avewright/proofwiki-math huggingface dataset:
Theorem: Sequentially Compact Metric Space is Compact

A sequentially compact metric space is compact.

Proof:
Let $\struct {A, d}$ be a sequentially compact metric space.
From Sequentially Compact Metric Space is Lindelöf, every open cover of $A$ has a countable subcover.
Let $C$ be an open cover $A$.
Extract from it a countable subcover $\set {U_1, U_2, \ldots}$.
{{AimForCont}} there exists no finite subcover of $C$.
Then, for all $n \in \N_{\ge 1}$, the set $\set {U_1, \ldots, U_n}$ does ''not'' cover $A$.
Hence it is possible to choose $x_n \in A$ such that:
:$x_n \notin U_1 \cup \cdots \cup U_n$.
Thus we construct an infinite sequence $\sequence {x_n}_{n \mathop \ge 1}$ of points of $A$.
By assumption $A$ is sequentially compact.
Thus $\sequence {x_n}_{n \mathop \ge 1}$ has a subsequence which converges to some $x \in A$.
But because the $U_i: i \ge 1$ forms a cover for $A$, there exists some $U_m$ such that $x \in U_m$.
Hence by Limit of Sequence is Accumulation Point, there is an infinite number of terms in the sequence $\sequence {x_i}$ which are contained in $U_m$.
But from the method of construction of $\sequence {x_i}$, each $U_n$ can contain only points $x_i$ with $i < n$.
That is, each $U_n$ can only contain a finite number of the terms of $\sequence {x_i}$.
This is a contradiction.
Thus the supposition that there exists no finite subcover of $C$ was false.
Hence the result.
{{qed}}
Theorem: Sequentially Compact Metric Space is Compact

A sequentially compact metric space is compact.

Proof:
Let $M = \struct {A, d}$ be a sequentially compact metric space.
Let $\UU$ be any open cover of $M$.
By Lebesgue's Number Lemma, there exists a Lebesgue number for $\UU$.
Let $\set {x_1, x_2, \ldots, x_n}$ be a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number.
This exists by Sequentially Compact Metric Space is Totally Bounded.
Let $\map {B_\epsilon} {x_i}$ be the open $\epsilon$-ball of $x_i$.
By definition of Lebesgue number, $\map {B_\epsilon} {x_i}$ is contained in some $U_i \in \UU$.
Since:
:$\ds M \subseteq \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i} \subseteq \bigcup_{i \mathop = 1}^n U_i$
we have a finite subcover $\set {U_1, U_2, \ldots, U_n}$ of $\UU$ for $M$.
Hence the result.
{{qed}}
Theorem: Sequentially Compact Metric Space is Compact

A sequentially compact metric space is compact.

Proof:
Follows directly from:
:Sequentially Compact Space is Countably Compact
:Countably Compact Metric Space is Compact
{{qed}}
Theorem: Compact Subspace of Hausdorff Space is Closed

Let $H = \struct {A, \tau}$ be a $T_2$ (Hausdorff) space.
Let $C$ be a compact subspace of $H$.
Then $C$ is closed in $H$.

Proof:
From Subspace of Hausdorff Space is Hausdorff, a subspace of a $T_2$ (Hausdorff) space is itself $T_2$ (Hausdorff).
Let $a \in A \setminus C$.
We are going to prove that there exists an open set $U_a$ such that $a \in U_a \subseteq A \setminus C$.
For any single point $x \in C$, the Hausdorff condition ensures the existence of disjoint open set $\map U x$ and $\map V x$ containing $a$ and $x$ respectively.
Suppose there were only a finite number of points $x_1, x_2, \ldots, x_r$ in $C$.
Then we could take $\ds U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$ and get $a \in U_a \subseteq A \setminus C$.
Now suppose $C$ is not finite.
The set $\set {\map V x: x \in C}$ is an open cover for $C$.
As $C$ is compact, it has a finite subcover, say $\set {\map V {x_1}, \map V {x_2}, \dotsc, \map V {x_r} }$.
Let $\ds U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$.
Then $U_a$ is open because it is a finite intersection of open sets.
Also, $a \in U_a$ because $a \in \map U {x_i}$ for each $i = 1, 2, \ldots, r$.
Finally, if $b \in U_a$ then for any $i = 1, 2, \ldots, r$ we have $b \in \map U {x_i}$.
Because $\ds C \subseteq \bigcup_{i \mathop = 1}^r \map V {x_i}$:
:$b \notin \map V {x_i}$, so $b \notin C$
Thus:
:$U_a \subseteq A \setminus C$.
Then:
:$\ds A \setminus C = \bigcup_{a \mathop \in A \mathop \setminus C} U_a$
So $A \setminus C$ is open.
It follows that $C$ is closed.
{{qed}}
Theorem: Compact Subspace of Hausdorff Space is Closed

