avewright/proofwiki-math huggingface dataset:
Theorem: Sufficient Condition for Square of Product to be Triangular

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $2 n^2 \pm 1 = m^2$ be a square number.
Then $\paren {m n}^2$ is a triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {m n}^2
      | r = \paren {2 n^2 \pm 1} \times n^2
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 n^2 \pm 1} \paren {2 n^2} } 2
      | c = 
}}
{{end-eqn}}
That is, either:
:$\paren {m n}^2 = \dfrac {\paren {2 n^2 - 1} \paren {2 n^2} } 2$
and so:
:$\paren {m n}^2 = T_{2 n^2 - 1}$
or:
:$\paren {m n}^2 = \dfrac {\paren {2 n^2} \paren {2 n^2 + 1} } 2$
and so:
:$\paren {m n}^2 = T_{2 n^2}$
Hence the result.
{{Qed}}
Theorem: Trivial Topological Space is Indiscrete

Let $T = \struct {S, \tau}$ be a trivial topological space.
Then $\tau$ is an indiscrete topology.

Proof:
By definition of trivial topological space, $S$ is a singleton.
That is, $S$ is a set containing exactly one element.
Suppose $S = \set x$ for some object $x$. 
Then the power set of $S$ is the set:
:$\powerset S = \set {\O, \set x}$
That is:
:$\powerset S = \set {\O, S}$
Let $\tau$ be a topology on $S$.
We have that $\tau$ is a subset of $\powerset S$.
Hence $\tau$ must equal one of the following sets:
:$\tau_1 = \O$
:$\tau_2 = \set \O$
:$\tau_3 = \set S$
or
:$\tau_4 = \set {\O, S}$.
By definition of a topology, $S \in \tau$.
Thus $\tau \ne \tau_1$ and $\tau \ne \tau_2$.
By Empty Set is Element of Topology, also $\O \in \tau$. 
Therefore $\tau \ne \tau_3$.
By Indiscrete Topology is Topology, $\tau_4$ is a topology on $S$.
Hence if $S$ is a singleton, the only possible topology on $S$ is the indiscrete topology. 
{{qed}}
Category:Trivial Topological Spaces
Category:Indiscrete Topology
8oipc5kxrhb20jc5r3pv57v8is0cxfg
Theorem: Existence of Divisor with Remainder between 2b and 3b

For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q$ and $r$ where $2 b \le r < 3 b$ such that:
:$a = q b + r$

Proof:
From the Division Theorem, we have that:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
So, with a view to where this is going, let $q$ and $r$ be renamed such that $a = q' b + r'$ with $0 \le r' < b$.
Then let $q' = q + 2$.
We have:
{{begin-eqn}}
{{eqn | l = a
      | r = q' b + r'
      | c = 
}}
{{eqn | r = \paren {q + 2} b + r'
      | c = 
}}
{{eqn | r = q b + 2 b + r'
      | c = 
}}
{{end-eqn}}
Setting $r = 2 b + r'$, it follows that:
:$2 b \le r < 3 b$
and the result follows.
{{qed}}
Theorem: Integer of form 6k + 5 is of form 3k + 2 but not Conversely

Let $n \in \Z$ be an integer of the form:
:$n = 6 k + 5$
where $k \in \Z$.
Then $n$ can also be expressed in the form:
:$n = 3 k + 2$
for some other $k \in \Z$.
However it is not necessarily the case that if $n$ can be expressed in the form:
:$n = 3 k + 2$
then it can also be expressed in the form:
:$n = 6 k + 5$

Proof:
{{begin-eqn}}
{{eqn | l = n
      | r = 6 k + 5
      | c = 
}}
{{eqn | r = 3 \paren {2 k} + 3 + 2
      | c = 
}}
{{eqn | r = 3 \paren {2 k + 1} + 2
      | c = for some $2 k + 1 \in \Z$
}}
{{end-eqn}}
Replacing $2 k + 1$ with $k$ gives the result.
{{qed|lemma}}
Now consider $n = 8$.
We have that:
:$8 = 3 \times 2 + 2$
and so can be expressed in the form:
:$n = 3 k + 2$
However:
:$8 = 6 \times 1 + 2$
and is in the form:
:$n = 6 k + 2$
and so cannot be expressed in the form:
:$n = 6 k + 5$
{{qed}}
Theorem: Odd Integer Modulo 4

Let $n$ be an odd integer.
Then $n$ can be expressed either as:
:$n = 4 k + 1$
or as:
:$n = 4 k + 3$

