avewright/proofwiki-math huggingface dataset:
Theorem: Smallest Non-Palindromic Number with Palindromic Square

$26$ is the smallest non-palindromic integer whose square is palindromic.

Proof:
Checking the squares of all non-palindromic integers in turn from $10$ upwards, until a palindromic integer is reached:
{{begin-eqn}}
{{eqn | l = 10^2
      | r = 100
}}
{{eqn | l = 12^2
      | r = 144
}}
{{eqn | l = 13^2
      | r = 169
}}
{{eqn | l = 14^2
      | r = 196
}}
{{eqn | l = 15^2
      | r = 225
}}
{{eqn | l = 16^2
      | r = 256
}}
{{eqn | l = 17^2
      | r = 289
}}
{{eqn | l = 18^2
      | r = 324
}}
{{eqn | l = 19^2
      | r = 361
}}
{{eqn | l = 20^2
      | r = 400
}}
{{eqn | l = 21^2
      | r = 441
}}
{{eqn | l = 23^2
      | r = 529
}}
{{eqn | l = 24^2
      | r = 576
}}
{{eqn | l = 25^2
      | r = 625
}}
{{eqn | l = 26^2
      | r = 676
}}
{{end-eqn}}
{{qed}}
Theorem: Numbers Partitioned into Six Hexagonal Numbers

The integers $11$ and $26$ cannot be represented by the sum of less than $6$ hexagonal numbers.

Proof:
Recall the sequence of hexagonal numbers:
{{:Hexagonal Number/Sequence}}
Hence:
{{begin-eqn}}
{{eqn | l = 11
      | r = 6 + 1 + 1 + 1 + 1 + 1
      | c = 
}}
{{eqn | l = 26
      | r = 6 + 6 + 6 + 6 + 1 + 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Integer as Sum of 27 Primes

Every positive integer greater than $1$ can be expressed as the sum of no more than $27$ primes.

Proof:
{{questionable|A far cry from the Goldbach Conjecture. Puzzling statement.}}
Theorem: Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.
Let $P = \struct {T, \precsim}$ be a continuous lattice subframe of $L$.
Then $P$ is algebraic lattice.

Proof:
By definition of algebraic ordered set:
:$L$ is up-complete.
By Up-Complete Lower Bounded Join Semilattice is Complete:
:$L$ is a complete lattice.
By definition:
:$P$ is closure system of $L$.
By Image of Operator Generated by Closure System is Set of Closure System
:$\map {\operatorname {operator} } P \sqbrk S = T$
By Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema
:$\map {\operatorname {operator} } P$ preserves directed suprema.
Thus by Image of Directed Suprema Preserving Closure Operator is Algebraic Lattice:
:$P$ is an algebraic lattice.
{{qed}}
Theorem: 27 is Smallest Number whose Period of Reciprocal is 3

$27$ is the smallest positive integer the decimal expansion of whose reciprocal has a period of $3$:
:$\dfrac 1 {27} = 0 \cdotp \dot 03 \dot 7$

Proof:
From Reciprocal of $27$:
{{:Reciprocal of 27}}
It can be determined by inspection of all smaller integers that this is indeed the smallest to have a period of $3$.
{{qed}}
Theorem: Long Period Prime/Examples/7

$7$ is the smallest long period prime:
:$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$

Proof:
From Reciprocal of $7$:
{{:Reciprocal of 7}}
The reciprocals of $1$, $2$, $4$ and $5$ do not recur:
{{begin-eqn}}
{{eqn | l = \frac 1 1
      | r = 1
      | c = 
}}
{{eqn | l = \frac 1 2
      | r = 0 \cdotp 5
      | c = 
}}
{{eqn | l = \frac 1 4
      | r = 0 \cdotp 25
      | c = 
}}
{{eqn | l = \frac 1 5
      | r = 0 \cdotp 2
      | c = 
}}
{{end-eqn}}
while those of $3$ and $6$ do recur, but with the non-maximum period of $1$:
{{begin-eqn}}
{{eqn | l = \frac 1 3
      | r = 0 \cdotp \dot 3
      | c = 
}}
{{eqn | l = \frac 1 6
      | r = 0 \cdotp 1 \dot 6
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sequence of Smallest Numbers whose Reciprocal has Period n