Let $H = \struct {A, \tau}$ be a $T_2$ (Hausdorff) space.
Let $C$ be a compact subspace of $H$.
Then $C$ is closed in $H$.

Proof:
For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$.
Consider an arbitrary point $x \in C^{\complement}$.
Define the set:
:$\ds \OO = \set {V \in \tau: \exists U \in \tau: x \in U \subseteq V^\complement}$
By Empty Intersection iff Subset of Complement:
:$U \subseteq V^\complement \iff U \cap V = \O$
Hence, by the definition of a Hausdorff space, it follows that $\OO$ is an open cover for $C$.
By the definition of a compact subspace, there exists a finite subcover $\FF$ of $\OO$ for $C$.
By the Principle of Finite Choice, there exists an $\FF$-indexed family $\family {U_V}_{V \mathop \in \FF}$ of elements of $\tau$ such that:
:$\forall V \in \FF: x \in U_V \subseteq V^\complement$
Define:
:$\ds U = \bigcap_{V \mathop \in \FF} U_V$
By General Intersection Property of Topological Space:
:$U \in \tau$
Clearly, $x \in U$.
We have that:
{{begin-eqn}}
{{eqn | l = U
      | o = \subseteq
      | r = \bigcap_{V \mathop \in \FF} \paren {V^\complement}
      | c = Set Intersection Preserves Subsets
}}
{{eqn | r = \paren {\bigcup \FF}^\complement
      | c = De Morgan's laws
}}
{{eqn | o = \subseteq
      | r = C^\complement
      | c = {{Defof|Cover of Set}} and Relative Complement inverts Subsets
}}
{{end-eqn}}
From Subset Relation is Transitive, we have that $U \subseteq C^\complement$.
Hence $C^\complement$ is a neighborhood of $x$.
From Set is Open iff Neighborhood of all its Points:
:$C^\complement \in \tau$
That is, $C$ is closed in $H$.
{{qed}}
Theorem: Continuous Image of Compact Space is Compact

Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant.

Proof:
By {{Corollary|Continuous Image of Compact Space is Compact|2}}, $f \sqbrk S$ is bounded.
By Supremum of Bounded Above Set of Reals is in Closure:
:$\map \sup {f \sqbrk S} \in \map \cl {f \sqbrk S}$
and by Infimum of Bounded Below Set of Reals is in Closure:
:$\map \inf {f \sqbrk S} \in \map \cl {f \sqbrk S}$
From Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact in $\R$.
From Non-Closed Set of Real Numbers is not Compact, it follows from the Rule of Transposition that $f \sqbrk S$ is closed in $\R$.
From Closed Set equals its Closure:
:$f \sqbrk S = \map \cl {f \sqbrk S}$
Hence the result that:
:$\map \sup {f \sqbrk S} \in f \sqbrk S$
and:
:$\map \inf {f \sqbrk S} \in f \sqbrk S$
{{qed}}
Theorem: Continuous Image of Compact Space is Compact

Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant.

