hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === Suppose $\left({S \times T, \circ}\right)$ is anticommutative and neither $\left({S, \circ_1}\right)$ nor $\left({T, \circ_2}\right)$ are. Then for some distinct $a, b \in S$: :$a \circ_1 b = b \circ_1 a$ Similarly, for some distinct $c, d \in T$: :$c \circ_2 d = d \circ_2 c$ Then we compute by definition of $\circ$ in the external direct product: :$\left({a,c}\right) \circ \left({b,d}\right) = \left({a\circ_1 b, c \circ_2 d}\right)$ :$\left({b,d}\right) \circ \left({a,c}\right) = \left({b\circ_1 a, d \circ_2 c}\right)$ So:  :$\left({a,c}\right) \circ \left({b,d}\right) = \left({b,d}\right) \circ \left({a,c}\right)$ which contradicts our assumption that $\left({S \times T, \circ}\right)$ was anticommutative. Hence at least one of $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ are anticommutative. {{qed|lemma}} === Necessary Condition === Suppose $\left({S \times T, \circ}\right)$ is not anticommutative.  Then for some $\left({a, b}\right), \left({c, d}\right) \in S \times T$: :$\left({a, b}\right) \circ \left({c, d}\right) = \left({c, d}\right) \circ \left({a, b}\right)$ which by the definition of $\circ$ implies: :$\left({a \circ_1 c, b \circ_2 d} \right) = \left({c \circ_1 a, d \circ_2 b}\right)$ Therefore we conclude: :$a \circ_1 c = c \circ_1 a$ :$b \circ_2 d = d \circ_2 b$ Hence, neither $\left({S, \circ_1}\right)$ nor $\left({T, \circ_2}\right)$ can be anticommutative. Hence if one of $\left({S, \circ_1}\right)$ or $\left({T, \circ_2}\right)$ is anticommutative, then $\left({S \times T, \circ}\right)$ must be as well. {{qed}} Category:Associativity Category:Direct Products 9d7azx3vvomx33vvszxd3pb18orou27
From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections. We now need to show they are homomorphisms. Let $g, h \in \struct {G, \circ}$ where $g = \tuple {g_1, g_2}$ and $h = \tuple {h_1, h_2}$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_1} {g \circ h}       | r = \map {\pr_1} {\tuple {g_1, g_2} \circ \tuple {h_1, h_2} }       | c =  }} {{eqn | r = \map {\pr_1} {\tuple {g_1 \circ_1 h_1, g_2 \circ_2 h_2} }       | c =  }} {{eqn | r = g_1 \circ_1 h_1       | c =  }} {{eqn | r = \map {\pr_1} g \circ_1 \map {\pr_1} h       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map {\pr_2} {g \circ h}       | r = \map {\pr_2} {\tuple {g_1, g_2} \circ \tuple {h_1, h_2} }       | c =  }} {{eqn | r = \map {\pr_2} {\tuple {g_1 \circ_1 h_1, g_2 \circ_2 h_2} }       | c =  }} {{eqn | r = g_2 \circ_2 h_2       | c =  }} {{eqn | r = \map {\pr_2} g \circ_2 \map {\pr_2} h       | c =  }} {{end-eqn}} and thus the morphism property is demonstrated for both $\pr_1$ and $\pr_2$. {{Qed}}
A specific instance of Projection is Epimorphism, where the algebraic structures in question are groups. {{qed}}
From Canonical Injection is Injection we have that the canonical injections are in fact injective. It remains to prove the morphism property. Let $x, y \in G_1$. Then: {{begin-eqn}} {{eqn | l = \map {\inj_1} {x \circ_1 y}       | r = \tuple {x \circ_1 y, e_2} }} {{eqn | r = \tuple {x \circ_1 y, e_2 \circ_2 e_2} }} {{eqn | r = \tuple {x, e_2} \circ \tuple {y, e_2} }} {{eqn | r = \map {\inj_1} x \circ \map {\inj_1} y }} {{end-eqn}} and the morphism property has been demonstrated to hold for $\inj_1$. Thus $\inj_1: \struct {G_1, \circ_1} \to \struct {G_1, \circ_1} \times \struct {G_2, \circ_2}$ has been shown to be an injective group homomorphism and therefore a group monomorphism. The same argument applies to $\inj_2$. {{Qed}}
