avewright/proofwiki-math huggingface dataset:
Theorem: Phi in the Pentagon

Let $ABCDE$ be a regular pentagon.
The chords and sides of $ABCDE$ form three kinds of similar triangles.
* I: Isosceles, similar to $\triangle ACD$.
* II: Isosceles, similar to $\triangle ABE$.
The sides of the first type are in the ratio $\phi : 1$, where $\phi$ is the golden mean.
The sides of the second type are in the ratio $1 : \phi$.
300px
{{:Euclid:Proposition/XIII/8}}

Proof:
{{hypothesis}} five sides of $ABCDE$ are equal.
Draw all five chords.
From symmetry, the chords are all equal.
$\leadsto$:
:$\triangle ACD$ is an isosceles triangle.
$\leadsto$:
:$\triangle BCD$ is an isosceles triangle
{{begin-eqn}}
{{eqn | l = \angle DBC
      | r = \angle CDB = \alpha
      | c = Isosceles Triangle has Two Equal Angles
}}
{{eqn | l = \angle ACD
      | r = \angle ADC
      | c = Isosceles Triangle has Two Equal Angles
}}
{{eqn | l = \triangle ACD
      | r = 3 \beta + 2 \alpha = 180 \degrees
      | c = Sum of Internal Angles of Triangle
}}
{{eqn | l = \triangle BCD
      | r = 1 \beta + 4 \alpha = 180 \degrees
      | c = Sum of Internal Angles of Triangle
}}
{{eqn | l = 2 \beta - 2 \alpha
      | r = 0
      | c = subtracting
}}
{{eqn | ll = \leadsto
      | l = \alpha
      | r = \beta
      | c =
}}
{{eqn | l = \angle CA'D
      | r = \angle BCD
      | c = Sum of Internal Angles of Triangle
}}
{{eqn | l = \angle B'A'E'
      | r = \angle CA'D
      | c = Vertical Angle Theorem
}}
{{eqn | l = \angle B'A'E' 
      | r = 3 \beta = \angle BCD
      | c = {{EuclidCommonNotionLink|1}}
}}
{{end-eqn}}
Hence $A'B'C'D'E'$ has all internal angles equal and is Similar to $ABCDE$.
{{begin-eqn}}
{{eqn | l = \alpha
      | r = \beta = \dfrac \pi 5 = 72^{\circ}
      | c = Sum of Internal Angles of Polygon
}}
{{end-eqn}}
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal:
:$BCDC'$ is a parallelogram.
Also by symmetry:
:$\triangle AD'C'$ is isosceles
{{begin-eqn}}
{{eqn | l = \triangle AD'C'
      | o = \sim
      | r = \triangle ACD
      | c = Equiangular Triangles are Similar
}}
{{end-eqn}}
Let the length of $D'C' = B'C' = 1$.
Let the length of $EB' = AD' = AC' = \phi$, where $\phi$ is to be determined.
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = \dfrac \phi 1
      | r = \dfrac {2 \phi + 1} {\phi + 1}
      | c =
}}
{{eqn | l = \phi^2
      | r = \phi + 1
      | c = rearranging
}}
{{eqn | l = \phi^2 - \phi - 1
      | r = 0
      | c =
}}
{{end-eqn}}
By the quadratic equation:
:$\phi = \dfrac{ 1 \pm \sqrt{5} } 2$
We take $\phi > 0$ so:
:$\phi = \dfrac{ 1 + \sqrt{5} } 2$
By definition of golden mean:
:$\phi$ is the golden mean
The length of the sides of the pentagon is $\phi + 1$.
By Equiangular Triangles are Similar:
:$\triangle ADE \sim \triangle B'ED$
$\leadsto$:
:$\dfrac {\phi + 1} { 2 \phi + 1 } = \dfrac {\phi} {\phi + 1}$
Working with the {{RHS}} and substituting from above:
:$\dfrac {\phi} {\phi + 1} = \dfrac {\phi} {\phi^2}$
:$= \dfrac 1 \phi$
The result follows.
{{qed}}
Theorem: Real Elementary Functions are Continuous

The elementary functions on the real numbers $\R$ are all continuous at every point of their domain.