Proof:
By Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact.
From Compact Metric Space is Complete and Compact Metric Space is Totally Bounded, $f \sqbrk S$ is complete and totally bounded.
A Totally Bounded Metric Space is Bounded.
Hence both the supremum and the infimum of $f \sqbrk S$ exist in $\R$.
Because $f \sqbrk S$ is  complete:
:$\sup f \sqbrk S \in f \sqbrk S$
and:
:$\inf f \sqbrk S \in f \sqbrk S$
{{explain|why?}}
{{qed}}
Theorem: Continuous Image of Compact Space is Compact

Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant.

Proof:
From Restriction of Mapping to Image is Surjection, the restriction of $f$ to $\Img f$ is a surjection.
The result follows from Compactness is Preserved under Continuous Surjection.
{{qed}}
Theorem: Continuous Image of Compact Space is Compact

Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant.

Proof:
Suppose $\UU$ is an open cover of $f \sqbrk {T_1}$ by sets open in $T_2$.
Because $f$ is continuous, it follows that $f^{-1} \sqbrk U$ is open in $T_1$ for all $U \in \UU$.
The set $\set {f^{-1} \sqbrk U: U \in \UU}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \in \UU$.
Because $T_1$ is compact, it has a finite subcover $\set {f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2}, \ldots, f^{-1} \sqbrk {U_r} }$.
It follows that $\set {U_1, U_2, \ldots, U_r}$ is a finite subcover of $f \sqbrk {T_1}$.
{{qed}}
Theorem: Closed Subspace of Compact Space is Compact

Let $T = \struct {S, \tau}$ be a compact topological space.
Let $C = \struct {H, \tau_H}$ be a subspace of $T$.
Let $C$ be closed in $T$.
Then $\struct {H, \tau}$ is compact.
That is, the property of being compact is weakly hereditary.

Proof:
Let $T$ be a compact space.
Let $C$ be a closed subspace of $T$.
Let $\UU$ be an open cover of $C$.
We have that $C$ is closed in $T$.
It follows by definition of closed that $T \setminus C$ is open in $T$.
So if we add $T \setminus C$ to $\UU$, we see that $\UU \cup \set {T \setminus C}$ is also an open cover of $T$.
As $T$ is compact, there exists a finite subcover of $\UU \cup \set {T \setminus C}$, say $\VV = \set {U_1, U_2, \ldots, U_r}$.
This covers $C$ by the fact that it covers $T$.
Suppose $T \setminus C$ is an element of $\VV$.
Then $T \setminus C$ may be removed from $\VV$, and the rest of $\VV$ still covers $C$.
Thus we have a finite subcover of $\UU$ which covers $C$.
Hence $C$ is compact.
{{qed}}
Theorem: Closure of Real Interval is Closed Real Interval

Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then $I^-$ is the closed real interval $\closedint a b$.

Proof:
There are four cases to cover:
:$(1): \quad$ Let $I = \openint a b$.
From Closure of Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
{{qed|lemma}}
:$(2): \quad$ Let $I = \hointr a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
{{qed|lemma}}
:$(3): \quad$ Let $I = \hointl a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
{{qed|lemma}}
:$(4): \quad$ Let $I = \closedint a b$.
From Closed Real Interval is Closed in Real Number Line:
:$I$ is closed in $\R$.
From Set is Closed iff Equals Topological Closure:
:$I^- = \closedint a b$
{{qed|lemma}}
Thus all cases are covered.
The result follows by Proof by Cases.
{{qed}}
Theorem: Closure of Real Interval is Closed Real Interval

Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then $I^-$ is the closed real interval $\closedint a b$.

Proof:
Let $I$ be one of the intervals as specified in the exposition.
Note that:
:$(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ {{iff}} every open set in $\R$ containing $x$ contains a point in $I$.
:$(2): \quad$ From Union of Open Sets of Metric Space is Open, every open set in $\R$ is a union of open intervals.
Thus we also have that $x \in I^-$ {{iff}} every open interval containing $x$ also contains a point in $I$.
This equivalence will be made use of throughout.
=== Lemma 1 ===
{{:Closure of Real Interval is Closed Real Interval/Proof 2/Lemma 1}}{{qed|lemma}}
=== Lemma 2 ===
{{:Closure of Real Interval is Closed Real Interval/Proof 2/Lemma 2}}{{qed|lemma}}
By the two lemmas proven above:
:$\closedint a b = I^-$
{{qed}}
Theorem: Topological Product of Compact Spaces

Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Then $T_1 \times T_2$ is compact {{iff}} both $T_1$ and $T_2$ are compact.