A specific instance of Canonical Injection is Monomorphism, where the algebraic structures in question are groups. {{qed}}
This is a specific instance of External Direct Product of Projection with Canonical Injection, where the algebraic structures in question are groups. {{qed}}
The canonical injection $\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ is defined as: :$\forall x \in G_1: \map {\inj_1} x = \tuple {x, e_2}$ Thus: :$\Img {\inj_1} = \set {\tuple {x, e_2}: x \in G_1}$ The second projection $\pr_2: \struct {G_1 \times G_2, \circ} \to \struct {G_2, \circ_2}$ is defined as: :$\forall \tuple {x, y} \in G_1 \times G_2: \map {\pr_2} {x, y} = y$ Thus by definition of kernel: :$\map \ker {\pr_2} = \map {\pr_2^{-1} } {e_2} = \set {\tuple {x, e_2}: x \in G_1}$ As can be seen: :$\Img {\inj_1} = \map \ker {\pr_2}$ Similarly: :$\Img {\inj_2} = \set {\tuple {e_1, y}: y \in G_2}$ :$\map \ker {\pr_1} = \map {\pr_1^{-1} } {e_1} = \set {\tuple {e_1, y}: y \in G_2}$ Hence the result. {{qed}}
From Image of Canonical Injection is Kernel of Projection: :$\Img {\inj_1} = \map \ker {\pr_2}$ :$\Img {\inj_2} = \map \ker {\pr_1}$ That is: :the image of the (first) canonical injection is the kernel of the second projection :the image of the (second) canonical injection is the kernel of the first projection. The domain of the projections is $G_1 \times G_2$, by definition. The result follows from Kernel is Normal Subgroup of Domain. {{qed}}
The direct product of $f_1$ and $f_2$ $f_1 \times f_2: g \to H_1 \times H_2$ is defined as: :$\forall g \in G: \map {\paren {f_1 \times f_2} } g = \tuple {\map {f_1} g, \map {f_2} g}$ From External Direct Product of Groups is Group, the group direct product $H_1 \times H_2$ is a group. It remains to be shown that $f_1 \times f_2$ fulfils the morphism property. Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | o =        | r = \map {\paren {f_1 \times f_2} } g * \map {\paren {f_1 \times f_2} } h       | c =  }} {{eqn | r = \tuple {\map {f_1} g, \map {f_2} g} * \tuple {\map {f_1} h, \map {f_2} h}       | c = {{Defof|Direct Product of Group Homomorphisms|Direct Product of $f_1$ and $f_2$}} }} {{eqn | r = \tuple {\map {f_1} g *_1 \map {f_1} h, \map {f_2} g *_2 \map {f_2} h}       | c = {{Defof|Group Direct Product|Group Direct Product}} $\struct {H_1 \times H_2, *}$ }} {{eqn | r = \tuple {\map {f_1} {g \circ h}, \map {f_2} {g \circ h} }       | c = $f_1$ and $f_2$ are Group Homomorphisms }} {{eqn | r = \map {\paren {f_1 \times f_2} } {g \circ h}       | c = {{Defof|Direct Product of Group Homomorphisms|Direct Product of $f_1$ and $f_2$}} }} {{end-eqn}} {{qed}}
The proof of $(2)$ follows exactly the same lines as the proof of $(1)$, so the proof of $(1)$ only will be shown here. === Matching Domains === From Domain of Composite Relation, the domain of $\pr_1 \circ \paren {f_1 \times f_2}$ is the domain of $f_1 \times f_2$. By definition of the direct product of $f_1$ and $f_2$, the domain of $f_1 \times f_2$ is $G$. Also by definition, the domain of $f_1$ is also $G$. Thus the domains of $\pr_1 \circ \paren {f_1 \times f_2}$ and $f_1$ are the same.  {{qed}} === Matching Codomains === From Codomain of Composite Relation, the codomain of $\pr_1 \circ \paren {f_1 \times f_2}$ is the codomain of $\pr_1$. By definition of the first projection $H_1 \times H_2$, the codomain of $\pr_1$ is $H_1$. Also by definition, the codomain of $f_1$ is also $H_1$. Thus the codomains of $\pr_1 \circ \paren {f_1 \times f_2}$ and $f_1$ are the same.  {{qed}} === Matching Images === Let $g \in G$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_1 \circ \paren {f_1 \times f_2} } g       | r = \map {\pr_1} {\map {\paren {f_1 \times f_2} } g}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\pr_1} {\map {f_1} g, \map {f_2} g}       | c = {{Defof|Direct Product of Group Homomorphisms|Direct Product of $f_1$ and $f_2$}} }} {{eqn | r = \map {f_1} g       | c = {{Defof|First Projection}} }} {{end-eqn}} So: :$g \in G \implies \map {\pr_1 \circ \paren {f_1 \times f_2} } g = \map {f_1} g$ and so Equality of Mappings has been demonstrated. {{qed}}
We will use the One-Step Subgroup Test. This is valid, as the Gaussian integers are a subset of the complex numbers. We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$. Let $a + b i, c + d i \in \Z \sqbrk i$. Then we have $-\paren {c + d i} = -c - d i$, and so: {{begin-eqn}} {{eqn | l = \paren {a + b i} + \paren {-\paren {c + d i} }       | r = \paren {a + b i} + \paren {- c - d i}       | c =  }} {{eqn | r = \paren {a + \paren {-c} } + \paren {b + \paren {-d} } i       | c =  }} {{eqn | r = \paren {a - c} + \paren {b - d} i       | c =  }} {{end-eqn}} We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain. Therefore by definition $\Z$ is a ring. So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence $\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$. So by the One-Step Subgroup Test, $\Z \sqbrk i$ is a subgroup of $\struct {\C, +}$. {{qed}}
We can express $a$ and $b$ as: :$a = \dfrac a 1, b = \dfrac b 1$ Thus from Mediant is Between: :$\dfrac a 1 < \dfrac {a + b} {1 + 1} < \dfrac b 1$ Hence $c = \dfrac {a + b} 2$ is an element of $\R$ between $a$ and $b$. {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f + g} + h} } x       | r = \paren {\map f x + \map g x} + \map h x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | r = \map f x + \paren {\map g x + \map h x}       | c = Real Addition is Associative }} {{eqn | r = \map {\paren {f + \paren {g + h} } } x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Addition Category:Real Addition Category:Associativity 02ubsew2vta6l1i2g9wpqboqmp953nu
By Coproduct is Unique, it suffices to verify that $\map M {A \sqcup B}$ is a coproduct for $\map M A$ and $\map M B$. By the UMP of $\map M A$, $\map M B$ and $\map M {A \sqcup B}$, we have the following commutative diagram: ::$\begin {xy} <0em, 5em>*+{N} = "N", <-5em,0em>*+{\map M A}         = "MA", <0em,0em>*+{\map M {A \sqcup B} } = "MAB", <5em,0em>*+{\map M B}          = "MB", <-5em,-5em>*+{A}         = "A", <0em,-5em>*+{A \sqcup B} = "AB", <5em,-5em>*+{B}          = "B", "A";"MA" **@{-} ?>*@{>}   ?*!/_.8em/{i_A}, "B";"MB" **@{-} ?>*@{>}   ?*!/^.8em/{i_B}, "AB";"MAB" **@{-} ?>*@{>} ?*!/_.8em/{i_{A \mathop \sqcup B} }, "A";"AB" **@{-} ?>*@{>} ?*!/^.8em/{i_1}, "B";"AB" **@{-} ?>*@{>} ?*!/_.8em/{i_2}, "MA";"MAB" **@{-} ?>*@{>} ?*!/_.8em/{j_1}, "MB";"MAB" **@{-} ?>*@{>} ?*!/^.8em/{j_2}, "MA";"N" **@{-} ?>*@{>}   ?*!/_.8em/{\bar f}, "MB";"N" **@{-} ?>*@{>}   ?*!/^.8em/{\bar g}, "MAB";"N" **@{--} ?>*@{>} ?*!