hcju/proofwikiqa huggingface dataset:
Let $A_1 \times A_1$ and $A_2 \times A_2$ be considered with the product topology. Let $F: A_1 \times A_1 \to A_2 \times A_2$ be the mapping defined as: :$\map F {x, y} = \tuple {\map f x, \map f y}$ By Projection from Product Topology is Continuous, we have that the (first and second) projections on $A_1 \times A_1$ are continuous. By Continuity of Composite Mapping and Continuous Mapping to Topological Product, it follows that $F$ is continuous. By Distance Function of Metric Space is Continuous and Continuity of Composite Mapping, it follows that $d_2 \circ F: A_1 \times A_1 \to \R$ is continuous. Let $\epsilon \in \R_{>0}$ be a strictly positive real number. By Continuity Defined from Closed Sets, the set: :$C = \paren {d_2 \circ F}^{-1} \sqbrk {\hointr \epsilon \to} = \set {\tuple {x, y} \in A_1 \times A_1: \map {d_2} {\map f x, \map f y} \ge \epsilon}$ is closed in $A_1 \times A_1$. By Topological Product of Compact Spaces, $A_1 \times A_1$ is compact. By Closed Subspace of Compact Space is Compact, $C$ is compact. By Distance Function of Metric Space is Continuous and Continuous Image of Compact Space is Compact, $d_1 \sqbrk C$ is compact. Therefore, $d_1 \sqbrk C$ has a smallest element $\delta$. {{explain}} By metric space axioms $(\text M 1)$ and $(\text M 4)$, we have that $\delta > 0$. By construction, it follows that: :$\forall x, y \in A_1: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map f x, \map f y} < \epsilon$ Hence, $f$ is uniformly continuous. {{qed}}
We have that $\R$ is a metric space under the usual (Euclidean) metric. We also have from the Heine-Borel Theorem that $\closedint a b$ is compact. So the result Heine-Cantor Theorem applies. {{qed}}
{{begin-eqn}} {{eqn | l = \sin \left({\frac \pi 2 - \theta}\right)       | r = \sin \frac \pi 2 \cos \theta - \cos \frac \pi 2 \sin \theta       | c = Sine of Difference }} {{eqn | r = 1 \times \cos \theta - 0 \times \sin \theta       | c = Sine of Right Angle and Cosine of Right Angle }} {{eqn | r = \cos \theta }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \sin {\frac \pi 2 - \theta}       | r = -\map \sin {\theta - \frac \pi 2}       | c = Sine Function is Odd }} {{eqn | r = \map \cos {\theta - \frac \pi 2 + \frac \pi 2}       | c = Cosine of Angle plus Right Angle }} {{eqn | r = \cos \theta }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \sin {\dfrac \pi 2 - \theta}       | r = \map \Im {e^{i \paren {\frac \pi 2 - \theta} } }       | c = Euler's Formula }} {{eqn | r = \map \Im {e^{i \frac \pi 2} e^{-i \theta} }       | c = Exponent Combination Laws }} {{eqn | r = \map \Im {\paren {\cos \dfrac \pi 2+i \sin \dfrac \pi 2} e^{-i \theta} }       | c = Euler's Formula }} {{eqn | r = \map \Im {i e^{-i \theta} }       | c = Cosine of Right Angle, Sine of Right Angle }} {{eqn | r = \map \Re {e^{-i \theta} }       | c =  }} {{eqn | r = \map \cos {-\theta}       | c = Euler's Formula }} {{eqn | r = \cos \theta       | c = Cosine Function is Even }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cos {\frac \pi 2 - \theta}       | r = \cos \frac \pi 2 \cos \theta + \sin \frac \pi 2 \sin \theta       | c = Cosine of Difference }} {{eqn | r = 0 \times \cos \theta + 1 \times \sin \theta       | c = Cosine of Right Angle and Sine of Right Angle }} {{eqn | r = \sin \theta }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos \left({\frac \pi 2 - \theta}\right)       | r = \cos \left({\theta - \frac \pi 2}\right)       | c = Cosine Function is Even }} {{eqn | r = \sin \left({\theta - \frac \pi 2 + \frac \pi 2}\right)       | c = Sine of Angle plus Right Angle }} {{eqn | r = \sin \theta }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cos {\dfrac \pi 2 - \theta}       | r = \frac 1 2 \paren {e^{i \paren {\frac \pi 2 - \theta} } + e^{-i \paren {\frac \pi 2 - \theta} } }       | c = Cosine Exponential Formulation }} {{eqn | r = \frac 1 2 \paren {e^{i \frac \pi 2} e^{-i \theta} + e^{-i \frac \pi 2} e^{i \theta} }       | c = Exponential of Sum: Complex Numbers }} {{eqn | r = \frac 1 2 \paren {\paren {\map \cos {\frac \pi 2} + i \, \map \sin {\frac \pi 2} } e^{-i \theta} + \paren {\map \cos {-\frac \pi 2} + i \, \map \sin {-\frac \pi 2} } e^{i \theta} }       | c = Euler's Formula }} {{eqn | r = \frac 1 2 \paren {i e^{-i \theta} - i e^{i \theta} }       | c = Cosine of Right Angle, Sine of Right Angle, Cosine Function is Even, Sine Function is Odd }} {{eqn | r = \frac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }       | c = {{Defof|Imaginary Unit}} }} {{eqn | r = \sin \theta       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}}
Let $g = f^{-1}$. We need to show that $g: T_2 \to T_1$ is continuous. For any $V \subseteq T_1$, we have $g^{-1} \left({V}\right) = f  \left({V}\right)$. We are to show that if $V$ is closed in $T_1$, then $g^{-1} \left({V}\right)$ is closed in $T_2$. Suppose $V$ is closed in $T_1$. Since $T_1$ is compact, $V$ is compact by Closed Subspace of Compact Space is Compact. So $f \left({V}\right)$ is compact from Continuous Image of Compact Space is Compact. Since $T_2$ is Hausdorff, $f  \left({V}\right)$ closed by Compact Subspace of Hausdorff Space is Closed. But $f  \left({V}\right) = g^{-1} \left({V}\right)$, so $g^{-1} \left({V}\right)$ is closed. From Continuity Defined from Closed Sets, it follows that $g$ is continuous. Thus by definition, $f$ is a homeomorphism. {{qed}}
=== $(1) \iff (2)$: No Separation iff No Union of Closed Sets === {{:Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets}}{{qed|lemma}} === $(1) \iff (4)$: No Separation iff No Clopen Sets === {{:Equivalence of Definitions of Connected Topological Space/No Separation iff No Clopen Sets}}{{qed|lemma}} === $(2) \implies (3)$: No Union of Closed Sets implies No Subsets with Empty Boundary === {{:Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary}}{{qed|lemma}} === $(3) \implies (4)$: No Subsets with Empty Boundary implies No Clopen Sets === {{:Equivalence of Definitions of Connected Topological Space/No Subsets with Empty Boundary implies No Clopen Sets}}{{qed|lemma}} === $(4) \implies (5)$: No Clopen Sets implies No Union of Separated Sets === {{:Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets}}{{qed|lemma}} === $(5) \implies (6)$: No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space === {{:Equivalence of Definitions of Connected Topological Space/No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space}}{{qed|lemma}} === $(6) \implies (1)$: No Continuous Surjection to Discrete Two-Point Space implies No Separation === {{:Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation}}{{qed}}
For proofs, see: * Reductio ad Absurdum/Sequent Form * Proof by Contradiction/Sequent Form
