nyuuzyou/wikis huggingface dataset:
Separation Axioms on Double Pointed Topology/T3 Axiom

== Theorem ==
Let $T_1 = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_1 \times D, \tau}$ be the double pointed topological space on $T_1$.
Then $T \times D$ is a $T_3$ space $T$ is also a $T_3$ space.
== Proof ==
Let $S' = S \times \set {a, b}$.
Let $F' \subseteq S'$ such that $F'$ is closed in $T \times D$.
Then $F' = F \times \set {a, b}$ or $F' = F \times \O$ by definition of the double pointed topology.
If $F' = F \times \O$ then $F' = \O$ from Cartesian Product is Empty iff Factor is Empty, and the result is trivial.
So suppose $F' = F \times \set {a, b}$.
From Open and Closed Sets in Multiple Pointed Topology it follows that $F$ is closed in $T$.
Let $y' = \tuple {y, q} \in \relcomp {S'} {F'}$.
Then $y \notin F$.
Suppose that $T$ is a $T_3$ space.
Then by definition:
:For any closed set $F$ of $T$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
Then $y' \in V \times \set {a, b}$ and $F' \subseteq U \times \set {a, b}$ and:
:$U \times \set {a, b} \cap V \times \set {a, b} = \O$
demonstrating that $T \times D$ is a $T_3$ space.
Now suppose that $T \times D$ is a $T_3$ space.
Then $\exists U', V' \in S': y' \in V'$ and $F' \subseteq U'$ such that $U' \cap V' = \O$.
As $D$ is the indiscrete topology it follows that:
:$U' = U \times \set {a, b}$
:$V' = V \times \set {a, b}$
for some $U, V \subseteq T$.
From Open and Closed Sets in Multiple Pointed Topology it follows that $U$ and $V$ are open in $T$.
As $U' \cap V' = \O$ it follows that $U \cap V = \O$.
It follows that $F$ and $y$ fulfil the conditions that make $T$ a $T_3$ space.
Hence the result.
== Sources ==
* : Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$
Category:Separation Axioms on Double Pointed Topology
Category:T3 Spaces
Separation Axioms on Double Pointed Topology/T4 Axiom

== Theorem ==
Let $T_1 = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_1 \times D, \tau}$ be the double pointed topological space on $T_1$.
Then $T \times D$ is a $T_4$ space $T$ is also a $T_4$ space.
== Proof ==
Let $S' = S \times \set {a, b}$.
Let $H' \subseteq S'$ such that $H$ is closed in $T \times D$.
Then $H' = H \times \set {a, b}$ or $H' = H \times \O$ by definition of the double pointed topology.
If $H' = H \times \O$ then $H' = \O$ from Cartesian Product is Empty iff Factor is Empty, and the result is trivial.
So suppose $H' = H \times \set {a, b}$.
From Open and Closed Sets in Multiple Pointed Topology it follows that $H$ is closed in $T$.
Suppose that $T$ is a $T_4$ space.
Then by definition:
:For any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
Then $A \times \set {a, b} \subseteq U \times \set {a, b}$ and $B \times \set {a, b} \subseteq V \times \set {a, b}$ and:
:$U \times \set {a, b} \cap V \times \set {a, b} = \O$
demonstrating that $T \times D$ is a $T_4$ space.
Now suppose that $T \times D$ is a $T_4$ space.
Then $\exists U', V' \in S': A' \subseteq U'$ and $B' \subseteq V'$ such that $U' \cap V' = \O$.
As $D$ is the indiscrete topology it follows that:
:$U' = U \times \set {a, b}$
:$V' = V \times \set {a, b}$
for some $U, V \subseteq T$.
From Open and Closed Sets in Multiple Pointed Topology it follows that $U$ and $V$ are open in $T$.
As $U' \cap V' = \O$ it follows that $U \cap V = \O$.
It follows that $A$ and $B$ fulfil the conditions that make $T$ a $T_4$ space.
Hence the result.
== Sources ==
* : Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$
Category:Separation Axioms on Double Pointed Topology
Category:T4 Spaces
Separation Axioms on Double Pointed Topology/T5 Axiom