Proof:
In turn:
:Real Polynomial Function is Continuous
:Real Trigonometric Functions are Continuous
:Real Logarithm Function is Continuous
:Real Exponential Function is Continuous
It remains to show that:
:All real functions that are compositions of the above
:All real functions obtained by adding, subtracting, multiplying and dividing any of the above types any finite number of times
are also continuous at every point of their domain.
{{finish}}
Theorem: Real Trigonometric Functions are Continuous

The trigonometric functions on the real numbers $\R$:
:sine
:cosine
:tangent
:cotangent
:secant
:cosecant
are all continuous at every point of their domain.

Proof:
In turn:
:Real Sine Function is Continuous
:Real Cosine Function is Continuous
:Real Tangent Function is Continuous
:Real Cotangent Function is Continuous
:Real Secant Function is Continuous
:Real Cosecant Function is Continuous
{{qed}}
Theorem: Continuous Function is not necessarily Differentiable

Let $f$ be a real function defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is continuous at $x_0$.
Then it is not necessarily the case that $f$ is differentiable at $x_0$.

Proof:
;Proof by Counterexample
Consider the real function $f: \R \to \R$ defined as:
:$\forall x \in \R: \map f x = \size x$
where $\size x$ denotes the absolute value function.
From Absolute Value Function is Continuous, $f$ is continuous on $\R$.
From Derivative of Absolute Value Function, $f$ is not differentiable at $x = 0$.
Hence $f$ is an example of a real function which at $x_0 = 0$ is continuous at $x_0$ but not differentiable at $x_0$.
{{qed}}
Theorem: Combination Theorem for Continuous Functions/Complex/Difference Rule

:$f - g$ is continuous on $S$.

Proof:
We have that:
:$\map {\paren {f - g} } x = \map {\paren {f + \paren {-g} } } x$
From Multiple Rule for Continuous Complex Functions:
:$-g$ is continuous on $S$.
From Sum Rule for Continuous Complex Functions:
:$f + \paren {-g}$ is continuous on $S$.
The result follows.
{{qed}}
Category:Combination Theorems for Continuous Functions
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Theorem: Real Function both Convex and Concave is Linear

Let $f$ be a real function which is both convex and concave.
Then $f$ is a linear function.

Proof:
Let $f$ be both convex and concave on a subset $S \subseteq \R$ of the real numbers $\R$.
Then by definition:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1
      | l = \map f {\alpha x + \beta y}
      | o = \le
      | r = \alpha \map f x + \beta \map f y
      | c = {{Defof|Convex Real Function}}
}}
{{eqn | q = \forall x, y \in S: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1
      | l = \map f {\alpha x + \beta y}
      | o = \ge
      | r = \alpha \map f x + \beta \map f y
      | c = {{Defof|Concave Real Function}}
}}
{{eqn | ll= \leadsto
      | l = \map f {\alpha x + \beta y}
      | r = \alpha \map f x + \beta \map f y
      | c = 
}}
{{end-eqn}}
{{ProofWanted|how to get there from here}}
Theorem: Equivalence of Definitions of Convex Hull

{{TFAE|def = Convex Hull}}
Let $V$ be a vector space over $\R$. 
Let $U \subseteq V$.

Proof:
=== Definition $(1)$ iff Definition $(3)$ ===
This is demonstrated in Convex Hull is Smallest Convex Set containing Set.
{{qed}}
Theorem: Euler Triangle Formula/Corollary

Let $\rho$ be the inradius of a triangle.
Then:
:$\dfrac 1 \rho = \dfrac 1 {R - d} + \dfrac 1 {R + d}$
where:
:$R$ is the circumradius
:$d$ is the distance between the incenter and the circumcenter.