Let $x$ be an even integer. Let $y = 2 n + 5$. Assume $y = x + 5$ is not an odd integer. Then: :$y = x + 5 = 2 n$ where $n \in \Z$. Then: {{begin-eqn}} {{eqn | l = x       | r = 2 n - 5       | c =  }} {{eqn | r = \paren {2 n - 6} + 1       | c =  }} {{eqn | r = 2 \paren {n - 3} + 1       | c =  }} {{eqn | r = 2 r + 1       | c = where $r = n - 3 \in \Z$ }} {{end-eqn}} Hence $x$ is odd. That is, it is false that $x$ is even. It follows by the Rule of Transposition that if $x$ is even, then $y$ is odd. {{qed}}
=== First assertion === Let $\displaystyle T_1 = \bigcup_{i \mathop  =1}^n V_i$, with the $V_i$ closed in $T_1$. Assume that for each $i$, the restriction $f \restriction_{V_i}$ is continuous. Then $f$ satisfies the hypotheses of Continuous Mapping on Finite Union of Closed Sets. Hence $f$ is continuous. {{qed}} === Second assertion === Let $\displaystyle T_1 = \bigcup_{i \mathop \in I} U_i$, with the $U_i$ open in $T_1$. Assume that for each $i \in I$, the restriction $f \restriction_{U_i}$ is continuous. Then $f$ satisfies the hypotheses of Continuous Mapping on Union of Open Sets. Hence $f$ is continuous. {{qed}}
From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive. Therefore, the following suffices: === Sufficient Condition === Suppose $S$ is an interval of $\R$. Suppose further that $A \mid B$ is a separation of $S$. Let $a \in A, b \in B$. {{WLOG}}, suppose that $a < b$. Since $a, b \in S$, and $S$ is an interval, $\closedint a b \subseteq S$. Let $A' = A \cap \closedint a b$ and $B' = B \cap \closedint a b$. Then: {{begin-eqn}} {{eqn | l = A' \cup B'       | r = \paren {A \cap \closedint a b} \cup \paren {B \cap \closedint a b} }} {{eqn | r = \paren {A \cup B} \cap \closedint a b       | c = Intersection Distributes over Union }} {{eqn | n = 1       | ll= \leadsto       | l = A' \cup B'       | r = \closedint a b       | c = Intersection with Subset is Subset }} {{end-eqn}} By the definition of a separation, both $A$ and $B$ are closed in $S$. Hence by Closed Set in Topological Subspace, $A'$ and $B'$ are also closed in $\closedint a b$. From Closed Set in Topological Subspace: Corollary, $A'$ and $B'$ are closed in $\R$. Now, since $B' \ne \O$, and $B$ is bounded below (by, for example, $a$), by the Continuum Property $b' := \map \inf {B'}$ exists, and $b' \ge a$. We have that $B'$ is closed in $\R$ Hence from Closure of Real Interval is Closed Real Interval: :$b' \in B'$ Since $a \in A'$ and $A \cap B = \O$, it follows that $b' > a$. Now let $A'' = A' \cap \closedint a {b'}$. Using the same argument as for $B'$, we have that $a'' = \map \sup {A''}$ exists, that $a'' \in A''$ and also $a'' < b'$. Now $\openint {a''} {b'} \cap A' = \O$ or $a''$ would not be an upper bound for $A''$. Similarly, $\openint {a''} {b'} \cap B' = \O$ or $b'$ would not be a lower bound for $B''$. Thus: :$\openint {a''} {b'} \cap \paren {A' \cup B'} = \O$ But since $a < a'' < b' < b$, we also have: :$\openint {a''} {b'} \subseteq \closedint a b$, and :$\openint {a''} {b'}$ is non-empty. So, there is an element $z \in \openint {a''} {b'}$, and hence in $\closedint a b$, which is not in $A' \cup B'$. This contradicts $(1)$ above, which says that we have $A' \cup B' = \closedint a b$. From this contradiction it follows that there can be no such separation $A \mid B$ on the interval $S$. Therefore, by definition, $S$ is connected. {{qed|lemma}} === Necessary Condition === Suppose $S$ is not an interval of $\R$. Then $\exists x, y \in S$ and $z \in \R \setminus S$ such that $x < z < y$. Consider the sets $S \cap \openint \gets z$ and $S \cap \openint z \to$. Then $S \cap \openint \gets z$ and $S \cap \openint z \to$ are open by definition of the subspace topology on $S$. Neither is empty because they contain $x$ and $y$ respectively. They are disjoint, and their union is $S$, since $z \notin S$. Therefore $S \cap \openint \gets z \mid S \cap \openint z \to$ is a separation of $S$. It follows by definition that $S$ is disconnected. {{qed}}