Let $\sequence {s_n}$ be the sequence defined as:
:$s_n$ is the smallest positive integer the decimal expansion of whose reciprocal has a period of $n$
for $n = 0, 1, 2, \ldots$
Then $\sequence {s_n}$ begins:
:$1, 3, 11, 27, 101, 41, 7, 239, 73, 81, 451, \ldots$
{{OEIS|A003060}}

Proof:
Demonstrated by inspection and calculation:
{{begin-eqn}}
{{eqn | l = \frac 1 1
      | r = 1 \cdotp 0
      | c = hence has a period of $0$
}}
{{eqn | l = \frac 1 3
      | r = 0 \cdotp \dot 3
      | c = Reciprocal of $3$
}}
{{eqn | l = \frac 1 {11}
      | r = 0 \cdotp \dot 0 \dot 9
      | c = Reciprocal of $11$
}}
{{eqn | l = \frac 1 {27}
      | r = 0 \cdotp \dot 03 \dot 7
      | c = Reciprocal of $27$
}}
{{eqn | l = \frac 1 {101}
      | r = 0 \cdotp \dot 009 \dot 9
      | c = Reciprocal of $101$
}}
{{eqn | l = \frac 1 {41}
      | r = 0 \cdotp \dot 0243 \dot 9
      | c = Reciprocal of $41$
}}
{{eqn | l = \frac 1 7
      | r = 0 \cdotp \dot 14285 \dot 7
      | c = Reciprocal of $7$
}}
{{eqn | l = \frac 1 {239}
      | r = 0 \cdotp \dot 00418 \, 4 \dot 1
      | c = Reciprocal of $239$
}}
{{eqn | l = \frac 1 {73}
      | r = 0 \cdotp \dot 01369 \, 86 \dot 3
      | c = Reciprocal of $73$
}}
{{eqn | l = \frac 1 {81}
      | r = 0 \cdotp \dot 01234 \, 567 \dot 9
      | c = Reciprocal of $81$
}}
{{eqn | l = \frac 1 {451}
      | r = 0 \cdotp \dot 00221 \, 7294 \dot 9
      | c = Reciprocal of $451$
}}
{{end-eqn}}
{{qed}}
Theorem: Numbers whose Cyclic Permutations of 3-Digit Multiples are Multiples

Let $n$ be a two-digit positive integer with the following property:
:Let $m$ be a $3$-digit multiple of $n$.
:Then any cyclic permutation of the digits of $m$ is also a multiple of $n$.
Then $n$ is either $27$ or $37$.

Proof:
Let $m$ be a multiple of $n$ with $3$ digits.
Then we have:
{{begin-eqn}}
{{eqn | l = n \times c
      | r = a_2 \times 10^2 + a_1 \times 10^1 + a_0
}}
{{end-eqn}}
Let us now cyclically permute the digits of $m$ by multiplying by $10$.
Then we have:
{{begin-eqn}}
{{eqn | l = 10 \times n \times c 
      | r = 10 \times \paren {a_2 \times 10^2 + a_1 \times 10^1 + a_0}
      | c = multiply original number by $10$
}}
{{eqn | r =  a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1
      | c = 
}}
{{eqn | r =  a_1 \times 10^2 + a_0 \times 10^1 + a_2 \times 10^0
      | c = $10^3$ and $10^0 \equiv 1 \pmod {n}$
}}
{{end-eqn}}
From the above, we see that:
:$n$ is a divisor of a cyclic permutation of $m$
{{iff}}:
:$n \divides \paren {10^3 - 1 }$
We now note that:
:$10^3 - 1 = 37 \times 27 = 37 \times 3^3$
Upon inspection, we see that the only $2$-digit factors are $27$ and $37$.
{{qed}}
Theorem: Sequence of Successive Longest Collatz Sequence Generators

The sequence of integers which generate a Collatz process which is longer than that of any smaller integers begins:
{{begin-eqn}}
{{eqn | l = 1
      | o = :
      | c = $0$ steps
}}
{{eqn | l = 2
      | o = :
      | c = $1$ step
}}
{{eqn | l = 3
      | o = :
      | c = $7$ steps
}}
{{eqn | l = 6
      | o = :
      | c = $8$ steps
}}
{{eqn | l = 7
      | o = :
      | c = $16$ steps
}}
{{eqn | l = 9
      | o = :
      | c = $19$ steps
}}
{{eqn | l = 18
      | o = :
      | c = $20$ steps
}}
{{eqn | l = 25
      | o = :
      | c = $23$ steps
}}
{{eqn | l = 27
      | o = :
      | c = $111$ steps
}}
{{end-eqn}}
{{OEIS|A006877}}