/_.8em/{\bar h}, \end{xy}$ Here (in the notation for free monoids): :$j_1 = \overline {\paren {i_{A \mathop \sqcup B} \circ i_1} }$ :$j_2 = \overline {\paren {i_{A \mathop \sqcup B} \circ i_2}}$ and $i_1$, $i_2$ are the injections for the coproduct. {{finish|I don't see how to continue formally with the machinery already developed; probably some preliminary results are needed for rigour's sake}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f + g} + h} } x       | r = \paren {\map f x + \map g x} + \map h x       | c = {{Defof|Pointwise Addition}} }} {{eqn | r = \map f x + \paren {\map g x + \map h x}       | c = Associative Law of Addition }} {{eqn | r = \map {\paren {f + \paren {g + h} } } x       | c = {{Defof|Pointwise Addition}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f + g} + h} } x       | r = \paren {\map f x + \map g x} + \map h x       | c = {{Defof|Pointwise Addition of Complex-Valued Functions}} }} {{eqn | r = \map f x + \paren {\map g x + \map h x}       | c = Complex Addition is Associative }} {{eqn | r = \map {\paren {f + \paren {g + h} } } x       | c = {{Defof|Pointwise Addition of Complex-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Addition Category:Complex Addition Category:Associativity 0af8d46jvxcja4hep7qyh7kib89npxf
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {f + g} } x       | r = \map f x + \map g x       | c = {{Defof|Pointwise Addition}} }} {{eqn | r = \map g x + \map f x       | c = Commutative Law of Addition }} {{eqn | r = \map {\paren {g + f} } x       | c = {{Defof|Pointwise Addition}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f \times g} \times h} } x       | r = \paren {\map f x \times \map g x} \times \map h x       | c = {{Defof|Pointwise Multiplication}} }} {{eqn | r = \map f x \times \paren {\map g x \times \map h x}       | c = Associative Law of Multiplication }} {{eqn | r = \map {\paren {f \times \paren {g \times h} } } x       | c = {{Defof|Pointwise Multiplication}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {f \times g} } x       | r = \map f x \times \map g x       | c = {{Defof|Pointwise Multiplication}} }} {{eqn | r = \map g x \times \map f x       | c = Commutative Law of Multiplication }} {{eqn | r = \map {\paren {g \times f} } x       | c = {{Defof|Pointwise Multiplication}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f + g} + h} } x       | r = \paren {\map f x + \map g x} + \map h x       | c = {{Defof|Pointwise Addition of Integer-Valued Functions}} }} {{eqn | r = \map f x + \paren {\map g x + \map h x}       | c = Integer Addition is Associative }} {{eqn | r = \map {\paren {f + \paren {g + h} } } x       | c = {{Defof|Pointwise Addition of Integer-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Addition Category:Integer Addition Category:Associativity 0q1vieyebiqhz85xnolw3cgesynngeg
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f + g} + h} } x       | r = \paren {\map f x + \map g x} + \map h x       | c = {{Defof|Pointwise Addition of Rational-Valued Functions}} }} {{eqn | r = \map f x + \paren {\map g x + \map h x}       | c = Rational Addition is Associative }} {{eqn | r = \map {\paren {f + \paren {g + h} } } x       | c = {{Defof|Pointwise Addition of Rational-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Addition Category:Rational Addition Category:Associativity knd7nophv1b3spk206fffapaau9jvv4
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f \times g} \times h} } x       | r = \paren {\map f x \times \map g x} \times \map h x       | c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}} }} {{eqn | r = \map f x \times \paren {\map g x \times \map h x}       | c = Integer Multiplication is Associative }} {{eqn | r = \map {\paren {f \times \paren {g \times h} } } x       | c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Multiplication Category:Integer Multiplication Category:Associativity 72mycc0rpmdxa6q0abqm87ex6d7nvho