Let $i: f \sqbrk {T_1} \to T_2$ be the inclusion mapping. Let $g: T_1 \to f \sqbrk {T_1}$ be the surjective restriction of $f$. Then $f = i \circ g$. Hence, by Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is continuous. We use a Proof by Contradiction. Suppose that $A \mid B$ is a partition of $f \sqbrk {T_1}$. Then it follows that $g^{-1} \sqbrk A \mid g^{-1} \sqbrk B$ is a partition of $T_1$. {{explain|Add the justification for the above - may already exist as a result, otherwise add a proof something like Preimage of Partition is Partition.}} {{qed}}
Suppose that $S_2 = f \left({S_1}\right)$ is not connected in $T_2$. Then by definition there exist open sets $U_2$ and $V_2$ in $T_2$ such that: :$S_2 \subseteq U_2 \cup V_2$ :$U_2 \cap V_2 \cap S_2 = \varnothing$ :$U_2 \cap S_2 \ne \varnothing$ :$V_2 \cap S_2 \ne \varnothing$ By hypothesis, $f: T_1 \to T_2$ is continuous. Thus $U_1 = f^{-1} \left({U_2}\right)$ and $V_1 = f^{-1} \left({V_2}\right)$ are open in $T_1$. We have that: :$U_2 \cap S_2 \ne \varnothing$ Therefore: :$\exists x \in S_1: f \left({x}\right) \in U_2$ Then: :$x \in f^{-1} \left({U_2}\right) = U_1$ and: :$x \in S_1$ so: :$U_1 \cap S_1 \ne \varnothing$ Similarly: :$V_1 \cap S_1 \ne \varnothing$ Suppose there exists $x \in S_1$ such that $x \in U_1 \cap V_1 \cap S_1$. Then: :$f \left({x}\right) \in U_2 \cap V_2 \cap S_2$ which is a contradiction. It follows that: :$U_1 \cap V_1 \cap S_1 = \varnothing$ Thus by definition $S_1$ is not connected in $T_1$. The result follows by the Rule of Transposition.
Let $A := \displaystyle \bigcup \AA$. Let $D = \set {0, 1}$, with the discrete topology. Let $f: A \to D$ be continuous. To show that $A$ is connected, we need to show that $f$ is not a surjection. Since each $C \in \AA$ is connected and the restriction $f \restriction_C$ is continuous: :$\map f C = \set {\map \epsilon C}$ where $\map \epsilon C = 0$ or $1$. But, for all $B, C \in \AA$: :$B \cap C \ne \O$ Hence $\map \epsilon B = \map \epsilon C$.  Thus $f$ is constant on $A$ as required. {{qed}}
==== Necessary Condition ==== Let $T_1 \times T_2$ be connected. By Projection from Product Topology is Continuous, $T_1$ and $T_2$ are continuous images under the projections $\operatorname{pr}_1$ and $\operatorname{pr}_2$. Hence by Continuous Image of Connected Space is Connected, $T_1$ and $T_2$ are connected. {{qed|lemma}} ==== Sufficient Condition ==== Suppose that $T_1$ and $T_2$ are connected. Let: : $I = T_2$ : $\forall y \in T_2: C_y = T_1 \times \left\{{y}\right\}$ : $B = \left\{{x_0}\right\} \times T_2$ for some fixed $x_0 \in T_1$. Each $C_y$ is homeomorphic to $T_1$ by Topological Product with Singleton. By Connectedness is a Topological Property, each $C_y$ is therefore connected. By the same argument, $B$ is also connected. Also: : $C_y \cap B = \left\{{\left({x_0, y}\right)}\right\}$ and hence is non-empty : $\displaystyle T_1 \times T_2 = B \cup \bigcup_{y \mathop \in T_2} C_y$. So by the corollary to Union of Connected Sets with Non-Empty Intersections is Connected, it follows that $T_1 \times T_2$ is connected.
