nyuuzyou/wikis huggingface dataset:
Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed/Proof 2

== Theorem ==
== Proof ==
Let $\pi : X \to X/N$ be the quotient mapping.
Let $\VV$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Let $\VV_N$ be the set of open neighborhoods of ${\mathbf 0}_{X/N}$.
Then, we have, from Expression for Closure of Set in Topological Vector Space: 
:$\ds N = \bigcap_{V \mathop \in \VV} \paren {N + V}$
From Preimage of Image of Linear Transformation, we have $\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$. 
From Kernel of Quotient Mapping, we then obtain $\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$.
Hence, we have:
:$\ds N^- = \bigcap_{V \mathop \in \VV} \pi^{-1} \sqbrk {\pi \sqbrk V}$ 
From Preimage of Intersection under Mapping: Family of Sets, we obtain:
:$\ds N^- = \pi^{-1} \sqbrk {\bigcap_{V \mathop \in \VV} \pi \sqbrk V}$ 
Since $\pi$ is surjective, we obtain:
:$\ds \pi \sqbrk {N^-} = \pi \sqbrk {\pi^{-1} \sqbrk {\bigcap_{V \mathop \in \VV} \pi \sqbrk V} } = \bigcap_{V \mathop \in \VV} \pi \sqbrk V$
From Open Set in Quotient Topological Vector Space, we have:
:$\VV_N = \set {\pi \sqbrk U : U \in \VV}$
Hence, we have:
:$\ds \pi \sqbrk {N^-} = \bigcap_{U \in \VV_N} U$
We argue that $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ $N^- = N$.
If $N^- = N$, then we immediately have $\pi \sqbrk {N^-} = \pi \sqbrk N = \set { {\mathbf 0}_{X/N} }$. 
If $N^- \ne N$, then there exists $x_0 \in N^- \setminus N$.
From Kernel of Quotient Mapping, we then have $\map \pi {x_0} \ne {\mathbf 0}_{X/N}$ and hence $\pi \sqbrk {N^-} \ne \set { {\mathbf 0}_{X/N} }$. 
Hence we have $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ $N^- = N$.
From Set is Closed iff Equals Topological Closure, we therefore have that $\pi \sqbrk {N^-} = \set { {\mathbf 0}_{X/N} }$ $N$ is closed.
That is, $N$ is closed :
:$\ds \bigcap_{U \in \VV_N} U = \set { {\mathbf 0}_{X/N} }$
From Characterization of Hausdorff Topological Vector Space, we conclude that $N$ is closed $X/N$ is Hausdorff.
Category:Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed
Quotient Topology is Topology

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $q_\RR: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Let $\tau_\RR$ be the quotient topology on $S / \RR$ by $q_\RR$.
Then $\tau_\RR$ is a topology on $S$.
== Proof ==
By definition of quotient topology, $\tau_\RR$ is the identification topology on $S / \RR$ with respect to $q_\RR$ and $\struct {S, \tau}$.
Identification Topology is Topology shows that $\tau_\RR$ is a topology on $S / \RR$.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Topology
Quotient Topology of Partition Topology is Discrete Space

== Theorem ==
Let $\PP$ be a partition of a set $S$.
Let $T = \struct {S, \tau}$ be the partition space formed from $\PP$.
Let $S / \PP$ be the quotient set of $S$ by $\PP$.
Then the quotient topology $\tau_{S / \PP}$ is a discrete topology.
== Proof ==
Let $\BB$ be the set defined as:
:$\BB = \set {\set A: A \in S / \PP}$
From Basis for Partition Topology, $\BB$ forms a basis for a partition space on $S$.
From Basis for Discrete Topology, $\BB$ forms a basis for the discrete topology on $S / \PP$.
Hence the result, by definition of quotient topology.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology: $1$
Category:Discrete Topology
Category:Partition Topology
Quotient Vector Space is Vector Space