Proof:
{{begin-eqn}}
{{eqn | l = d^2
      | r = R \paren {R - 2 \rho}
      | c = Euler Triangle Formula
}}
{{eqn | ll= \leadsto
      | l = d^2
      | r = R^2 - 2 R \rho
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 R \rho
      | r = R^2 - d^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \rho
      | r = \dfrac {\paren {R + d} \paren {R - d} } {2 R}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 \rho
      | r = \dfrac {2 R} {\paren {R + d} \paren {R - d} }
      | c = 
}}
{{eqn | r = \dfrac 1 {R - d} + \dfrac 1 {R + d}
      | c = 
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Leonhard Paul Euler}}
Category:Euler Triangle Formula
pbzvyl4wz9ds1ds682fjtq3zf9conth
Theorem: Condition for Cross-Ratio to be Harmonic

Let $A$, $B$, $C$ and $D$ be points on a straight line.
Let their '''cross-ratio''' $\set {A, B; C, D}$ be a harmonic ratio.
Then:
:$\set {A, B; C, D} = \set {B, A; C, D}$

Proof:
Let $\set {A, B; C, D}$ form a harmonic ratio.
We have:
{{begin-eqn}}
{{eqn | l = \set {A, B; C, D}
      | r = -1
      | c = {{Defof|Harmonic Ratio}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac {AC \cdot DB} {AD \cdot CB}
      | r = -1
      | c = {{Defof|Cross-Ratio of Points on Line}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {-CA} \cdot \paren {-BD} } {\paren {-DA} \cdot \paren {-BC} }
      | r = -1
      | c = employing the fact that $AC$, $DB$ etc. are directed line segments
}}
{{eqn | ll= \leadsto
      | l = \dfrac {CA \cdot BD} {DA \cdot BC}
      | r = -1
      | c = all those $-1$ factors cancel out
}}
{{eqn | ll= \leadsto
      | l = \dfrac {DA \cdot BC} {CA \cdot BD}
      | r = -1
      | c = taking reciprocal of both sides
}}
{{eqn | ll= \leadsto
      | l = \dfrac {BC \cdot DA} {BD \cdot CA}
      | r = -1
      | c = Real Multiplication is Commutative
}}
{{eqn | ll= \leadsto
      | l = \set {B, A; C, D}
      | r = -1
      | c = {{Defof|Cross-Ratio of Points on Line}}
}}
{{eqn | ll= \leadsto
      | l = \set {A, B; C, D}
      | r = \set {B, A; C, D}
      | c = as both are equal to $-1$
}}
{{end-eqn}}
{{qed}}
Theorem: Predecessor is Infimum

Let $\struct{S, \preceq}$ be an ordered set.
Let $x, y \in S$.
Then:
:$x \preceq y$ {{iff}} $x = \map \inf {x, y}$
where $\map \inf {x, y}$ is the infimum of $x$ and $y$

Proof:
This is the dual statement of Successor is Supremum by Dual Pairs (Order Theory).
The result follows from the Duality Principle.
{{qed}}
Category:Orderings
Category:Infima
gtst9yw29m63l4pccqo3fj81owkfysa
Theorem: Successor is Supremum

Let $\struct {S, \preceq}$ be an ordered set.
Let $x, y \in S$.
Then:
:$x \preceq y$ {{iff}} $y = \map \sup {x, y}$
where $\map \sup {x, y}$ is the supremum of $x$ and $y$

Proof:
=== Necessary Condition ===
Let $x \preceq y$.
By {{Ordering-axiom|1}}:
:$y \preceq y$
Hence $y$ is an upper bound of $\set {x, y}$ by definition.
Let $z$ be an upper bound of $\set {x, y}$.
By definition of upper bound:
:$y \preceq z$
Hence $y$ is the supremum of $x$ and $y$ by definition.
{{qed|lemma}}
Theorem: Cubical Parabola has Point of Inflection

The '''cubical parabola''' has a point of inflection.
When the equation for the '''cubical parabola''' is in the form $y = a x^3$, this point of inflection is the origin.

Proof:
{{begin-eqn}}
{{eqn | l = y
      | r = a x^3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d y} {\d x}
      | r = 3 a x^2
      | c = Derivative of Power
}}
{{end-eqn}}
By definition, there is a point of inflection at a point where $x = \xi$ {{iff}} the first derivative has either a local maximum or a local minimum at $\xi$.
We note that $3 a x^2 \ge 0$ is the equation for the parabola.
From Square of Non-Zero Real Number is Strictly Positive, we note that:
:$3 a x^2 = 0 \iff x = 0$
and:
:$\forall x \in \R: x \ne 0 \implies x^2 > 0$
So:
:if $a > 0$ then $3 a x^2$ has a local minimum at $x = 0$
:if $a < 0$ then $3 a x^2$ has a local maximum at $x = 0$.
Hence the result.
{{qed}}
Theorem: Euler Theorem on Curvature of Surface

Let $S$ be a surface in space.
The principal directions of the '''curvature''' of $S$ are perpendicular to each other.