Let $D$ be the discrete space $\left\{{0, 1}\right\}$. Let $f: K \to D$ be an arbitrary continuous mapping. From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous. We have that: :$H$ is connected :$f \restriction_H$ is continuous Thus by definition of connected set: :$f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$  {{WLOG}}, let $f \left({H}\right) = \left\{{0}\right\}$. {{AimForCont}} $\exists k \in K: f \left({k}\right) = 1$. By definition of discrete space, $\left\{{1}\right\}$ is open in $D$. Hence by definition of continuous mapping: : $f^{-1} \left({\left\{{1}\right\}}\right)$ is open in $K$. Let $K$ be given the subspace topology. Then for some $U$ open in $T$: : $f^{-1} \left({\left\{{1}\right\}}\right) = K \cap U$ We have that: : $k \in f^{-1} \left({\left\{{1}\right\}}\right) \subseteq U$ and: :$k \in H^-$ By definition of topology: : $\exists x \in H \cap U$ As $x \in H$, we have that: :$f \left({x}\right) = 0$ But because $x \in H \cap U \subseteq K \cap U = f^{-1} \left({\left\{{1}\right\}}\right)$: : $f \left({x}\right) = 1$ This contradicts the definition of mapping. Thus by Proof by Contradiction, $f: K \to D$ can not be a surjection. Thus $K$ is connected. {{qed}}
Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T$ whose underlying set is $K$. Let $\map {\cl_K} H$ denote the closure of $H$ in $K$. From Closure of Subset in Subspace: :$\map {\cl_K} H = K \cap H^-$ By hypothesis: :$K \subseteq H^-$ and so by Intersection with Subset is Subset: :$\map {\cl_K} H = K$ Let $D$ be the discrete space $\set {0, 1}$. Let $f: K \to D$ be any continuous mapping. From Continuity of Composite with Inclusion, the restriction $f \restriction_H$ is continuous. SWe have that: :$H$ is connected :$f \restriction_H$ is continuous Thus by definition of connected set: :$f \sqbrk H = \set 0$ or $f \sqbrk H = \set 1$  {{WLOG}}, let $f \sqbrk H = \set 0$. From Continuity Defined by Closure: :$f \sqbrk {\map {\cl_K} H} \subseteq \map {\cl_K} {f \sqbrk H} = \set 0^-$ where $\set 0^-$ denotes the closure of $\set 0$ in $D$. As $D$ is the discrete space, it follows from Set in Discrete Topology is Clopen that $\set 0$ is closed in $D$. Thus by Set is Closed iff Equals Topological Closure: :$\set 0^- = \set 0$ That is, $f \sqbrk K = \set 0$. Thus $K$ is connected by definition. {{qed}}
Let $D$ be the discrete space $\set {0, 1}$. Let $T$ be path-connected. Let $f: T \to D$ be a continuous surjection. Let $x, y \in T: \map f x = 0, \map f y = 1$. Let $I \subset \R$ be the closed real interval $\closedint 0 1$. Let $g: I \to T$ be a path from $x$ to $y$. Then by Continuity of Composite Mapping it follows that $f \circ g: I \to D$ is a continuous surjection. This contradicts the connectedness of $I$ as proved in Subset of Real Numbers is Interval iff Connected. {{explain|Why does that follow? Explain what the chain of steps is}} Hence the result. {{qed}}