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $N$ be a linear subspace of $X$.
Define: 
:$X/N = \set {x + N : x \in X}$
where $x + N$ is the Minkowski sum of $x$ and $N$.
Define: 
:$\paren {x + N} +_{X/N} \paren {y + N} = \paren {x + y} + N$
for $x, y \in X$, and:
:$\alpha \circ_{X/N} {x + N} = \paren {\alpha x} + N$
for $\alpha \in K$ and $x \in X$. 
Then $\paren {X/N, +_{X/N}, \circ_{X/N} }_K$ is a vector space.
== Proof ==
=== Lemma ===
=== Vector Addition is Well-Defined ===
To show that $+_{X/N}$ is well-defined, we show that if $x, x', y, y' \in N$ are such that: 
:$x + N = x' + N$
and:
:$y + N = y' + N$
we have:
:$\paren {x + y} + N = \paren {x' + y'} + N$
Since $x + N = x' + N$, we have:
:$x' \in x + N$
and so there exists $z \in N$ such that $x + z = x'$. 
Then $z = x' - x$, and so $x' - x \in N$. 
Similarly, we have that $y' - y \in N$.
So we have $\paren {x' + y'} - \paren {x + y} \in N$, since $N$ is a linear subspace of $X$, so that: 
:$N = \paren {x' + y'} - \paren {x + y} + N$
by the lemma.
So that: 
:$\paren {x + y} + N = \paren {x' + y'} + N$
=== Scalar Multiplication is Well-Defined ===
Let $\alpha \in K$. 
Suppose that $x, x' \in N$ are such that: 
:$x + N = x' + N$
We need to show that: 
:$\alpha x + N = \alpha x' + N$
As argued in proving vector addition is well-defined, we have that $x' - x \in N$.
Since $N$ is a linear subspace of $X$, we have: 
:$\alpha \paren {x' - x} = \alpha x' - \alpha x \in N$ 
So from the lemma, we have: 
:$N = \alpha x' - \alpha x + N$ 
So that:
:$\alpha x + N = \alpha x' + N$ 
=== Proof of ===
Let $\mathbf x, \mathbf y \in X/N$.
Then there exists $x, y \in N$ such that $\mathbf x = x + N$ and $\mathbf y = y + N$.
Then we have: 
=== Proof of ===
Let $\mathbf x, \mathbf y, \mathbf z \in X/N$.
Then there exists $x, y, z \in N$ such that $\mathbf x = x + N$, $\mathbf y = y + N$ and $\mathbf z = z + N$.
Then we have: 
=== Proof of ===
Let $0_{X/N} = 0_X + N = N$. 
We prove that $0_{X/N}$ satisfies the demand of .
Let $\mathbf x \in X/N$.
Then there exists $x \in X$ such that $\mathbf x = x + N$. 
We have: 
=== Proof of ===
Let $\mathbf x \in X/N$.
Then there exists $x \in X$ such that $\mathbf x = x + N$. 
Set $-\mathbf x = -x + N$.
We then have: 
=== Proof of ===
Let $\lambda, \mu \in K$. 
Let $\mathbf x \in X/N$.
Then there exists $x \in X$ such that $\mathbf x = x + N$.
Then we have: 
=== Proof of ===
Let $\lambda \in K$.
Let $\mathbf x, \mathbf y \in X/N$.
Then there exists $x, y \in N$ such that $\mathbf x = x + N$ and $\mathbf y = y + N$. 
We then have: 
=== Proof of ===
Let $\lambda, \mu \in K$.
Let $\mathbf x \in X/N$.
Then there exists $x \in X$ such that $\mathbf x = x + N$. 
We have: 
=== Proof of ===
Let $\mathbf x \in X/N$.
Then there exists $x \in X$ such that $\mathbf x = x + N$.
Then we have: 
Category:Quotient Vector Spaces
Category:Quotient Vector Space is Vector Space
Quotient Vector Space is Vector Space/Lemma