Proof:
By the Division Theorem, $n$ can be expressed as:
:$n = 4 k + r$
where:
:$k, r \in \Z$
:$0 \le r < 4$
That is, one of the following holds:
{{begin-eqn}}
{{eqn | l = n
      | r = 4 k
}}
{{eqn | l = n
      | r = 4 k + 1
}}
{{eqn | l = n
      | r = 4 k + 2
}}
{{eqn | l = n
      | r = 4 k + 3
}}
{{end-eqn}}
Of these:
{{begin-eqn}}
{{eqn | l = n
      | r = 4 k
}}
{{eqn | r = 2 \paren {2 k}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = n
      | r = 4 k + 2
}}
{{eqn | r = 2 \paren {2 k + 1}
}}
{{end-eqn}}
and so are even by definition.
Then:
{{begin-eqn}}
{{eqn | l = n
      | r = 4 k + 1
}}
{{eqn | r = 2 \paren {2 k} + 1
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = n
      | r = 4 k + 3
}}
{{eqn | r = 2 \paren {2 k + 1} + 1
}}
{{end-eqn}}
and so are odd by definition.
{{qed}}
Theorem: Cube Modulo 9

Let $x \in \Z$ be an integer.
Then one of the following holds:
{{begin-eqn}}
{{eqn | l = x^3
      | o = \equiv
      | r = 0 \pmod 9
      | c = 
}}
{{eqn | l = x^3
      | o = \equiv
      | r = 1 \pmod 9
      | c = 
}}
{{eqn | l = x^3
      | o = \equiv
      | r = 8 \pmod 9
      | c = 
}}
{{end-eqn}}

Proof:
Let $x$ be an integer.
There are three cases to consider:
:$(1): \quad x \equiv 0 \pmod 3$: we have $x = 3 k$
:$(2): \quad x \equiv 1 \pmod 3$: we have $x = 3 k + 1$
:$(3): \quad x \equiv 2 \pmod 3$: we have $x = 3 k + 2$
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = x
      | r = 3 k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^3
      | r = \paren {3 k}^3
      | c = 
}}
{{eqn | r = 3^3 k^3
      | c = 
}}
{{eqn | r = 9 \times \paren {3 k^3}
      | c =  
}}
{{eqn | ll= \leadsto
      | l = x^3
      | o = \equiv
      | r = 0 \pmod 9
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | n = 2
      | l = x
      | r = 3 k + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^3
      | r = \paren {3 k + 1}^3
      | c = 
}}
{{eqn | r = \paren {3 k}^3 + 3 \paren {3 k}^2 + 3 \paren {3 k} + 1
      | c = 
}}
{{eqn | r = 9 \times \paren {3 k^3 + 3 k^2 + k} + 1
      | c =  
}}
{{eqn | ll= \leadsto
      | l = x^3
      | o = \equiv
      | r = 1 \pmod 9
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | n = 3
      | l = x
      | r = 3 k + 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^3
      | r = \paren {3 k + 2}^3
      | c = 
}}
{{eqn | r = \paren {3 k}^3 + 3 \times 2 \paren {3 k}^2 + 3 \times 2^2 \paren {3 k} + 2^3
      | c = 
}}
{{eqn | r = 9 \times \paren {3 k^3 + 6 k^2 + 4 k} + 8
      | c =  
}}
{{eqn | ll= \leadsto
      | l = x^3
      | o = \equiv
      | r = 8 \pmod 9
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Number which is Square and Cube Modulo 7

Let $n \in \Z$ be an integer.
Let $n$ be both a square and a cube at the same time.
Then either:
:$n \equiv 0 \pmod 7$
or:
:$n \equiv 1 \pmod 7$