Let $R \subset \mathbb{SO} \left({3}\right)$ be the group generated by the $\pi$ and $\dfrac {2 \pi} 3$ rotations around different axes. The elements: :$\psi = \begin{pmatrix} -\tfrac 1 2 & \tfrac{\sqrt 3} 2 & 0 \\ -\tfrac{\sqrt 3} 2 & -\tfrac 1 2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$ :$\phi = \begin{pmatrix} -\cos \left({\vartheta}\right) & 0 & \sin \left({\vartheta}\right) \\ 0 & -1 & 0 \\ \sin \left({\vartheta}\right) & 0 & \cos \left({\vartheta}\right) \\ \end{pmatrix}$ form a basis for $R$ for some $\vartheta$. We have: :$\psi^3 = \phi^2 = \mathbf I_3$ where $\mathbf I_3$ is the identity mapping in $\mathbb R^3$. Therefore $\forall r \in R: r \ne \mathbf I_3, \psi, \phi, \psi^2$ there exists some $n \in \N$ and some set of numbers $m_k \in \left\{{1, 2}\right\}, 1 \le k \le n$ such that $r$ can written as one of the following: :$\displaystyle \text{(a)}: r = \prod_{k \mathop = 1}^n \phi \psi^{m_k}$ :$\displaystyle \text{(b)}: r = \psi^{m_1} \left({\prod_{k \mathop = 2}^n \phi \psi^{m_k}}\right) \phi$ :$\displaystyle \text{(c)}: r = \left({\prod_{k \mathop = 1}^n \phi \psi^{m_k}}\right) \phi$ :$\displaystyle \text{(d)}: r = \psi^{m_1} \left({\prod_{k \mathop = 2}^n \phi \psi^{m_k}}\right)$ Now we fix $\vartheta$ such that $\mathbf I_3$ cannot be written in any of the ways $\text{(a)}$, $\text{(b)}$, $\text{(c)}$, $\text{(d)}$.  The action of $R$ on $\mathbb S^2$ will leave two points unchanged for each element of $R$ (the intersection of the axis of rotation and the sphere, to be exact).   Since $R$ is finitely generated, it is a countable group. Therefore the set of points of $\mathbb S^2$ which are unchanged by at least one element of $R$ is also countable.   We call this set $D \subset \mathbb S^2$, so that $R$ acts freely on $\mathbb S^2 - D$. By Set of Orbits forms Partition, this partitions $\mathbb S^2 - D$ into orbits. By the Axiom of Choice, there is a set $X$ containing one element of each orbit.   For any $r \in R$, let $X_r$ be the action of $r$ on $X$. We have: :$\displaystyle \mathbb S^2 - D = \bigcup_{r \mathop \in R} X_r$ Define the sets $A, B, C$ to be the smallest sets satisfying  :$X \subseteq A$ :$\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi} \subset B, A, C, \text{respectively.}$ :$\text{If } X_r \subset A, B, C, \text{ then } X_{r\psi} \subset B, C, A, \text{respectively.}$ :$\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi^2} \subset C, A, B, \text{respectively.}$ These sets are defined due to the uniqueness of the properties $\text{(a)}$ to $\text{(d)}$ Also, $A, B, C, B \cup C$ are congruent since they are rotations of each other, namely: :$A_\psi = B, B_{\psi^2} = C, A_\phi = B \cup C$ Hence we have constructed the sets $A, B, C, D$ of the theorem. {{qed}} {{AoC}} Whether you view this result as a veridical paradox or an antinomy depends your acceptance or otherwise of the Axiom of Choice. {{namedfor|Felix Hausdorff|cat = Hausdorff}}
:350 px Let $ABC$ be a triangle We can extend $BA$ past $A$ into a straight line. There exists a point $D$ such that $DA = CA$. Therefore, from Isosceles Triangle has Two Equal Angles: :$\angle ADC = \angle ACD$ Thus by Euclid's fifth common notion: :$\angle BCD > \angle BDC$ Since $\triangle DCB$ is a triangle having $\angle BCD$ greater than $\angle BDC$, this means that $BD > BC$. But: :$BD = BA + AD$ and: :$AD = AC$ Thus: :$BA + AC > BC$ A similar argument shows that $AC + BC > BA$ and $BA + BC > AC$. {{qed}} {{Euclid Note|20|I|It is a geometric interpretation of the triangle inequality.}}