Proof:
=== Necessary Condition ===
Let $T_1 \times T_2$ be compact.
From Projection from Product Topology is Continuous, both $T_1$ and $T_2$ are the continuous images of $T_1 \times T_2$ under the projections $\pr_1$ and $\pr_2$ respectively.
So both $T_1$ and $T_2$ are compact by Continuous Image of Compact Space is Compact.
{{qed|lemma}}
Theorem: Heine-Cantor Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ be compact.
Let $f: A_1 \to A_2$ be a continuous mapping.
Then $f$ is uniformly continuous.

Proof:
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
For all $x \in A_1$, define:
:$\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$
Define:
:$\CC = \set {\map {B_{\delta} } x: x \in A_1, \, \delta \in \map \Delta x}$
where $\map {B_{\delta} } x$ denotes the open $\delta$-ball of $x$ in $M_1$.
From the definition of continuity, it follows that $\CC$ is a cover for $A_1$.
From Open Ball of Metric Space is Open Set, it therefore follows that $\CC$ is an open cover for $A_1$.
By the definition of a compact metric space, there exists a finite subcover $\set {\map {B_{\delta_1} } {x_1}, \map {B_{\delta_2} } {x_2}, \ldots, \map {B_{\delta_n} } {x_n} }$ of $\CC$ for $A_1$.
Define:
:$\delta = \min \set {\delta_1, \delta_2, \ldots, \delta_n}$
Let $x, y \in A_1$ satisfy $\map {d_1} {x, y} < \delta$.
By the definition of a cover, there exists a $k \in \set{1, 2, \ldots, n}$ such that $\map {d_1} {x, x_k} < \delta_k$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {y, x_k}
      | o = \le
      | r = \map {d_1} {y, x} + \map {d_1} {x, x_k}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | o = <
      | r = \delta + \delta_k
      | c = {{Metric-space-axiom|3}}
}}
{{eqn | o = \le
      | r = 2 \delta_k
}}
{{end-eqn}}
By the definition of $\map \Delta {x_k}$, it follows that:
:$\map {d_2} {\map f x, \map f {x_k} } < \dfrac \epsilon 2$
:$\map {d_2} {\map f y, \map f {x_k} } < \dfrac \epsilon 2$
Hence:
{{begin-eqn}}
{{eqn | l = \map {d_2} {\map f x, \map f y}
      | o = \le
      | r = \map {d_2} {\map f x, \map f {x_k} } + \map {d_2} {\map f {x_k}, \map f y}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2
      | c = {{Metric-space-axiom|3}}
}}
{{eqn|r = \epsilon
}}
{{end-eqn}}
The result follows from the definition of uniform continuity.
{{qed}}
Theorem: Heine-Cantor Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ be compact.
Let $f: A_1 \to A_2$ be a continuous mapping.
Then $f$ is uniformly continuous.

Proof:
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 Composite of Continuous Mappings is Continuous and Continuous Mapping to Product Space, it follows that $F$ is continuous.
By Distance Function of Metric Space is Continuous and Composite of Continuous Mappings is Continuous, 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-axiom|1}} and {{Metric-space-axiom|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}}
Theorem: Continuous Function on Closed Real Interval is Uniformly Continuous

Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $\closedint a b$.

Proof:
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}}
Theorem: Continuous Function on Closed Real Interval is Uniformly Continuous

Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $\closedint a b$.