Proof:
Let $n = r^2 = s^3$ for some $r, s \in \Z$.
Then:
:$n = \paren {m^2}^3 = \paren {m^3}^2 = m^6$
for some $m \in \Z$
There are $7$ cases to consider:
:$(0): \quad m \equiv 0 \pmod 7$: we have $m = 7 k$
:$(1): \quad m \equiv 1 \pmod 7$: we have $m = 7 k + 1$
:$(2): \quad m \equiv 2 \pmod 7$: we have $m = 7 k + 2$
:$(3): \quad m \equiv 3 \pmod 7$: we have $m = 7 k + 3$
:$(4): \quad m \equiv 4 \pmod 7$: we have $m = 7 k + 4$
:$(5): \quad m \equiv 5 \pmod 7$: we have $m = 7 k + 5$
:$(6): \quad m \equiv 6 \pmod 7$: we have $m = 7 k + 6$
Using Congruence of Powers throughout, we make use of:
:$x \equiv y \pmod 7 \implies x^j \equiv y^j \pmod 7$
for $j \in \Z_{>0}$.
First the easy cases:
{{begin-eqn}}
{{eqn | n = 0
      | l = m
      | o = \equiv
      | r = 0 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m^6
      | o = \equiv
      | r = 0 \pmod 7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 1
      | l = m
      | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m^6
      | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{end-eqn}}
It is sufficient to investigate the congruence modulo $7$ of the integers from $2$ to $6$.
We have:
{{begin-eqn}}
{{eqn | n = 2
      | l = 2^3
      | o = \equiv
      | r = 8 \pmod 7
      | c = 
}}
{{eqn | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2^6
      | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 3
      | l = 3^2
      | o = \equiv
      | r = 9 \pmod 7
      | c = 
}}
{{eqn | o = \equiv
      | r = 2 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 3^6 = \paren {3^2}^3
      | o = \equiv
      | r = 1 \pmod 7
      | c = from $(2)$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 4
      | l = 4^2
      | o = \equiv
      | r = 16 \pmod 7
      | c = 
}}
{{eqn | o = \equiv
      | r = 2 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4^6 = \paren {4^2}^3
      | o = \equiv
      | r = 1 \pmod 7
      | c = from $(2)$
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 5
      | l = 5^2
      | o = \equiv
      | r = 25 \pmod 7
      | c = 
}}
{{eqn | o = \equiv
      | r = 4 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 5^6 = \paren {4^2}^3
      | o = \equiv
      | r = 2^3 = 8 \pmod 7
      | c = from $(4)$
}}
{{eqn | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 6
      | l = 6^2
      | o = \equiv
      | r = 36 \pmod 7
      | c = 
}}
{{eqn | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 6^6 = \paren {6^2}^3
      | o = \equiv
      | r = 1 \pmod 7
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Division Theorem/Half Remainder Version

For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $-\dfrac {\size b} 2 \le r < \dfrac {\size b} 2$:
:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, -\dfrac {\size b} 2 \le r < \dfrac {\size b} 2$

Proof:
{{ProofWanted|boring}}
Theorem: Khinchin's Law

Let $P$ be a population.
Let $P$ have mean $\mu$ and finite variance. 
Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of random variables forming a random sample from $P$.
Let: 
:$\ds {\overline X}_n = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:${\overline X}_n \xrightarrow p \mu$
where $\xrightarrow p$ denotes convergence in probability.

Proof:
Let $\sigma$ be the standard deviation of $P$.
By the definition of convergence in probability, we aim to show that: 
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} = 1$
for all real $\epsilon > 0$.
Let $\epsilon > 0$ be a real number.
By Variance of Sample Mean:
:$\var {{\overline X}_n} = \dfrac {\sigma^2} n$
By the Bienaymé-Chebyshev Inequality, we have for real $k > 0$: 
:$\map \Pr {\size { {\overline X}_n - \mu} \ge \dfrac {k \sigma} {\sqrt n}} \le \dfrac 1 {k^2}$
As $\sigma > 0$ and $n > 0$, we can set: 
:$k = \dfrac {\sqrt n} {\sigma} \epsilon$
This gives: 
:$\map \Pr {\size {{\overline X}_n - \mu} \ge \epsilon} \le \dfrac {\sigma^2} {n \epsilon^2}$
We therefore have: 
{{begin-eqn}}
{{eqn | l = \map \Pr {\size { {\overline X}_n - \mu} < \epsilon}
      | r = 1 - \map \Pr {\size { {\overline X}_n - \mu} \ge \epsilon}
}}
{{eqn | o = \ge
      | r = 1 - \frac {\sigma^2} {n \epsilon^2}
}}
{{end-eqn}}
So: 
:$1 - \dfrac {\sigma^2} {n \epsilon^2} \le \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} \le 1$
We have: 
:$\ds \lim_{n \mathop \to \infty} \paren {1 - \dfrac {\sigma^2} {n \epsilon^2} } = 1$
and: 
:$\ds \lim_{n \mathop \to \infty} 1 = 1$
So by the Squeeze Theorem: 
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \epsilon} = 1$
for all real $\epsilon > 0$.
{{qed}}
Theorem: Variance of Linear Combination of Random Variables/Corollary

Let $X$ and $Y$ be independent random variables.