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f \times g} \times h} } x       | r = \paren {\map f x \times \map g x} \times \map h c       | c = {{Defof|Pointwise Multiplication of Complex-Valued Functions}} }} {{eqn | r = \map f x \times \paren {\map g x \times \map h x}       | c = Complex Multiplication is Associative }} {{eqn | r = \map {\paren {f \times \paren {g \times h} } } x       | c = {{Defof|Pointwise Multiplication of Complex-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Multiplication Category:Complex Multiplication Category:Associativity kfe1n81c681r4ikiwaaib3n8jwwir5d
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f \times g} \times h} } x       | r = \paren {\map f x \times \map g x} \times \map h x       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{eqn | r = \map f x \times \paren {\map g x \times \map h x}       | c = Real Multiplication is Associative }} {{eqn | r = \map {\paren {f \times \paren {g \times h} } } x       | c = {{Defof|Pointwise Multiplication of Real-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Multiplication Category:Real Multiplication Category:Associativity k8ti1lf9e0gcts9yzi2m24a3crfpvf9
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {\paren {f \times g} \times h} } x       | r = \paren {\map f x \times \map g x} \times \map h x       | c = {{Defof|Pointwise Multiplication of Rational-Valued Functions}} }} {{eqn | r = \map f x \times \paren {\map g x \times \map h x}       | c = Rational Multiplication is Associative }} {{eqn | r = \map {\paren {f \times \paren {g \times h} } } x       | c = {{Defof|Pointwise Multiplication of Rational-Valued Functions}} }} {{end-eqn}} {{qed}} Category:Pointwise Multiplication Category:Rational Multiplication Category:Associativity bufaz948k0tqlqxxyso14zg62al4bih
In the first case $A$ is a not a subset of $B$. So there exists an $x \in A$ such that $x \notin B$. By the definition of union: :$\forall X: x \in A \implies x \in A \cup X$ Hence the solution set is empty. {{qed|lemma}} {{proof wanted|formalising the second case}} Category:Set Theory Category:Set Union lmm3ffkr3xa94ugxps44pm26n5ifmlj
Let $\left({G, \circ}\right)$ be a non-abelian group whose identity is $e$. For a group $\left({G, \circ}\right)$ to be non-abelian, we require: : $\exists x, y \in G: x \circ y \ne y \circ x$ Suppose $x \circ y \in \left\{ {x, y, e}\right\}$. : $x \circ y = e \implies y \circ x = e$ and $\left({G, \circ}\right)$ is abelian. {{WLOG}}, suppose $x \circ y = x$. {{begin-eqn}} {{eqn | l = x \circ y = x       | o = \implies       | r = y = e       | c =  }} {{eqn | o = \implies       | r = y \circ x = x       | c =  }} {{eqn | o = \implies       | r = x \circ y = y \circ x       | c =  }} {{end-eqn}} and again, $\left({G, \circ}\right)$ is abelian. Similarly for $x \circ y = y$. Again, the same applies if $y \circ x \in \left\{ {x, y, e}\right\}$. So, if $x \circ y \ne y \circ x$, then; : $x \circ y$ and $y \circ x$ must be different elements : $x \circ y$ and $y \circ x$ must both different from $e, x$ and $y$. Thus, in a non-abelian group, there needs to be at least $5$ elements: :$e, x, y, x \circ y, y \circ x$ {{qed}}