== Theorem ==
Let $T_1 = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_1 \times D, \tau}$ be the double pointed topological space on $T_1$.
Then $T \times D$ is a $T_5$ space $T$ is also a $T_5$ space.
== Proof ==
=== Necessary Condition ===
Suppose that $T \times D$ is a $T_5$ space.
Let $A, B \subseteq T$ be two separated sets.
First, we will show that $A \times D$ and $B \times D$ are also separated.
To this end, observe that:
Mutatis mutandis, also:
:$\paren {B \times D}^- \cap \paren {A \times D} = \O$
Thus $A \times D$ and $B \times D$ are separated.
Since $T \times D$ is assumed to be a $T_5$ space, find open sets $U, V$ such that:
:$U \cap V = \O$
:$A^- \times D \subseteq U$
:$B^- \times D \subseteq V$
By Open Sets of Double Pointed Topology, there are open sets $U', V'$ of $\tau$ such that:
:$U = U' \times D$
:$V = V' \times D$
We have, furthermore:
From Cartesian Product is Empty iff Factor is Empty, $U' \cap V' = \O$ since $D$ is non-empty.
By Cartesian Product of Subsets, we also infer:
:$A^- \subseteq U'$
:$B^- \subseteq V'$
It follows that $T$ is a $T_5$ space.
=== Sufficient Condition ===
Suppose that $T$ is a $T_5$ space.
Let $X, Y \subseteq T \times D$ be separated sets.
By Closure of Subset of Double Pointed Topological Space, we know that:
:$X^- = {\pr_1 \sqbrk X}^- \times D$
:$Y^- = {\pr_1 \sqbrk Y}^- \times D$
where $\pr_1$ is the first projection.
Suppose now that for some $s \in S$, one has:
:$s \in {\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y$
Then it would be that for some $d \in D$:
:$\tuple {s, d} \in Y$
However, by the expression for $X^-$ above, also $\tuple {s, d} \in X^-$.
This is a contradiction, for $X$ and $Y$ were assumed separated.
Thus:
:${\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y = \O$
and analogously, we derive:
:$\pr_1 \sqbrk X \cap {\pr_1 \sqbrk Y}^- = \O$
that is, $\pr_1 \sqbrk X$ and $\pr_1 \sqbrk Y$ are separated.
Since $T$ is a $T_5$ space, we find open sets $U, V$ of $\tau$ such that:
:$U \cap V = \O$
:${\pr_1 \sqbrk X}^- \subseteq U$
:${\pr_1 \sqbrk Y}^- \subseteq V$
By Cartesian Product of Subsets, we have:
:$X^- = {\pr_1 \sqbrk X}^- \times D \subseteq U \times D$
:$Y^- = {\pr_1 \sqbrk Y}^- \times D \subseteq V \times D$
By Open Sets of Double Pointed Topology, $U \times D$ and $V \times D$ are also open sets in $S \times D$.
Finally, we compute:
That is, $T \times D$ is a $T_5$ space.
== Sources ==
* : Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$
Category:Separation Axioms on Double Pointed Topology
Category:T5 Spaces
Separation Properties Not Preserved by Expansion

== Theorem ==
These separation properties are not generally preserved under expansion:
:$T_3$ Space
:Regular Space
:$T_4$ Space
:Completely Regular Space
:$T_5$ Space
:Normal Space
:Completely Normal Space
== Proof ==
Let $\struct {\R, \tau_1}$ be the set of real numbers under the usual (Euclidean) topology.
Let $\struct {\R, \tau_2}$ be the indiscrete rational extension of $\struct {\R, \tau_1}$.
From Metric Space fulfils all Separation Axioms, $\struct {\R, \tau_1}$ is:
:$T_3$ Space
:Regular Space
:$T_4$ Space
:Completely Regular Space
:$T_5$ Space
:Normal Space
:Completely Normal Space
But we have:
:Indiscrete Rational Extension of Real Number Line is not $T_3$ Space
:Indiscrete Rational Extension of Real Number Line is not $T_4$ Space
:Indiscrete Rational Extension of Real Number Line is not $T_5$ Space
By definition, $\struct {\R, \tau_2}$ is an expansion of $\struct {\R, \tau_1}$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Separation Axioms
Separation Properties Preserved by Expansion