Proof:
{{ProofWanted}}
{{Namedfor|Leonhard Paul Euler|cat = Euler}}
Theorem: Volume of Cylinder/Height and Base Area

Let $\CC$ be a cylinder such that:
:the bases of $\CC$ have area $A$
:the height of $\CC$ is $h$.
The volume $\VV$ of $\CC$ is given by the formula:
:$\VV = A h$

Proof:
:600px
Consider a cylinder $C$ whose base is of area $A$ and whose height is $h$.
Consider a cuboid $K$ whose height is $h$ and whose base is also $A$.
Let $C$ be positioned with its base in the same plane as the base of $K$.
By Cavalieri's Principle $C$ and $K$ have the same volume.
From Volume of Cuboid, $K$ has volume given by:
:$V_K = A h$
Hence the result.
{{qed}}
Theorem: Finite Connected Simple Graph with Size One Less than Order has no Circuits

Let $G$ be a finite connected simple graph of order $n$.
Let the size of $T$ be $n - 1$.
Then $G$ has no circuits.

Proof:
Let $G$ be a connected simple graph of order $n$ with $n - 1$ edges.
By definition:
:the order of a finite tree is how many nodes it has
:the size of a finite tree is how many edges it has.
{{AimForCont}} $G$ contains at least one circuit.
From Condition for Edge to be Bridge:
:there exists at least one edge $e$ in $G$ which is not a bridge.
Let $G' := G \setminus \set e$ be $G$ after the edge deletion of $e$.
As $e$ is not a bridge, $G'$ is a graph which is:
:connected
:has $n$ vertices
:$n - 2$ edges.
Let us try and construct this connected simple graph $G'$ with $n$ vertices and $n - 2$ edges.
We start with the edgeless graph $N_n$, and add edges till $G'$ is connected.
We pick any two vertices of $N_n$, label them $u_1$ and $u_2$ for convenience, and use one edge to connect them, labelling that edge $e_1$.
We pick any other vertex, label it $u_3$, and use one edge to connect it to either $u_1$ or $u_2$, labelling that edge $e_2$.
We pick any other vertex, label it $u_4$, and use one edge to connect it to either $u_1, u_2$ or $u_3$, labelling that edge $e_3$.
We continue in this way, until we pick a vertex, label it $u_{n - 1}$, and use one edge to connect it to either $u_1, u_2, \ldots, u_{n - 2}$, labelling that edge $e_{n - 2}$.
That was the last of our edges, and the last vertex still has not been connected.
Therefore a simple graph with $n$ vertices and $n - 2$ edges cannot be connected.
That is, $G'$ cannot exist.
This contradicts our supposition that $G$ contains at least one circuit.
Hence, by Proof by Contradiction, $G$ has no circuits.
{{qed}}
Category:Finite Graphs
Category:Connectedness (Graph Theory)
Category:Simple Graphs
Category:Circuits (Graph Theory)
tuzrpx1twqr8t9v52qpgfu3np5a3utj
Theorem: Characteristics of Finite Tree/Condition 1

:$T$ is a finite tree of order $n$ {{iff}} $T$ has $n - 1$ edges and has no circuits.

Proof:
First we recall that by definition of a tree, $T$ is:
:a simple connected graph
:with no circuits.
Theorem: Characteristics of Finite Tree/Condition 2

:$T$ is a finite tree of order $n$ {{iff}} $T$ has $n - 1$ edges and is connected.

Proof:
This is an instance of:
:Finite Connected Simple Graph is Tree iff Size is One Less than Order.
{{qed}}
Category:Characteristics of Finite Tree
bgl3f96rpnzfu3eih511kjz9pomd9bv
Theorem: Characteristics of Finite Tree/Condition 3

:$T$ is a finite tree {{iff}} two arbitrary vertices of $T$ are connected by exactly one path.