Proof:
Missing complementary orbits:
{{begin-eqn}}
{{eqn | l = 4
      | o = :
      | c = $2$ steps
}}
{{eqn | l = 5
      | o = :
      | c = $5$ steps
}}
{{eqn | l = 8
      | o = :
      | c = $3$ steps
}}
{{eqn | l = 10
      | o = :
      | c = $6$ steps
}}
{{end-eqn}}
{{Proofread}}
Theorem: Boundary of Compact Closed Set is Compact

Let $X$ be a topological space.
Let $K\subset X$ be a compact subspace of $X$.
Let $K$ be closed in $X$.
Then its boundary $\partial K$ is compact.

Proof:
By Boundary of Set is Closed, $\partial K$ is closed in $X$.
By Set is Closed iff it Contains its Boundary, $\partial K \subset K$.
By Closed Set in Topological Subspace, $\partial K$ is closed in $K$.
By Closed Subspace of Compact Space is Compact, $\partial K$ is compact.
{{qed}}
Theorem: Discrete Subgroup of Hausdorff Group is Closed

Let $G$ be a Hausdorff topological group.
Let $H$ be a discrete subgroup of $G$.
Then $H$ is closed in $G$.

Proof:
Let $g \in \overline H$ be in the closure of $H$.
We will show that $g \in H$.
{{AimForCont}} $g \notin H$.
Let $e$ be the identity of $G$.
Because $H$ is discrete, there exists an open set $U \subset G$ such that $U \cap H = \set e$.
Then $V = U \cap U^{-1}$ is an open neighborhood of $e$ in $G$.
By Right and Left Regular Representations in Topological Group are Homeomorphisms, $g V$ is a neighborhood of $g$ in $G$.
By definition of closure, $g V \cap H \ne \O$.
Let $h \in g V \cap H$.
Because $g \ne h$ and $G$ is Hausdorff, there exists a neighborhood $W$ of $g$ in $G$ with $h \notin W$.
Then $g V \cap W$ is a neighborhood of $g$.
Let $k \in g V \cap W \cap H$.
Then $k \ne h$.
Then
{{begin-eqn}}
{{eqn | l = k^{-1} h
      | o = \in
      | r = \paren {g V \cap H}^{-1} \cdot \paren {g V \cap H}
      | c = 
}}
{{eqn | r = \paren {g V}^{-1}g V \cap H
      | c = 
}}
{{eqn | r = V^{-1} V \cap H
      | c = 
}}
{{eqn | r = V \cap H
      | c = 
}}
{{eqn | o = \subset
      | r =  U \cap H
      | c = 
}}
{{end-eqn}}
Bu assumption, $k^{-1} h = e$, so $k = h$, which is a contradiction.
Thus, by Proof by Contradiction, $g \in H$.
Thus $\overline H = H$.
Thus $H$ is closed.
{{qed}}
Theorem: Boundary of Compact Set in Hausdorff Space is Compact

Let $T = \struct {S, \tau}$ be a Hausdorff topological space.
Let $K \subset S$ be a compact subspace of $T$.
Then its boundary $\partial K$ is compact.

Proof:
By Compact Subspace of Hausdorff Space is Closed, $K$ is closed in $T$.
By Boundary of Compact Closed Set is Compact, $\partial K$ is compact.
{{qed}}
Category:Set Boundaries
Category:Compact Topological Spaces
Category:Closed Sets
Category:Hausdorff Spaces
07a2hroov7qiiahtgttwg0xpq8tzk2t
Theorem: Group is Hausdorff iff Discrete Subgroups are Closed

A topological group is Hausdorff {{iff}} its discrete subgroups are closed.

Proof:
Follows directly from:
:Discrete Subgroup of Hausdorff Group is Closed
:Topological Group is Hausdorff iff Identity is Closed
{{qed}}
Theorem: Group is Hausdorff iff has Closed Discrete Subgroup

A topological group is Hausdorff {{iff}} it has a closed discrete subgroup.