== Theorem ==
These separation properties are preserved under expansion:
:$T_0$ (Kolmogorov) Space
:$T_1$ (Fréchet) Space
:$T_2$ (Hausdorff) Space
:$T_{2 \frac 1 2}$ (Completely Hausdorff) Space
== Proof ==
Let $S$ be a set.
Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be topological spaces based on $S$ such that $\tau_2$ is an expansion of $\tau_1$.
That is, let $\tau_1$ and $\tau_2$ be topologies on $S$ such that $\tau_1 \subseteq \tau_2$.
Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$.
From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed.
We also have Identity Mapping is Bijection.
So we can directly apply:
:$T_0$ (Kolmogorov) Space is Preserved under Closed Bijection
:$T_1$ (Fréchet) Space is Preserved under Closed Bijection
:$T_2$ (Hausdorff) Space is Preserved under Closed Bijection
:$T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Closed Bijection
and hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Separation Axioms
Separation Properties Preserved in Subspace

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T_H$ be a subspace of $T$.
If $T$ has one of the following properties then $T_H$ has the same property:
:$T_0$ (Kolmogorov) Property
:$T_1$ (Fréchet) Property
:$T_2$ (Hausdorff) Property
:$T_{2 \frac 1 2}$ (Completely Hausdorff) Property
:$T_3$ Property
:$T_{3 \frac 1 2}$ Property
:$T_5$ Property
That is, the above properties are all hereditary.
=== Corollary ===
== Proof ==
:$T_0$ Property is Hereditary
:$T_1$ Property is Hereditary
:$T_2$ Property is Hereditary
:Completely Hausdorff Property is Hereditary
:$T_3$ Property is Hereditary
:$T_3 \frac 1 2$ Property is Hereditary
:$T_5$ Property is Hereditary
== $T_4$ Space ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Topological Subspaces
Category:Separation Axioms
Separation Properties Preserved in Subspace/Corollary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T_H$ be a subspace of $T$.
If $T$ has one of the following properties then $T_H$ has the same property:
:Regular Property
:Tychonoff (Completely Regular) Property
:Completely Normal Property
That is, the above properties are all hereditary.
== Proof ==
A regular space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3$ space.
Hence from $T_0$ Property is Hereditary and $T_3$ Property is Hereditary it follows that the property of being a regular space is also hereditary.
A Tychonoff (completely regular) space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3 \frac 1 2$ space.
Hence from $T_0$ Property is Hereditary and $T_3 \frac 1 2$ Property is Hereditary it follows that the property of being a Tychonoff (completely regular) space is also hereditary.
A completely normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_5$ space.
Hence from $T_1$ Property is Hereditary and $T_5$ Property is Hereditary it follows that the property of being a completely normal space is also hereditary.
== $T_4$ Space ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Topological Subspaces
Category:Separation Axioms
Separation Properties Preserved in Subspace/T4 Space

== Theorem ==
Of all the separation axioms, the $T_4$ axiom differs from the others.
It does not necessarily hold that a subspace of a $T_4$ space is also a $T_4$ space, unless that subspace is closed.
This is demonstrated in the result $T_4$ Property is not Hereditary.
However, it is the case that the $T_4$ property is weakly hereditary.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Topological Subspaces
Category:T4 Spaces
Separation Properties Preserved under Topological Product