== Lemma ==
Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $N$ be a linear subspace of $X$.
Let $z \in N$.
Then: 
:$z + N = N$
== Proof ==
Since $N$ is a linear subspace, we have: 
:$z + N \subseteq N$
Conversely, let $x \in N$.
Then since $z \in N$ and $N$ is a linear subspace, we have $x - z \in N$. 
Then: 
:$x = z + \paren {x - z} \in z + N$ 
So that: 
:$N \subseteq z + N$
giving:
:$N = z + N$
Category:Quotient Vector Space is Vector Space
Quotient and Remainder to Number Base

== Theorem ==
Let $n \in \Z: n > 0$ be an integer.
Let $n$ be expressed in base $b$:
:$\ds n = \sum_{j \mathop = 0}^m {r_j b^j}$
that is:
:$n = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
Then:
:$\ds \floor {\frac n b} = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1}_b$
:$n \bmod b = r_0$
where:
:$\floor {\, \cdot \,}$ denotes the floor function;
:$n \bmod b$ denotes the modulo operation.
=== General Result ===
== Proof ==
From the Quotient-Remainder Theorem, we have:
:$\exists q, r \in \Z: n = q b + r$
where $0 \le b < r$.
We have that:
Hence we can express $n = q b + r$ where:
:$\ds q = \sum_{j \mathop = 1}^m {r_j b^{j - 1} }$
:$r = r_0$
where:
:$\ds \sum_{j \mathop = 1}^m {r_j b^{j - 1} } = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1}_b$
The result follows from the definition of the modulo operation.
== Example ==
This result is oftenSo often, in some business applications, that the author of this page is utterly sick of it, truth be told. used in computer algorithms for converting a date (in $yyyymmdd$ format) into a date object with separate day, month and year.
Performing the above "mod and div" operations on $20100209$, we get:
== Notes ==
Category:Number Theory
Quotient and Remainder to Number Base/General Result

== Theorem ==
Let $n \in \Z: n > 0$ be an integer.
Let $n$ be expressed in base $b$:
:$\ds n = \sum_{j \mathop = 0}^m {r_j b^j}$
that is:
:$n = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
Let $0 \le s \le m$.
Then:
:$\floor {\dfrac n {b^s} } = \sqbrk {r_m r_{m - 1} \ldots r_{s + 1} r_s}_b$
and:
:$\ds n \mod {b^s} = \sum_{j \mathop = 0}^{s - 1} {r_j b^j} = \sqbrk {r_{s - 1} r_{s - 2} \ldots r_1 r_0}_b$
where:
:$\floor {\, \cdot \,}$ denotes the floor function
:$n \mod b$ denotes the modulo operation.
== Proof ==
We have:
where:
In particular:
:$\sqbrk {r_m r_{m - 1} \ldots r_{s + 1} r_s}_b \le \dfrac n {b^s} < \sqbrk {r_m r_{m - 1} \ldots r_{s + 1} r_s}_b + 1$
By definition of floor function:
:$\floor {\dfrac n {b^s} } = \sqbrk {r_m r_{m - 1} \ldots r_{s + 1} r_s}_b$
Finally:
Category:Number Theory
Quotient is Primitive Recursive