Proof:
{{AimForCont}} $f$ is not uniformly continuous. 
Then there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ there are points $x ,y \in \closedint a b$ for which:
:$\size {x - y} < \delta$
and:
:$\size {\map f x - \map f y} \ge \epsilon$
We will now choose for each $k \in \N$ numbers $x_k, y_k \in \closedint a b$ such that:
:$\size {x_k - y_k} < \dfrac 1 k$
and:
:$\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$
We have that the sequence $\sequence {x_k}$ is bounded in $\closedint a b$.
So, by the Bolzano-Weierstrass Theorem there exists a convergent subsequence $\sequence {x_{k_j} }$ whose limit lies in $\closedint a b$.
Let this limit be denoted $x_0$.
Now we have:
:$\size {x_0 - y_{k_j} } \le \size {x_0 - x_{k_j} } + \size {x_{k_j} - y_{k_j} } \le \size {x_0 - x_{k_j} } + \dfrac 1 {k_j}$
Therefore the sequence $\sequence {y_{k_j} }$ also converges to $x_0$.
{{explain|Make it clear that the above is a definition of sequential continuity in order for the next statement to follow<br/>In fact, we should eb able to make use of the fact that a continuous function is already proven to be sequentially continuous, and if it does not so exist, add the above as a proof.}}
Because $f$ is sequentially continuous at $x_0 \in \closedint a b$, we have:
:$\ds \lim_{j \mathop \to \infty} \map f {x_{k_j} } = \map f {x_0} = \lim_{j \mathop \to \infty} \map f {y_{k_j} }$
This is however, a contradiction, as $\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$ for all $k$ and thus all $k_j$. 
Therefore $f$ is uniformly continuous.
{{qed}}
Theorem: Sine of Complement equals Cosine

:$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \map \sin {\frac \pi 2 - \theta}
      | 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}}
Theorem: Sine of Complement equals Cosine

:$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$

Proof:
{{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}}
Theorem: Sine of Complement equals Cosine

:$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$

Proof:
{{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}}
Theorem: Sine of Complement equals Cosine

:$\map \sin {\dfrac \pi 2 - \theta} = \cos \theta$

Proof:
:300px
Let $\angle xOP$ and $\angle QOy$ be complementary.
Then:
:$\angle xOP = \angle QOy$
Hence:
:the projection of $OP$ on the $x$-axis
equals:
:the projection of $OQ$ on the $y$-axis.
Hence the result.
{{qed}}
Theorem: Cosine of Complement equals Sine

:$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$

Proof:
{{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}}
Theorem: Cosine of Complement equals Sine

:$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {\frac \pi 2 - \theta}
      | r = \map \cos {\theta - \frac \pi 2}
      | c = Cosine Function is Even
}}
{{eqn | r = \map \sin {\theta - \frac \pi 2 + \frac \pi 2}
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \sin \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Complement equals Sine

:$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$

Proof:
{{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 = Euler's Cosine Identity
}}
{{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 = Euler's Sine Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Complement equals Sine

:$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$

Proof:
:300px
Let $\angle xOP$ and $\angle QOy$ be complementary.
Then:
:$\angle xOP = \angle QOy$
Hence:
:the projection of $OP$ on the $x$-axis
equals:
:the projection of $OQ$ on the $y$-axis.
Hence the result.
{{qed}}
Theorem: Continuous Bijection from Compact to Hausdorff is Homeomorphism

Let $T_1$ be a compact space.
Let $T_2$ be a Hausdorff space.
Let $f: T_1 \to T_2$ be a continuous bijection.
Then $f$ is a homeomorphism.

Proof:
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 $g^{-1} \sqbrk V = \map f V$.
We are to show that if $V$ is closed in $T_1$, then $g^{-1} \sqbrk V$ 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 $f \sqbrk V$ is compact from Continuous Image of Compact Space is Compact.
Since $T_2$ is Hausdorff, $f \sqbrk V$ is closed by Compact Subspace of Hausdorff Space is Closed.
But $f \sqbrk V = g^{-1} \sqbrk V$, so $g^{-1} \sqbrk V$ is closed.
From Continuity Defined from Closed Sets, it follows that $g$ is continuous.
Thus by definition, $f$ is a homeomorphism.
{{qed}}