Proof:
=== Necessary Condition ===
Follows directly from Topological Group is Hausdorff iff Identity is Closed.
{{explain|How?}}
{{qed|lemma}}
Theorem: Closure of Subgroup is Group

Let $G$ be a topological group.
Let $H\leq G$ be a subgroup.
Let $\overline H$ denote its closure.
Then $\overline H$ is a subgroup of $G$.

Proof:
We use the One-Step Subgroup Test.
Because $H \subset \overline H$, $\overline H$ is non-empty.
Let $a, b \in \overline H$.
Let $U$ be a neighborhood of $a b^{-1}$.
Let the mapping $f: G\times G \to G$ be defined as:
:$\map f {x, y} = x y^{-1}$
By definition of topological group, $f$ is continuous.
By definition of product space, there exist neighborhoods $A, B$ of $a, b$ respectively such that:
:$A \times B \subset f^{-1} \sqbrk U$
By assumption, there exist $x \in A \cap H$ and $y \in B \cap H$.
Then:
:$x y^{-1} \subset U \cap H$
Because $U$ was arbitrary:
:$a b^{-1} \in \overline H$
By One-Step Subgroup Test, $\overline H$ is a subgroup of $G$.
{{qed}}
Category:Topological Groups
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Theorem: Image of Group Homomorphism is Hausdorff Implies Kernel is Closed

Let $G$ and $H$ be topological groups.
Let $f: G \to H$ be a morphism.
Let its image $\Img f$ be Hausdorff.
Then its kernel $\map \ker f$ is closed in $G$.

Proof:
By Image of Group Homomorphism is Subgroup, $\Img f$ is a group.
Let $e$ be the identity of $H$.
By Topological Group is Hausdorff iff Identity is Closed, $\set e$ is closed in $\Img f$.
Because $f$ is continuous, $\map \ker f = \map {f^{-1} } e$ is closed in $G$.
{{qed}}
Category:Topological Groups
Category:Hausdorff Spaces
5zyfd3xvxrgt9mw0fn3u8ovtws5b1kz
Theorem: Group is Connected iff Subgroup and Quotient are Connected

Let $G$ be a topological group.
Let $H \le G$ be a subgroup.
{{TFAE}}
:$(1): \quad G$ is connected
:$(2): \quad H$ is connected and the left quotient space $G / H$ is connected
:$(3): \quad H$ is connected and the right quotient space $G / H$ is connected.

Proof:
{{ProofWanted}}
Category:Topological Groups
Category:Connected Topological Spaces
Category:Quotient Spaces (Topology)
99kmfihh1whhdxtg8jiqafyg9130q7z
Theorem: Group Acts by Homeomorphisms Implies Projection on Quotient Space is Open

Let $G$ be a group acting by homeomorphisms on a topological space $X$.
Then the projection map $\pi: X \to X / G$ is open.

Proof:
Let $U \subset X$ be open.
We have to show that $\pi \sqbrk U$ is open.
By definition of quotient topology, this is the case {{iff}} $\pi^{-1} \sqbrk {\pi \sqbrk U}$ is open.
By definition of saturation under group action:
:$\ds \pi^{-1} \sqbrk {\pi \sqbrk U} = \bigcup_{g \mathop \in G} g U$
Because $G$ acts by homeomorphisms, $\pi^{-1} \sqbrk {\pi \sqbrk U}$ is open.
{{qed}}
Theorem: Neighborhood iff Contains Neighborhood

Let $X$ be a topological space.
Let $x\in X$.
Let $V\subset X$ be a subset.
Then the following are equivalent:
:$V$ is a neighborhood of $x$ in $X$
:$V$ contains a neighborhood of $x$ in $X$

Proof:
Follows directly from the definition of neighborhood and Subset Relation is Transitive.
{{qed}}
Theorem: Even Perfect Number ends in 6 or 28 preceded by Odd Digit

Let $n$ be an even perfect number.
Then $n$ ends either in $6$ or $28$ preceded by an odd digit.