== Theorem ==
Let $m, n \in \N$ be natural numbers.
Let us define the function $\operatorname {quot}: \N^2 \to \N$:
:$\map {\operatorname {quot} } {n, m} = \begin{cases}
\text{the quotient when } n \text{ is divided by } m & : m \ne 0 \\
0 & : m = 0
\end{cases}$
where the $\text {quotient}$ and $\text {remainder}$ are as defined in the Division Theorem:
:If $n = m q + r$, where $0 \le r < m$, then $q$ is the quotient.
Then $\operatorname {quot}$ is primitive recursive.
== Proof ==
We note that if $m \ne 0$ and $n = m q + r$, we have:
:$\dfrac n m = q + \frac r m$
Also note that $\dfrac n m$ and $\dfrac r m$ are rational numbers and not necessarily natural numbers.
Indeed, we have:
:$0 \le \dfrac r m < 1$
So if $m > 0$ then $\map {\operatorname {quot} } {n, m}$ is the floor $\floor {\dfrac n m}$ of $\dfrac n m$.
So we have:
:$\map {\operatorname {quot} } {n, m} = \begin{cases}
0 & : m = 0 \\
\floor {\dfrac n m} & : m \ne 0
\end{cases}$
Then we see that for $m \ne 0$:
:$\floor {\dfrac {n + 1} m} = \begin {cases}
\floor {\dfrac n m} + 1 & : m \divides \paren {n + 1} \\
\floor {\dfrac n m} & : \text {otherwise}
\end {cases}$
So for $m \ne 0$:
:$\map {\operatorname {quot} } {n + 1, m} = \begin{cases}
\map {\operatorname {quot} } {n, m} + 1 & : \map \rem {n + 1, m} = 0 \\
\map {\operatorname {quot} } {n, m} & : \map \rem {n + 1, m} \ne 0
\end{cases}$
Now note that:
:$\map {\overline {\sgn} } {\map \rem {n + 1, m} } = \begin{cases}
1 & : \map \rem {n + 1, m} = 0 \\
0 & : \map \rem {n + 1, m} \ne 0
\end{cases}$
So the $\operatorname {quot}$ is defined as:
:$\map {\operatorname {quot} } {0, m} = 0$
:$\map {\operatorname {quot} } {n + 1, m} = \map \sgn m \map {\operatorname {quot} } {n, m} + \map {\overline {\sgn} } {\map \rem {n + 1, m} }$
(note that the factor $\map \sgn m$ is needed to cover the case where $m = 0$).
Thus $\operatorname {quot}$ is obtained by primitive recursion (over the first variable, which is allowed by Permutation of Variables of Primitive Recursive Function) from the primitive recursive functions:
:Signum function $\sgn$
:Remainder $\rem$
:Addition
:Multiplication.
So it follows that $\operatorname {quot}$ is primitive recursive.
Category:Primitive Recursive Functions
Quotient of C*-Algebra by Closed Ideal is C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $I$ be a closed ideal of $A$.
Let $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ be the quotient normed algebra of $A$ modulo $I$.
Define $\ast : A/I \to A/I$ by:
:$\paren {a + I}^\ast = a^\ast + I$
for each $a \in I$. 
Then $\struct {A/I, \ast, \norm {\, \cdot \,}_{A/I} }$ is a $\text C^\ast$-algebra.
== Proof ==
We first show that $\ast$ is well-defined. 
That is, we show that:
:$a + I = b + I$ implies that $a^\ast + I = b^\ast + I$.
That is:
:$b - a \in I$ implies that $b^\ast - a^\ast \in I$.
From Closed Ideal of C*-Algebra is Self-Adjoint, $I$ is self-adjoint. 
Hence, if $b - a \in I$ then we have $b^\ast - a^\ast = \paren {b - a}^\ast \in I$ from $(\text C^\ast 2)$ in the definition of an involution.
So $a^\ast + I = b^\ast + I$.
From Sufficient Condition for C* Identity, it is enough to show:
:$\norm {a + I}^2_{A/I} \le \norm {a a^\ast + I}_{A/I}$
Let $B_I$ be the closed unit ball of $I$.
Let $B_I^+ \subseteq B_I$ be the set of positive elements of $A$ contained in $B_I$.
Let $\struct {A_+, \ast, \norm {\, \cdot \,} }$ be the unitization of $A$.
We then have:
From Element of C*-Algebra is Positive iff Positive in Unitization, $\tuple {x, 0}$ is positive in $A_+$ for each $x \in B_I^+$.
From Spectrum of Element of Banach Algebra is Bounded, we therefore have $\map {\sigma_{A_+} } {\tuple {x, 0} } \subseteq \closedint 0 1$.
From the Spectral Mapping Theorem for Polynomials, we have $\map {\sigma_{A_+} } {\tuple {0, 1} - \tuple {x, 0} } \subseteq \closedint 0 1$.
Then, from Spectral Radius of Normal Element of C*-Algebra Equal to Norm we have $\norm {\tuple {0, 1} - \tuple {x, 0} }_\ast \le 1$.
We therefore have:
Hence:
:$\norm {a + I}^2_{A/I} \le \norm {a a^\ast + I}_{A/I}$
as desired.
== Sources ==
* : Theorem $5.6$
Category:Quotient C*-Algebras
Quotient of Cauchy Sequences is Metric Completion