Proof:
This is an instance of:
:Path in Tree is Unique.
{{qed}}
Category:Characteristics of Finite Tree
t8zmbtb1155wc42mbeamf4bjkq0pgle
Theorem: Characteristics of Finite Tree/Condition 4

:$T$ is a finite tree {{iff}} $T$ has no circuits, but adding one edge creates a cycle.

Proof:
=== Sufficient Condition ===
Let $T = \struct {V, E}$ be a finite tree.
Then by definition $T$ is connected and has no circuits.
Let $u, v \in V$ be any two vertices of $T$.
Let $P = \tuple {u, u_1, u_2, \ldots, u_{n - 1}, v}$ be a path from $u$ to $v$.
Let a new edges $\set {u, v}$ be added.
Then $\tuple {u, u_1, u_2, \ldots, u_{n - 1}, v, u}$ is now a cycle, which is by definition also a circuit, in $T$.
Note that this applies even when $P = \tuple {u, v}$: $\tuple {u, v, u}$ is still a cycle in $T$, but now $T$ is a multigraph.
{{qed|lemma}}
Theorem: Simple Graph with no Circuits and Size One Less than Order is Connected

Let $G$ be a simple graph of order $n$ such that $G$ has no circuits.
Let the size of $G$ be $n - 1$.
Then $T$ is connected.

Proof:
{{AimForCont}} $G$ is not connected.
Then $G$ has at least $2$ components.
By the Pigeonhole Principle, at least one of those components has at least as many edges as vertices.
Hence by Finite Connected Simple Graph is Tree iff Size is One Less than Order, that component is not a tree.
Hence as that component is itself connected but not a tree by definition it has at least one circuit.
But {{hypothesis}} $G$ has no circuits.
The result follows by Proof by Contradiction.
{{qed}}
Category:Simple Graphs
Category:Connectedness (Graph Theory)
51lbac1r7nprcj9r2x9btyxy44o0xgf
Theorem: Propositional Calculus is Decidable

The '''propositional calculus''' is a decidable system.

Proof:
The Method of Truth Tables is an effective procedure for determining the validity of propositional formulas with respect to boolean interpretations.
{{finish}}
Theorem: Join Semilattice has Smallest Element iff has Identity

Let $\struct{S, \vee, \preceq}$ be a join semilattice.
Let $s \in S$.
Then:
:$s$ is the smallest element of $S$
{{iff}}
:$s$ is the identity in $\struct{S, \vee}$.

Proof:
By definition of smallest element:
:$s$ is the smallest element of $S$ 
{{iff}}:
:$\forall t \in S : s \preceq t$
We have:
{{begin-eqn}}
{{eqn | q = \forall t \in S
      | l = s \preceq t
      | o = \iff
      | r = t = s \vee t
      | c = Successor is Supremum
}}
{{eqn | o = \iff
      | r = t = t \vee s
      | c = {{Semilattice-axiom|2}}
}}
{{end-eqn}}
By definition of identity:
:$s$  is the identity in $\struct{S, \vee}$
{{iff}}
:$\forall t \in S :s \vee t = t = t \vee s$
The result follows.
{{qed}}
Category:Join Semilattices
k4366mssixeu3px21ct60v19inoyv38
Theorem: Join Semilattice is Dual to Meet Semilattice

Let $\struct {S, \preceq}$ be an ordered set.
The following are dual statements:
:$\struct{S, \vee, \preceq}$ is a join semilattice, where $\vee$ is the  join for all $a,b \in S$
:$\struct{S, \wedge, \preceq}$ is a meet semilattice, where $\wedge$ is the  meet for all $a,b \in S$

Proof:
By definition of join semilattice:
:$\struct{S, \vee, \preceq}$ is a join semilattice 
{{iff}}:
:$\forall a,b \in S : a \vee b \in S$, where $a \vee b$ is the  join  of $a$ and $b$
The dual of this statement is:
:$\forall a,b \in S : a \wedge b \in S$, where $a \wedge b$ is the  meet  of $a$ and $b$
by Dual Pairs (Order Theory).
By definition of meet semilattice:
:$\struct{S, \wedge, \preceq}$ is a meet semilattice 
{{qed}}
Theorem: Meet Semilattice has Greatest Element iff has Identity

Let $\struct{S, \wedge, \preceq}$ be a meet semilattice.
Let $s \in S$.
Then:
:$s$ is the greatest element of $S$
{{iff}}
:$s$ is the identity in $\struct{S, \wedge}$.