It follows from Trivial Group is Cyclic Group and Prime Group is Cyclic that groups of order less than $4$ are cyclic. Therefore, by Cyclic Group is Abelian, all groups of order less than $4$ are abelian. Let $G$ have order $4$. From Order of Element Divides Order of Finite Group, every element of $G$ has order that divides $4$. Therefore any element of $G$ has order $1$, $2$ or $4$. Let $g \in G$ have order $4$. Then $G$ is generated by $g$. Therefore $G$ is cyclic, and therefore abelian by Cyclic Group is Abelian. Let $G$ be such that it has no element of order $4$. From Identity is Only Group Element of Order 1, all the elements apart from its identity must have order $2$. That is, $G$ is a Boolean group. By Boolean Group is Abelian, $G$ is abelian. The result follows. {{qed}}
Follows directly from the Total Probability Theorem: :$\ds \map \Pr A = \sum_i \map \Pr {A \mid B_i} \map \Pr {B_i}$ and Bayes' Theorem: :$\map \Pr {B_i \mid A} = \dfrac {\map \Pr {A \mid B_i} \map \Pr {B_i} } {\map \Pr A}$ {{qed}} {{Namedfor|Thomas Bayes|cat = Bayes}}
Let $\mathbf P^{\text{op}}$ be the dual category of $\mathbf P$. From Dual of Order Category, it is the order category corresponding to the dual ordering $\succeq$. From Dual Pairs (Order Theory), it follows that in $\mathbf P^{\text{op}}$: :$r = \inf \left\{{p, q}\right\}$ where $\inf$ denotes infimum. By Infimum is Product in Order Category, $r$ is the product of $p$ and $q$ in $\mathbf P^{\text{op}}$. By Dual Pairs (Category Theory), $r$ is the coproduct of $p$ and $q$ in $\mathbf P$. {{qed}}
By definition of isolation point: :$\exists \delta \in \R_{> 0}: \openint {x - \delta} {x + \delta} \cap A = \set x$ Pick any $\epsilon > 0$. We have that for any $z \in \openint {x - \delta} {x + \delta} \cap A = \set x$: :$\size {\map f z - \map f x} = \size {\map f x - \map f x} = 0 < \epsilon$. This satisfies the condition of continuity at $x$, and the limit in this case is trivially equal to $\map f x$. {{qed}} Category:Continuity Category:Real Functions d2fk3sfbiu72cv0yiqghjrolfvg95m7
=== Definition by Limits implies $\epsilon$-$\delta$ Definition === Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that: :$(1): \quad$ The limit of $\map f x$ as $x \to a$ exists :$(2): \quad \ds \lim_{x \mathop \to a} \map f x = \map f a$ Let $\ds \lim_{x \mathop \to a} \map f x$. Then by the $\epsilon$-$\delta$ definition of limit: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, L} < \epsilon$ By $(2)$:  :$\ds \lim_{x \mathop \to a} \map f x = \map f a$ that is: :$\map f a = L$ Then: :$0 = \map {d_1} {a, a}$ and so: :$\map {d_2} {\map f a, L} = 0 < \epsilon$ Thus the definition by limits implies the $\epsilon$-$\delta$ definition. {{qed|lemma}} === $\epsilon$-$\delta$ Definition implies Definition by Limits === Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ where $\R_{>0}$ denotes the set of all strictly positive real numbers. There are two possibilities for $\map {d_1} {x, a}$: $(\text a): \quad 0 < \map {d_1} {x, a} < \delta$ Then by the $\epsilon$-$\delta$ definition of limit: :$\ds \lim_{x \mathop \to a} \map f x = \map f a$ $(\text b): \quad \map {d_1} {x, a} = 0$ That is, $x = a$. Hence: :$\map f x = \map f a$ and so again: :$\ds \lim_{x \mathop \to a} \map f x = \map f a$ Thus the $\epsilon$-$\delta$ definition implies the definition by limits. {{qed}}
Suppose that for morphisms $x,y: Z \to E$, it holds that: :$e \circ y = e \circ x$ Putting $z = e \circ x$, the following commutative diagram applies: ::$\begin{xy}\xymatrix{  E   \ar[r]^*{e}  &  C   \ar[r]<2pt>^*{f}   \ar[r]<-2pt>_*{g}  &  D  \\  Z   \ar[u]<2pt>^*{x}   \ar[u]<-2pt>_*{y}   \ar[ur]_*{z} }\end{xy}$ It follows that $f \circ z = g \circ z$. Since $e$ is an equalizer, there thus exists a unique $u: Z \to E$ with: :$z = e \circ u$ Hence $x = u = y$, and it follows that $e$ is a monomorphism. {{qed}}