Proof:
By the Theorem of Even Perfect Numbers:
:$n = 2^{p - 1} \paren {2^p - 1}$
where $p$ is prime.
With the exception of $6 = 2^1 \paren {2^2 - 1}$ and $28 = 2^2 \paren {2^3 - 1}$:
:$p$ is odd and $p > 4$.
We claim that:
:$n$ ends in $\phantom 0 6$ preceded by an odd digit if $p \equiv 1 \pmod 4$
:$n$ ends in $28$ preceded by an odd digit if $p \equiv 3 \pmod 4$
These statements are equivalent to:
:$n \equiv \phantom 0 16 \pmod {\phantom 0 20}$ if $p \equiv 1 \pmod 4$
:$n \equiv 128 \pmod {200}$ if $p \equiv 3 \pmod 4$
By Powers of 16 Modulo 20, we have:
:$2^{4 n} = 16^n \equiv 16 \pmod {20}$ for $n \ge 1$
Theorem: Open Subset of Locally Connected Space is Locally Connected

Let $T = \struct {S, \tau}$ be a locally connected topological space.
Let $U \subseteq S$ be open in $T$.
Then $U$ is locally connected.

Proof:
{{MissingLinks}}
Theorem: Component of Locally Connected Space is Open

Let $T = \left({S, \tau}\right)$ be a locally connected topological space.
Let $G$ be a component of $T$.
Then $G$ is open.

Proof:
By definition of locally connected space, $T$ has a basis of connected sets in $T$.
Thus $S$ is a union of open connected sets in $T$.
By Components are Open iff Union of Open Connected Sets, the components of $T$ are open.
{{qed}}
Theorem: Path Component of Locally Path-Connected Space is Open

Let $T = \struct {S, \tau}$ be a locally path-connected topological space.
Let $G$ be a path component of $T$.
Then $G$ is open in $T$.

Proof:
By definition of locally path-connected, $T$ has a basis of path-connected set.
Thus $S$ is a union of open path-connected sets of $T$.
By Path Components are Open iff Union of Open Path-Connected Sets, the path components of $T$ are open in $T$.
{{qed}}
Theorem: Connected and Locally Path-Connected Implies Path-Connected

Let $T = \struct {S, \tau}$ be a connected and locally path-connected topological space.
Then $T$ is path-connected.

Proof:
By:
:Path Component of Locally Path-Connected Space is Closed
:Path Component of Locally Path-Connected Space is Open
the path components of $T$ are clopen.
Because $T$ is connected, every path component equals $S$.
That is, $T$ is path-connected.
{{qed}}
Category:Connected Topological Spaces
Category:Path-Connected Spaces
Category:Locally Path-Connected Spaces
8ojhnr50gdgax0me1cgk8oayacrw2az
Theorem: Components are Open iff Union of Open Connected Sets

Let $T = \struct {S, \tau}$ be a topological space.
{{TFAE}}
:$(1): \quad$ The components of $T$ are open.
:$(2): \quad S$ is a union of open connected sets of $T$.

Proof:
=== Condition (1) implies Condition (2) ===
{{:Components are Open iff Union of Open Connected Sets/Components are Open implies Space is Union of Open Connected Sets}}{{qed|lemma}}
Theorem: Path Components are Open iff Union of Open Path-Connected Sets

Let $T = \struct {S, \tau}$ be a topological space.
{{TFAE}}
:$(1): \quad$ The path components of $T$ are open.
:$(2): \quad S$ is a union of open path-connected sets of $T$.

Proof:
=== Condition (1) implies Condition (2) ===
{{:Path Components are Open iff Union of Open Path-Connected Sets/Path Components are Open implies Space is Union of Open Path-Connected Sets}}{{qed|lemma}}
Theorem: Even Perfect Number is Hexagonal

All perfect numbers which are even are hexagonal.

Proof:
Let $a$ be an even perfect number.
From the Theorem of Even Perfect Numbers, $a$ is in the form $2^{p - 1} \paren {2^p - 1}$ where $2^p - 1$ is prime.
Thus:
{{begin-eqn}}
{{eqn | l = a
      | r = \paren {2^p - 1} 2^{p - 1}
      | c = 
}}
{{eqn | r =  2^{p - 1} \paren {2 \times 2^{p - 1} - 1}
      | c = 
}}
{{eqn | r = n \paren {2 n - 1}
      | c = where $n = 2^{p - 1}$
}}
{{end-eqn}}
The result follows from Closed Form for Hexagonal Numbers.
{{qed}}
Theorem: Hexagonal Number is Triangular Number