Proof:
From Variance of Linear Combination of Random Variables, we have: 
:$\var {a X + b Y} = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y}$
where $\cov {X, Y}$ is the covariance of $X$ and $Y$. 
From Covariance of Independent Random Variables is Zero: 
:$2 a b \, \cov {X, Y} = 0$
The result follows.
{{qed}}
Category:Variance
8jc0sfnr7tjd01olvha76a5y2a6lz98
Theorem: Singleton is Connected in Topological Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then the singleton $\set x$ is connected.

Proof:
Let $A = \set x$.
From definition $3$ of a connected set, $A$ is connected in $T$ {{iff}} the subspace $\struct {A, \tau_A}$ is a connected space.
From Topology on Singleton is Indiscrete Topology, $\tau_A$ is the indiscrete topology.
From Indiscrete Space is Connected, $\struct {A, \tau_A}$ is a connected space.
{{qed}}
Category:Connected Sets (Topology)
Category:Singletons
7xswkp26igzflelyfg72bpqwjliuy1f
Theorem: Equivalence of Definitions of Component/Lemma 1

:$C$ is connected in $T$ and $C \in \CC_x$.

Proof:
From Singleton is Connected in Topological Space, $\set{x}$ is a connected set of $T$ containing $x$.
It follows that $x \in C$.
From Union of Connected Sets with Common Point is Connected, $C$ is a  connected set of $T$.
Hence $C \in \CC_x$.
{{qed}}
Category:Equivalence of Definitions of Component
g3treh4at5mzazjpq27gz0l3xcydn7l
Theorem: Variance of Sample Mean

Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.
Let: 
:$\ds \overline X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then: 
:$\var {\overline X} = \dfrac {\sigma^2} n$

Proof:
{{begin-eqn}}
{{eqn | l = \var {\overline X}
      | r = \var {\frac 1 n \sum_{i \mathop = 1}^n X_i}
}}
{{eqn | r = \frac 1 {n^2} \sum_{i \mathop = 1}^n \var {X_i}
      | c = repeated application of {{Corollary|Variance of Linear Combination of Random Variables}}
}}
{{eqn | r = \frac 1 {n^2} \sum_{i \mathop = 1}^n \sigma^2
}}
{{eqn | r = \frac {\sigma^2 n} {n^2}
      | c = as $\ds \sum_{i \mathop = 1}^n 1 = n$
}}
{{eqn | r = \frac {\sigma^2} n
}}
{{end-eqn}}
{{qed}}
Category:Variance
Category:Inductive Statistics
81orhay5yae0e8hdivfjbwn0i1flmdv
Theorem: Product of Divisors is Divisor of Product

Let $a, b, c, d \in \Z$ be integers such that $a, c \ne 0$.
Let $a \divides b$ and $c \divides d$, where $\divides$ denotes divisibility.
Then:
:$a c \divides b d$

Proof:
By definition of divisibility:
{{begin-eqn}}
{{eqn | q = \exists k_1 \in \Z
      | l = b
      | r = a k_1
}}
{{eqn | q = \exists k_2 \in \Z
      | l = d
      | r = c k_2
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = b d
      | r = \paren {a k_1} \paren {c k_2}
      | c = 
}}
{{eqn | r = k_1 k_2 \paren {a c}
      | c = 
}}
{{end-eqn}}
and the result follows by definition of divisibility.
{{Qed}}
Theorem: Common Divisor Divides Integer Combination/General Result

Let $c$ be a common divisor of a set of integers $A := \set {a_1, a_2, \dotsc, a_n}$.
That is:
:$\forall x \in A: c \divides x$
Then $c$ divides any integer combination of elements of $A$:
:$\forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$
Theorem: One is Common Divisor of Integers

Let $a, b \in \Z$ be integers.
Then $1$ is a common divisor of $a$ and $b$.

Proof:
From One Divides all Integers:
:$1 \divides a$
and:
:$1 \divides b$
where $\divides$ denotes divisibility.
The result follows by definition of common divisor.
{{Qed}}
Theorem: Set of Common Divisors of Integers is not Empty

Let $a, b \in \Z$ be integers.
Let $S$ be the set of common divisors of $a$ and $b$.
Then $S$ is not empty.