:250px Given a triangle $ABC$ and a point $D$ inside it. We can construct lines connecting $A$ and $B$ to $D$, and then extend the line $AD$ to a point $E$ on $BC$. In $\triangle ABE$, $AB + AE>BE$. Then, $AB + AC = AB + AE + EC > BE + EC$ by Euclid's second common notion. Similarly, $CE + ED > CD$, so $CE + EB = CE + ED + DB > CD + DB$. Thus, $AB + AC > BE + EC > CD + DB$. {{qed}} {{Euclid Note|21|I}}
Since $F$ is the center of the circle with radius $FD$, it follows from {{EuclidDefLink|I|15|Circle}} that $DF = KF$, so $a = KF$ by Euclid's first common notion. Since $G$ is the center of the circle with radius $GH$, it follows from {{EuclidDefLink|I|15|Circle}} that $GH = GK$, so $c = GK$ by Euclid's first common notion. $FG = b$ by construction. Therefore the lines $FK$, $FG$, and $GK$ are, respectively, equal to the lines $a$, $b$, and $c$, so $\triangle FGK$ is indeed the required triangle. {{qed}} {{Euclid Note|22|I}} Note that the condition required of the lengths of the segments is the equality shown in Proposition $20$: Sum of Two Sides of Triangle Greater than Third Side.  Thus, this is a necessary condition for the construction of a triangle. When {{AuthorRef|Euclid}} first wrote the proof of this proposition in {{BookLink|The Elements|Euclid}}, he neglected to prove that the two circles described in the construction actually do intersect, just as he did in Proposition $1$: Construction of Equilateral Triangle.
Since all three sides of the triangles are equal, the interior angles of the triangles are also equal. Thus, $\angle GAF = \angle ECD$, with $\angle GAF$ at the point $A$ on the straight line $AB$. {{qed}} {{Euclid Note|23|I|The careful reader will note that Proposition $22$: Construction of Triangle from Given Lengths does not directly create the necessary triangle at point $A$, but rather at a distance $CE$ from point $A$.  However, a slight modification of the construction produces the desired triangle at the desired location.}} This theorem is one of the two attributed to {{AuthorRef|Oenopides of Chios}} (the other being {{EuclidPropLink|prop = 12|book = I|title = Perpendicular through Given Point}}).
Let $a \in U$. Let $H \subseteq U$ be the subset of points in $U$ which can be joined to $a$ by a path in $U$. Let $K = U \setminus H$. Let $x \in H$. Then: : $\exists \epsilon > 0: B_\epsilon \left({x}\right) \subseteq U$ where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$. Given any $y \in B_\epsilon \left({x}\right)$, there is a (straight line) path $g$ in $B_\epsilon \left({x}\right) \subseteq U$ connecting $x$ and $y$. But since $x \in H$, there is a path $f$ in $U$ joining $a$ to $x$. From Joining Paths makes Another Path, traversing $f$, and then $g$ forms a path from $a$ to $y$. It follows that $y \in H$, and therefore $B_\epsilon \left({x}\right) \subseteq H$. Thus $H$ is open. By a similar argument, $K$ is also shown to be open: If $x \in K$, then $B_\epsilon \left({x}\right) \subseteq U$ for some $\epsilon > 0$. If any point in $B_\epsilon \left({x}\right)$ can be joined to $a$ by a path in $U$, then so could $x$. It is clear that $H \cap K = \varnothing$ and $H \cup K = U$ by definition of set difference. As, trivially, $a \in H$, we have $H \ne \varnothing$. Knowing that $U$ is connected, it follows that $K = \varnothing$, and $H = U$. Hence the result. {{qed}}