), M_2 = \left({X_2, d_2}\right), M_3 = \left({X_3, d_3}\right)$ be metric spaces. Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be continuous mappings. Then their composite $g \circ f: M_1 \to M_3$ is continuous.
Mapping from L1 Space to Real Number Space is Continuous
Let $\left({\R, d}\right)$ be the real number line under the usual metric $d$. Let $X$ be the set of continuous real functions $f: \left[{a \,.\,.\, b}\right] \to \R$. Let $d_1$ be the $L^1$ metric on $X$. Let $I: X \to \R$ be the real-valued function defined as: :$\displaystyle \forall f \in X: I \left({f}\right) := \int_a^b f \left({t}\right) \ \mathop d t$ Then the mapping: :$I : \left({X, d_1}\right) \to \left({\R, d}\right)$ is continuous.