Proof:
This is the dual statement of Join Semilattice has Smallest Element iff has Identity by Dual Pairs (Order Theory).
The result follows from the Duality Principle.
{{qed}}
Category:Meet Semilattices
a4bqwqvlw5fgems8dbfq6rz4d2iwn3f
Theorem: Joins of Homeomorphic Spaces are Homeomorphic

Let $X, Y, X', Y'$ be topological spaces.
Suppose that:
:$X \sim X'$
:$Y \sim Y'$
where $\sim$ denotes homeomorphic spaces.
Then:
:$(X \ast Y) \sim (X' \ast Y')$
where $\ast$ denotes the join.

Proof:
By definition of homeomorphic, let:
:$\phi_X : X \to X'$
:$\phi_Y : Y \to Y'$
be homeomorphisms.
Additionally, let:
:$I_{\closedint 0 1} : \closedint 0 1 \to \closedint 0 1$
be the identity mapping.
By Identity Mapping is Homeomorphism, we have that $I_{\closedint 0 1}$ is a homeomorphism.
Let $\phi : X \times Y \times \closedint 0 1 \to X' \times Y' \times \closedint 0 1$ be defined as:
:$\map \phi {x, y, t} = \tuple {\map {\phi_X} x, \map {\phi_Y} y, \map {I_{\closedint 0 1} } t}$
By Products of Homeomorphic Spaces are Homeomorphic:
:$\phi$ is a homeomorphism.
Consider the equivalence relations:
:$\RR \subseteq \paren {X \times Y \times \closedint 0 1}^2$
:$\RR' \subseteq \paren {X' \times Y' \times \closedint 0 1}^2$
each induced by the mapping:
:$\map R {x, y, t} = \begin{cases} x & : t = 0 \\ \tuple {x, y, t} & : t \in \openint 0 1 \\ y & : t = 1 \end{cases}$
which is taken over the corresponding domain and codomain as in the definition of join.
By Quotients of Homeomorphic Spaces are Homeomorphic, the result will follow if we can show:
:$\forall a, b \in X \times Y \times \closedint 0 1: \map \RR {a, b} \iff \map {\RR'} {\map \phi a, \map \phi b}$
By definition of the induced relation, it suffices to show that:
:$\map R {x, y, t} = \map R {x', y', t'} \iff \map R {\map {\phi_X} x, \map {\phi_Y} y, t} = \map R {\map {\phi_X} {x'}, \map {\phi_Y} {y'}, t'}$
We have:
{{begin-eqn}}
{{eqn | l = \map R {x, y, t} = \map R {x', y', t'}
      | o = \iff
      | r = \paren {x = x' \land t = t' = 0}
}}
{{eqn | o =
      | ro = \lor
      | r = \paren {x = x' \land y = y' \land t = t' \in \openint 0 1}
}}
{{eqn | o =
      | ro = \lor
      | r = \paren {y = y' \land t = t' = 1}
      | c = Definition of $R$
}}
{{eqn | o = \iff
      | r = \paren {\map {\phi_X} x = \map {\phi_X} {x'} \land t = t' = 0}
}}
{{eqn | o =
      | ro = \lor
      | r = \paren {\map {\phi_X} x = \map {\phi_X} {x'} \land \map {\phi_Y} y = \map {\phi_Y} {y'} \land t = t' \in \openint 0 1}
}}
{{eqn | o =
      | ro = \lor
      | r = \paren {\map {\phi_Y} y = \map {\phi_Y} {y'} \land t = t' = 1}
      | c = as $\phi_X$ and $\phi_Y$ are bijections
}}
{{eqn | o = \iff
      | r = \map R {\map {\phi_X} x, \map {\phi_Y} y, t} = \map R {\map {\phi_X} {x'}, \map {\phi_Y} {y'}, t'}
}}
{{end-eqn}}
Therefore, by the remarks above:
{{explain|Which "remarks above"? Best to refer back to earlier statements by label.}}
:$\paren {X \ast Y} \sim \paren {X' \ast Y'}$
{{qed}}
Category:Homeomorphisms (Topological Spaces)
on2hqs10q5hrw2ydrd0umfjv51hpi3x
Theorem: Cones on Homeomorphic Spaces are Homeomorphic

Let $X, Y$ be topological spaces.
Suppose:
:$X \sim Y$
denoting that $X$ is homeomorphic to $Y$.
Then:
:$C X \sim C Y$
where $C X$ denotes the cone on $X$.