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences in $R$.
Let $\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\norm {\, \cdot \,}: \CC \,\big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \,\big / \NN$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN} = \lim_{n \mathop \to \infty} \norm{x_n}$
Let $d'$ be the metric induced by $\struct {\CC \,\big / \NN, \norm {\, \cdot \,} }$
Let $\phi: R \to \CC \,\big / \NN$ be the mapping from $R$ to the quotient ring $\CC \,\big / \NN$ defined by:
:$\forall a \in R: \map \phi a = \sequence {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {a, a, a, \ldots}$.
Then:
:$\struct {\CC \,\big / \NN, d'}$ is the metric completion of $\struct {R,d}$
and:
:$\map \phi R$ is a dense subset of $\CC \,\big / \NN$
== Proof ==
By the definition of the metric induced by a norm then:
:a sequence $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$ $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, d}$.
So $\CC$ is the set of Cauchy sequences in $\struct {R, d}$.
Let $\sim$ be the equivalence relation on $\CC$ defined by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Let $\tilde {\CC} = \CC / \sim$ denote the set of equivalence classes under $\sim$.
For $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the equivalence class containing $\sequence {x_n}$.
=== Lemma 1 ===
Let $\tilde d: \tilde {\CC} \times \tilde {\CC} \to \R_{\ge 0}$ be the metric defined by:
:$\ds \map {\tilde d} {\eqclass {x_n} {}, \eqclass {y_n} {} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
By Completion of a Metric Space then:
:$\struct {\tilde {\CC}, \tilde d}$ is the metric completion of $\struct {R, d}$.
=== Lemma 2 ===
Let $\tilde \phi: R \to \tilde {\CC}$ be the mapping defined by:
:$\map {\tilde \phi} a = \eqclass {a, a, a, \dotsc} {}$
where $\eqclass {a, a, a, \dotsc} {}$ denotes the equivalence class containing the constant sequence $\sequence {a, a, a, \dotsc}$.
By Completion of a Metric Space then:
:$\map {\tilde \phi} R$ is a dense subset of $\tilde {\CC}$.
By Lemma 1 then:
:$\forall \sequence {x_n} \in \CC: \sequence {x_n} + \NN = \eqclass {x_n} {}$
In particular:
:$\forall a \in R: \sequence {a, a, a, \dotsc} + \NN = \eqclass {a, a, a, \dotsc} {}$
That is:
:$\forall a \in R: \map \phi a = \map {\tilde \phi} a$
Hence $\phi = \tilde \phi$.
The result follows.
== Sources ==
* : $\S 3.2$: Completions: Proposition $3.2.12$
* : $\S 1.3$: Construction of the completion of a normed field: Theorem $1.22$
Category:Normed Division Rings
Category:Complete Metric Spaces
Category:Completion of Normed Division Ring
Quotient of Cauchy Sequences is Metric Completion/Lemma 1