Proof:
From One is Common Divisor of Integers:
:$1$ is a common divisor of $a$ and $b$.
Thus, whatever $a$ and $b$ are:
:$1 \in S$
The result follows by definition of empty set.
{{Qed}}
Theorem: Equivalence of Definitions of Path Component

{{TFAE|def = Path Component|view = path component|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
=== Equivalence Class ===
{{:Definition:Path Component/Equivalence Class}}
=== Union of Path-Connected Sets ===
{{:Definition:Path Component/Union of Path-Connected Sets}}
=== Maximal Path-Connected Set ===
{{:Definition:Path Component/Maximal Path-Connected Set}}

Proof:
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text { is path-connected in } T}$
Let $C = \bigcup \CC_x$.
Theorem: GCD of Integer and its Negative

Let $a \in \Z$ be an integer.
Then:
:$\gcd \set {a, -a} = \size a$
where:
:$\gcd$ denotes greatest common divisor
:$\size a$ denotes the absolute value of $a$.

Proof:
From Integer Divisor Results, the divisors of $a$ include $a$ itself.
From Integer Divides its Negative, $a \divides \paren {-a}$.
Thus we have:
:$a \divides a$
and:
:$a \divides -a$
and so:
:$\gcd \set {a, -a} \ge \size a$
From Absolute Value of Integer is not less than Divisors, there is no divisor of $a$ which is greater than $a$.
That is:
:$\gcd \set {a, -a} \le \size a$
Hence the result.
{{qed}}
Category:Greatest Common Divisor
02pl9s60w8szjcfxx0ggf68m2jgxvj1
Theorem: Equivalence of Definitions of Path Component/Lemma

:$C$ is path-connected in $T$ and  $C \in \CC_x$.

Proof:
From Point is Path-Connected to Itself, $\set x$ is a path-connected subset of $T$ containing $x$.
It follows that $x \in C$.
From Union of Path-Connected Sets with Common Point is Path-Connected, $C$ is a  path-connected subset of $T$.
Hence $C \in \CC_x$.
{{qed}}
Category:Equivalence of Definitions of Path Component
or6j9ytz9x5c9y11an2a6y3xq99lrif
Theorem: Square Divides Product of Multiples

Let $a, b, c, \in \Z$ be integers.
Let:
:$a \divides b, a \divides c$
where $\divides$ denotes divisibility.
Then:
:$a^2 \divides b c$

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = a
      | o = \divides
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists k_1 \in \Z
      | l = k_1 a
      | r = b
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = a
      | o = \divides
      | r = c
      | c =
}}
{{eqn | ll= \leadsto
      | q = \exists k_2 \in \Z
      | l = k_2 a
      | r = c
      | c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = b c
      | r = \paren {k_1 a} \paren {k_2 a}
      | c =
}}
{{eqn | r = \paren {k_1 k_2} a^2
      | c =
}}
{{eqn | ll= \leadsto
      | l = a^2
      | o = \divides
      | r = b c
      | c = {{Defof|Divisor of Integer}}
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of Path Component/Equivalence Class equals Union of Path-Connected Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in T$.
Let $\CC_x = \leftset {A \subseteq S : x \in A \land A}$ is path-connected in $\rightset T$.
Let $C = \bigcup \CC_x$
Let $\sim$ be the equivalence relation defined by:
:$y \sim z$ {{iff}} $y$ and $z$ are path-connected in $T$.
Let $C'$ be the equivalence class of $\sim$ containing $x$.
Then $C = C'$.

Proof:
{{begin-eqn}}
{{eqn | l = y \in C'
      | o = \leadstoandfrom
      | r = x \text { is path-connected to } y \text { in } T
      | c = Definition of $\sim$
}}
{{eqn | o = \leadstoandfrom
      | r = \exists B \text{ a connected set of } T, x \in B, y \in B
      | c = Points are Path-Connected iff Contained in Path-Connected Set
}}
{{eqn | o = \leadstoandfrom
      | r = \exists B \in \CC_x : y \in B
      | c = Equivalent definition
}}
{{eqn | o = \leadstoandfrom
      | r = y \in \bigcup \CC_x
      | c = {{Defof|Union of Set of Sets}}
}}
{{eqn | o = \leadstoandfrom
      | r = y \in C
      | c = Definition of $C$
}}
{{end-eqn}}
The result follows.
Theorem: Image of Path is Path-Connected Set

Let $T = \struct {S, \tau}$ be a topological space.
Let $I \subset \R$ be the closed real interval $\closedint a b$.
Let $\gamma: I \to S$ be a path.
Then:
:$\map \gamma I$ is a path-connected set of $T$.