Let $H_n$ be the $n$th hexagonal number.
Then $H_n$ is the $2 n - 1$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = H_n
      | r = n \paren {2 n - 1}
      | c = Closed Form for Hexagonal Numbers
}}
{{eqn | r = \frac {2 n \paren {2 n - 1} } 2
      | c = 
}}
{{eqn | r = \frac {m \paren {m + 1} } 2
      | c = for $m = 2 n - 1$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Sum of Reciprocals of Divisors of Perfect Number is 2

Let $n$ be a perfect number.
Then:
:$\ds \sum_{d \mathop \divides n} \dfrac 1 d = 2$
That is, the sum of the reciprocals of the divisors of $n$ is equal to $2$.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{d \mathop \divides n} d
      | r = \map {\sigma_1} n
      | c = {{Defof|Divisor Sum Function}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 n \sum_{d \mathop \divides n} d
      | r = \dfrac {\map {\sigma_1} n} n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{d \mathop \divides n} \frac d n
      | r = \dfrac {\map {\sigma_1} n} n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sum_{d \mathop \divides n} \frac 1 d
      | r = \dfrac {\map {\sigma_1} n} n
      | c = 
}}
{{end-eqn}}
The result follows by definition of perfect number:
{{:Definition:Perfect Number/Definition 4}}{{qed}}
Theorem: Even Perfect Number except 6 is Congruent to 1 Modulo 9

Let $n$ be an even perfect number, but not $6$.
Then:
:$n \equiv 1 \pmod 9$

Proof:
From Theorem of Even Perfect Numbers:
:$n = 2^{p - 1} \paren {2^p - 1} = \dfrac {2^p \paren {2^p - 1} } 2$
where $p$ is prime.
From Odd Power of 2 is Congruent to 2 Modulo 3:
:$2^p \equiv 2 \pmod 3$
for odd $p$.
Thus:
{{begin-eqn}}
{{eqn | l = n
      | r = \dfrac {\paren {3 k + 2} \paren {3 k + 1} } 2
      | c = for some $k \in \Z_{>0}$
}}
{{eqn | r = \frac {9 k^2 + 9 k + 2} 2
      | c = 
}}
{{eqn | r = 9 \frac {k \paren {k + 1} } 2 + 1
      | c = 
}}
{{eqn | r = 9 T_k + 1
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
So $n$ is $1$ more than $9$ times the $k$th triangular number for some $k$.
That is:
:$n \equiv 1 \pmod 9$
When $n = 6$ the situation is different.
We have:
{{begin-eqn}}
{{eqn | l = n
      | r = \dfrac {2^2 \paren {2^2 - 1} } 2
      | c = 
}}
{{eqn | r = \dfrac {\paren {3 k + 1} \paren {3 k} } 2
      | c = where $k = 1$
}}
{{eqn | o = \equiv
      | r = 0 \pmod 3
      | c = 
}}
{{end-eqn}}
and so the result does not hold.
{{qed}}
Theorem: Congruence Modulo 3 of Power of 2

Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$2^n \equiv \paren {-1}^n \pmod 3$
where $\equiv$ denotes congruence.
That is:
:$\exists q \in \Z: 2^n = 3 q + \paren {-1}^n$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$2^n \equiv \paren {-1}^n \pmod 3$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = 2^0
      | r = 1
      | c = 
}}
{{eqn | r = \paren {-1}^0
      | c = 
}}
{{eqn | o = \equiv
      | r = \paren {-1}^0
      | rr= \pmod 3
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
Theorem: Odd Power of 2 is Congruent to 2 Modulo 3

Let $n \in \Z_{\ge 0}$ be an odd positive integer.
Then:
:$2^n \equiv 2 \pmod 3$

Proof:
From Congruence Modulo 3 of Power of 2:
:$2^n \equiv \paren {-1}^n \pmod 3$
We have that $n$ is odd.
Hence:
{{begin-eqn}}
{{eqn | l = 2^n
      | o = \equiv
      | r = -1
      | rr= \pmod 3
      | c = 
}}
{{eqn | o = \equiv
      | r = 3 - 1
      | rr= \pmod 3
      | c = 
}}
{{eqn | o = \equiv
      | r = 2
      | rr= \pmod 3
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Number Theory
Category:2
Category:3
7z9oqafotjpzea7v785tizd4gmegimi
Theorem: Open Set in Open Subspace

Let $T = \struct{X, \tau}$ be a topological space.
Let $T_U = \struct{U, \tau_U}$ be a subspace of $T$ where $U \subseteq X$ is open.
Let $V \subseteq U$ be a subset.
Then $V$ is open in $T_U$ {{iff}} $V$ is open in $T$.