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\sim$ be the equivalence relation on $\CC$ defined by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Let $\tilde \CC = \CC / \sim$ denote the set of equivalence classes under $\sim$.
For $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the equivalence class containing $\sequence {x_n}$.
Then:
:$\quad \CC \,\big / \NN = \tilde \CC$ 
== Proof ==
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $\CC$.
Then:
Hence:
:$\sequence {x_n}$ and $\sequence {y_n}$ belong to the same equivalence class in $\CC \,\big / \NN$ $\sequence {x_n}$ and $\sequence {y_n}$ belong to the same equivalence class in $\tilde \CC$.
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Completion of Normed Division Ring
Quotient of Cauchy Sequences is Metric Completion/Lemma 2

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $d$ be the metric induced by $\struct {R, \norm {\, \cdot \,} }$.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\norm {\, \cdot \,}: \CC \,\big / \NN \to \R_{\ge 0}$ be the norm on the quotient ring $\CC \,\big / \NN$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN} = \lim_{n \mathop \to \infty} \norm {x_n}$
Let $d'$ be the metric induced by $\struct {\CC \,\big / \NN, \norm {\, \cdot \,} }$
Let $\sim$ be the equivalence relation on $\CC$ defined by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Let $\tilde {\CC} = \CC / \sim$ denote the set of equivalence classes under $\sim$.
For $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the equivalence class containing $\sequence {x_n}$.
Let $\tilde d: \tilde {\CC} \times \tilde {\CC} \to \R_{\ge 0}$ be the metric defined by:
:$\map {\tilde d} {\eqclass {x_n} {}, \eqclass {y_n} {} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
Then:
:$\quad d' = \tilde d$ 
== Proof ==
By Lemma 1 of Quotient of Cauchy Sequences is Metric Completion we have that:
:$\CC \,\big / \NN = \tilde {\CC}$
Let $\eqclass {x_n} {}$ and $\eqclass {x_n} {}$ be equivalence classes in $\CC \,\big / \NN = \tilde {\CC}$.
Then:
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Completion of Normed Division Ring
Quotient of Commutative Ring by Nilradical is Reduced

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {N, +, \circ}$ denote the nilradical of $R$.
The quotient ring $R / N$ is a reduced ring.
== Proof ==
From Nilpotent Elements of Commutative Ring form Ideal, $\struct {N, +, \circ}$ is an ideal of $\struct {R, +, \circ}$.
Hence the quotient ring $R / N$ is defined.
By definition of the ideal of $\struct {R, +, \circ}$, $N$ is the zero of $R / N$.
Let $\paren {x + N}^n \in N$.
Then:
:$x^n \in N$
and so:
:$\paren {x^n}^m = 0_R$
for some $m \in \Z_{>0}$.
Hence:
:$x \in N$
and so:
:$x + N = N$
That is, if a (left) coset of $N$ is nilpotent in $R / N$ that (left) coset is $N$ itself.
The result follows.
== Sources ==
* : Chapter $9$: Rings: Exercise $11 \ \text {(ii)}$
Category:Nilpotent Ring Elements
Category:Commutative Rings
Category:Ideal Theory
Quotient of Complex Conjugates

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\overline z$ be the complex conjugate of the complex number $z$.
Then:
:$\overline {\paren {\dfrac {z_1} {z_2} } } = \dfrac {\paren {\overline {z_1} } } {\paren {\overline {z_2} } }$
for $z_2 \ne 0$.
== Proof ==
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $56 \ \text{(a)}$
* 
Category:Complex Conjugates
Quotient of Divisible Module is Divisible

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a divisible left $R$-module.
Let $N \subseteq M$ be an $R$-submodule.
Then the quotient module $M / N$ is divisible.
== Proof ==
Let $r \in R$ be a regular element of $R$.
Hence by definition $r$ is not a zero divisor of $R$.
Let $\eqclass m {} \in M / N$ be an arbitrary element represented by $m \in M$.
Since $M$ is divisible, there exists some $m' \in M$ such that $m = r m'$.
By definition of scalar multiplication on the quotient module $M / N$:
:$r \eqclass m {} = \eqclass {m'} {}$
It follows, that $M / N$ is divisible.
Category:Homological Algebra
Category:Module Theory
Quotient of F-Space by Closed Linear Subspace is F-Space