Proof:
From Path-Connected iff Path-Connected to Point, $\map \gamma I$ is a path-connected set {{iff}} every point of $\map \gamma I$ is path-connected to a common point.
It is shown that every point of $\map \gamma I$ is path-connected to $\map \gamma a$.
Let $H = \map \gamma I$.
Let $x = \map \gamma a$.
From Point is Path-Connected to Itself, $x$ is path-connected to itself.
Let $y$ be any element of $H \setminus \set x$.
By assumption:
:$\exists t \in \hointl a b : \map \gamma t = y$
From Continuity of Composite with Inclusion, $\mathbin \gamma {\restriction_{\closedint a t} } : \closedint a t \to H$ is a continuous mapping.
It follows that $\mathbin \gamma {\restriction_{\closedint a t} }$ is a path joining $x$ to $y$.
The result follows.
{{qed}}
Category:Path-Connected Sets
ncff27p6790yzwwgy5abid06l1k5bth
Theorem: Acceleration of Particle moving in Circle at Constant Speed

Let $P$ be a particle moving in a circular path $C$ at a constant speed.
Then the acceleration of $P$, denoted by $\mathbf a \in \R$, is given as:
:$\mathbf a = -\dfrac {\size {\mathbf v}^2 \mathbf r} {\size {\mathbf r}^2}$
where:
:$\mathbf v$ is the instantaneous velocity of $P$
:$\mathbf r$ is the vector whose magnitude equals the length of the radius of $C$ and whose direction is from the center of $C$ to $P$
:$\size {\, \cdot \,}$ denotes the magnitude of a vector.

Proof:
Let the center of $C$ be placed at the origin of a two-dimensional coordinate system where the circular path is in the $xy$ plane.
Using the parametric equation of a circle, we can write the position of $P$ as the vector-valued function
:$\ds \map {\mathbf r} t = \begin{bmatrix} r \map \cos {\omega t} \\ r \map \sin {\omega t} \end{bmatrix}$
where $r$ and $\omega$ are constants and $t$ is time.
This path is a circle because its distance from the origin is constant.
{{begin-eqn}}
{{eqn | l = \size {\map {\mathbf r} t}
      | r = \sqrt {r^2 \map {\cos^2} {\omega t} + r^2 \map {\sin^2} {\omega t} }
      | c = {{Defof|Magnitude}}
}}
{{eqn | r = \size r \sqrt {\map {\cos^2} {\omega t} + \map {\sin^2} {\omega t} }
}}
{{eqn | r = \size r
      | c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
The velocity $\mathbf v$ of $P$ is given by
{{begin-eqn}}
{{eqn | l = \map {\mathbf v} t
      | r = \frac {\d \mathbf r} {\d t}
      | c = {{Defof|Velocity}}
}}
{{eqn | r = \begin{bmatrix} -r \omega \map \sin {\omega t} \\ r \omega \map \cos {\omega t} \end{bmatrix}
      | c = Derivative of Cosine Function and Derivative of Sine Function
}}
{{end-eqn}}
The speed of $P$ is constant.
{{begin-eqn}}
{{eqn | l = \size {\map {\mathbf v} t}
      | r = \sqrt{r^2 \omega^2 \map {\sin^2} {\omega t} + r^2 \omega^2 \map {\cos^2} {\omega t} }
      | c = {{Defof|Magnitude}}
}}
{{eqn | r = \size r \size \omega \sqrt{\map {\sin^2} {\omega t} + \map {\cos^2} {\omega t} }
}}
{{eqn | r = \size r \size \omega
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll = \leadsto
      | l  = \size \omega
      | r  = \frac {\size {\mathbf v} } {\size r}
}}
{{end-eqn}}
The acceleration $\map {\mathbf a} t$ is given by
{{begin-eqn}}
{{eqn | l = \map {\mathbf a} t
      | r = \frac {\d \mathbf v} {\d t}
      | c = {{Defof|Acceleration}}
}}
{{eqn | r = \begin{bmatrix} -r \omega^2 \map \cos {\omega t} \\ -r \omega^2 \map \sin {\omega t} \end{bmatrix}
      | c = Derivative of Sine Function and Derivative of Cosine Function
}}
{{eqn | r = -\omega^2 \mathbf r
      | c = Definition of $\map {\mathbf r} t$ above
}}
{{eqn | r = -\frac {\size {\mathbf v}^2 \mathbf r} {r^2}
      | c = $\size \omega = \dfrac {\size {\mathbf v} } {\size r}$
}}
{{eqn | r = -\frac {\size {\mathbf v}^2 \mathbf r} {\size {\mathbf r}^2}
      | c = $\size r = \size {\mathbf r}$
}}
{{end-eqn}}
{{qed}}