Proof:
=== Necessary Condition ===
Let $V$ be open in $T$.
By Intersection with Subset is Subset, $V\cap U = V$.
By definition of topological subspace, $V$ is open in $T_U$.
Theorem: Equivalence of Definitions of Locally Compact Hausdorff Space

{{TFAE|def = Locally Compact Hausdorff Space|view = locally compact Hausdorff space}}
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.

Proof:
=== $1$ implies $2$ ===
Let $x \in S$.
Let $K$ be a compact neighborhood of $x$.
Let $\BB$ be the set of compact neighborhoods of $x$.
It is shown that $\BB$ is a neighborhood basis of $x$.
Let $U$ be a neighborhood of $x$.
We have to show that $U$ contains a compact neighborhood of $x$.
By Neighborhood in Topological Subspace, $U\cap K$ is a neighborhood of $x$ in $K$.
By Neighborhood in Compact Hausdorff Space Contains Compact Neighborhood, there is a compact neighborhood $V$ of $x$ in $K$ such that $V \subset U \cap K$.
From Neighborhood in Topological Subspace there exists a neighborhood $W$ of $x$ in $T$ such that $W \cap K = V$. 
From Intersection of Neighborhoods in Topological Space is Neighborhood we have $W \cap K = V$ is a neighborhood of $x$ in $T$.
From Compact in Subspace is Compact in Topological Space then $V$ is compact in $T$.
Hence $V \in \BB$.
The result follows.
{{qed|lemma}}
Theorem: Neighborhood in Compact Hausdorff Space Contains Compact Neighborhood

Let $X$ be a compact Hausdorff topological space.
Let $x\in X$.
Let $U$ be a neighborhood of $x$.
Then $U$ contains a compact neighborhood of $x$.

Proof:
By definition of neighborhood, there exists an open set $V$ with $x\in V\subset U$.
Then $X \setminus V$ is closed.
By Compact Hausdorff Space is $T_4$, there exist disjoint open sets $A, B$ such that:
:$X \setminus V \subset A$
:$x \in B$
Then:
:$X \setminus A$ is compact by Closed Subspace of Compact Space is Compact
:$B \subset X \setminus A$ by Empty Intersection iff Subset of Relative Complement
:$X \setminus A \subset X \setminus \paren{X \setminus V}$ by Relative Complement inverts Subsets
:$X \setminus \paren{X \setminus V} = V$ by Relative Complement of Relative Complement
Thus:
:$x \in B \subset X \setminus A \subset V$
so $X \setminus A$ is a compact neighborhood of $x$ contained in $V$.
{{qed}}
Theorem: Image of Idempotent and Directed Suprema Preserving Mapping is Complete Lattice

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $f:S \to S$ be a mapping that is idempotent and preserves directed suprema.
Let $F = \struct {f \sqbrk S, \precsim}$ be an ordered subset of $L$.
Then $F$ inherits directed suprema and is complete lattice.

Proof:
We will prove that
:$F$ inherits directed suprema.
Let $D$ be a directed subset of $f \sqbrk S$ such that
:$D$ admits a supremum in $L$.
By definition of ordered subset:
:$D$ is directed in $L$.
By definition of mapping preserves directed suprema:
:$\map {\sup_L} {f \sqbrk D} = \map f {\sup_L D}$
By definition of idempotent mapping:
:$f \sqbrk D = D$
Thus by definition of image of set:
:$\sup_L D \in f \sqbrk S$
{{qed|lemma}}
By definition of idempotent mapping:
:$f \sqbrk S = \set {x \in S: \map f x = x}$
By Directed Suprema Preserving Mapping is Increasing:
:$f$ is an increasing mapping.
Thus by Image under Increasing Mapping equal to Special Set is Complete Lattice:
:$F$ is complete lattice.
{{qed}}