Proof:
Let $T$ be the trivial topological space used in the definition of cone.
We have:
:$T \sim T$ by Homeomorphism Relation is Equivalence
:$X \sim Y$ {{hypothesis}}
Then, by Joins of Homeomorphic Spaces are Homeomorphic:
:$C X \sim C Y$
{{qed}}
Category:Homeomorphisms (Topological Spaces)
2nvzextr68ikw555lumqm0f4yb0dku7
Theorem: Compact Convex Sets with Nonempty Interior are Homeomorphic

Let $n \in \N_{> 0}$.
Let $T, T' \subseteq \R^n$ be compact convex subsets of real Euclidean $n$-space.
Then, $T$ is homeomorphic to $T'$.

Proof:
By Boundary of Compact Convex Set with Nonempty Interior is Homeomorphic to Sphere:
:$\partial T \sim \Bbb S^{n - 1}$
:$\partial T' \sim \Bbb S^{n - 1}$
Thus, by Homeomorphism Relation is Equivalence:
:$\partial T \sim \partial T'$
Hence, by Cones on Homeomorphic Spaces are Homeomorphic:
:$C \partial T \sim C \partial T'$
But, by Compact Convex Set with Nonempty Interior is Homeomorphic to Cone on Boundary:
:$T \sim C \partial T$
:$T' \sim C \partial T'$
Therefore, by Homeomorphism Relation is Equivalence:
:$T \sim T'$
{{qed}}
Category:Homeomorphisms (Topological Spaces)
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Theorem: First Order ODE/y' = x

The first order ODE:
:$\dfrac {\d y} {\d x} = x$
has the general solution:
:$y = \dfrac {x^2} 2 + C$

Proof:
{{begin-eqn}}
{{eqn | l = \dfrac {\d y} {\d x}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \rd y
      | r = \int x \rd x
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac {x^2} 2 + C
      | c = Primitive of Constant, Primitive of Power
}}
{{end-eqn}}
{{qed}}
Theorem: Fixed-Point Property is Topological

Let $T \sim T'$ be homeomorphic topological spaces.
Suppose every continuous $f : T \to T$ has a fixed point.
Then, every continuous $g : T' \to T'$ also has a fixed point.
{{improve|It's customary on {{ProofWiki}} to utilise the terminology of the title in the page, otherwise a casual reader would say "What has being continuous got to do with being topological?"}}

Proof:
Let $\phi : T \to T'$ be a homeomorphism.
Let $g : T' \to T'$ be an arbitrary continuous mapping.
Consider $f_g : T \to T$ defined as:
:$\map {f_g} x = \map {\phi^{-1}} {\map g {\map \phi x}}$
By definition of homeomorphism and Composite of Continuous Mappings is Continuous:
:$f_g$ is continuous
Therefore, {{hypothesis}}, there is a fixed point $x_0 \in T$ of $f_g$.
We then have:
{{begin-eqn}}
{{eqn | l = \map {f_g} {x_0}
      | r = x_0
      | c = {{Defof|Fixed Point}}
}}
{{eqn | ll = \leadsto
      | l = \map \phi {\map {f_g} {x_0} }
      | r = \map \phi {x_0}
}}
{{eqn | ll = \leadsto
      | l = \map \phi {\map {\phi^{-1} } {\map g {\map \phi {x_0} } } }
      | r = \map \phi {x_0}
      | c = Definition of $f_g$
}}
{{eqn | ll = \leadsto
      | l = \map g {\map \phi {x_0} }
      | r = \map \phi {x_0}
      | c = {{Defof|Inverse Mapping}}
}}
{{end-eqn}}
Therefore, $\map \phi {x_0}$ is a fixed point of $g$.
As $g$ was arbitrary, the result holds.
{{qed}}
Category:Fixed Point Theorems
Category:Topological Properties
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