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, d}$ be an $F$-space.
Let $N$ be a closed linear subspace of $X$. 
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $d_N$ be the quotient metric on $X/N$ induced by $d$.
Then $\struct {X/N, d_N}$ is an $F$-Space.
== Proof ==
Let $\pi : X \to X/N$ be a quotient mapping.
Let $\tau_N$ be the quotient topology of $X$ modulo $N$.
From Quotient Metric on Vector Space induces Quotient Topology, $d_N$ induces $\tau_N$ and $d_N$ is an invariant metric. 
From Quotient Topological Vector Space is Topological Vector Space, $\struct {X/N, \tau_N}$ is a topological vector space.
To establish that $\struct {X/N, d_N}$ is an $F$-space, we only need to show that each Cauchy sequence in $\struct {X/N, d_N}$ converges.
Let $\sequence {u_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {X/N, d_N}$.
Pick $n_1 \in \N$ so that:
:$\map {d_N} {u_n, u_m} < 2^{-1}$ for all $n, m \ge n_1$.
Now for each $j \ge 2$, inductively pick $n_j > n_{j - 1}$ such that:
:$\map {d_N} {u_n, u_m} < 2^{-j}$ for all $n, m \ge n_j$
Then for each $j \in \N$, we have:
:$\map {d_N} {u_{n_j}, u_{n_{j + 1} } } < 2^{-j}$
For each $j \in \N$, pick $v_j \in X$ such that $\map \pi {v_j} = u_{n_j}$.
Then for $j \ge 2$, we have: 
:$\ds \inf_{z \mathop \in N} \map d {v_j - v_{j + 1}, z} < 2^{-j}$
So there exists $z_j \in N$ such that:
:$\map d {v_j - v_{j + 1}, z_j} < 2^{-j}$ 
Then since $d$ is invariant, we have:
:$\map d {v_j, v_{j + 1} + z_j} < 2^{-j}$
Set $x_1 = v_1$ and $x_j = v_j + z_{j - 1}$ for $j \ge 2$.
Then, we have:
:$\map d {x_j, x_{j + 1} } < 2^{-j}$ for each $j \in \N$
and:
:$\map \pi {x_j} = \map \pi {v_j} = u_{n_j}$
from Kernel of Quotient Mapping.
Now, by , we have, for each $m > n$:
:$\ds \map d {x_n, x_m} \le \sum_{j \mathop = n}^{m - 1} \map d {x_j, x_{j + 1} } < \sum_{j \mathop = n}^{m - 1} 2^{-j}$
From Sum of Infinite Geometric Sequence, we have:
:$\ds \sum_{j \mathop = 0}^\infty 2^{-j}$ converges.
From Cauchy's Convergence Criterion for Series, it follows that for each $\epsilon > 0$ there exists $N \in \N$ such that for $m > n \ge 0$, we have:
:$\ds \sum_{j \mathop = n}^{m - 1} 2^{-j} < \epsilon$
so that:
:$\map d {x_n, x_m} < \epsilon$
Hence $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence.
Since $\struct {X, d}$ is an $F$-space, it is in particular a complete metric space.
So $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$.
From the definition of the quotient topology, $\pi$ is continuous.
From Continuous Mapping is Sequentially Continuous, $\pi$ is sequentially continuous.
So $\map \pi {x_j} \to \map \pi x$ as $j \to \infty$.
That is, $u_{n_j} \to \map \pi x$ as $j \to \infty$.
From Convergent Subsequence of Cauchy Sequence, it follows that $\sequence {u_n}_{n \mathop \in \N}$ converges to $\map \pi x$.
Hence every Cauchy sequence in $\struct {X/N, d_N}$ converges.
Hence $\struct {X/N, d_N}$ is a complete metric space.
Hence $\struct {X/N, d_N}$ is an $F$-space.
Category:F-Spaces
Category:Quotient Topological Vector Spaces