== Theorem ==
Let $\mathbb S = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod_{i \mathop \in I} \struct{S_i, \tau_i}$ be the product space of $\mathbb S$.
Then $T$ has one of the following properties each of $\struct {S_i, \tau_i}$ has the same property:
:$T_0$ (Kolmogorov) Property
:$T_1$ (Fréchet) Property
:$T_2$ (Hausdorff) Property
:$T_{2 \frac 1 2}$ (Completely Hausdorff) Property
:$T_3$ Property
:$T_{3 \frac 1 2}$ Property
If $T = \struct {S, \tau}$ has one of the following properties then each of $\struct {S_i, \tau_i}$ has the same property:
:$T_4$ Property
:$T_5$ Property
but the converse does not necessarily hold.
=== Corollary ===
== Proof ==
:Product Space is $T_0$ iff Factor Spaces are $T_0$
:Product Space is $T_1$ iff Factor Spaces are $T_1$
:Product Space is $T_2$ iff Factor Spaces are $T_2$
:Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff
:Product Space is $T_3$ iff Factor Spaces are $T_3$
:Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$
:Factor Spaces are $T_4$ if Product Space is $T_4$
:Factor Spaces are $T_5$ if Product Space is $T_5$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Separation Axioms
Category:Product Spaces
Separation Properties Preserved under Topological Product/Corollary

== Theorem ==
Let $\SS = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds T = \struct {S, \tau} = \prod_{i \mathop \in I} \struct {S_i, \tau_i}$ be the product space of $\SS$.
$T = \struct {S, \tau}$ has one of the following properties each of $\struct {S_i, \tau_i}$ has the same property:
:Regular Property
:Tychonoff (Completely Regular) Property
If $T = \struct {S, \tau}$ has one of the following properties then each of $\struct {S_i, \tau_i}$ has the same property:
:Normal Property
:Completely Normal Property
but the converse does not necessarily hold.
== Proof ==
A regular space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3$ space.
Hence from:
:Product Space is $T_0$ iff Factor Spaces are $T_0$
and
:Product Space is $T_3$ iff Factor Spaces are $T_3$
it follows that $T$ is a regular space each of $\struct {S_i, \tau_i}$ is a regular space.
A Tychonoff (completely regular) space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_{3 \frac 1 2}$ space.
Hence from:
:Product Space is $T_0$ iff Factor Spaces are $T_0$
and
:Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$
it follows that $T$ is a Tychonoff space each of $\struct {S_i, \tau_i}$ is a Tychonoff space.
A normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_4$ space.
Hence from:
:Product Space is $T_1$ iff Factor Spaces are $T_1$
and
:Factor Spaces are $T_4$ if Product Space is $T_4$
it follows that if $T$ is a normal space then each of $\struct {S_i, \tau_i}$ is a normal space.
A completely normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_5$ space.
Hence from:
:Product Space is $T_1$ iff Factor Spaces are $T_1$
and
:Factor Spaces are $T_5$ if Product Space is $T_5$
it follows that if $T$ is a completely normal space then each of $\struct {S_i, \tau_i}$ is a completely normal space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Separation Axioms
Category:Product Spaces
Separation Properties in Open Extension of Particular Point Topology

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton or a doubleton.
Let $T^*_{\bar q} = \struct {S^*_q, \tau^*_{\bar q} }$ be an open extension space of $T$.
Then:
:$T^*_{\bar q}$ is a $T_0$ (Kolmogorov) space.
:$T^*_{\bar q}$ is a $T_4$ (space.
:$T^*_{\bar q}$ is not a $T_1$ (Fréchet) space.
:$T^*_{\bar q}$ is not a $T_5$ (space.
== Proof ==
We have that a Particular Point Space is $T_0$.
Then from Condition for Open Extension Space to be $T_0$ Space, it follows that $T^*_{\bar q}$ is a $T_0$ (Kolmogorov) space.
We have directly that:
:An Open Extension Topology is not $T_1$.
:An Open Extension Topology is $T_4$.
Finally, we have that a Particular Point Topology with three points or more is not $T_4$.
From $T_5$ Space is $T_4$ Space, it follows that $T$ is not a $T_5$ space.
It follows from Condition for Open Extension Space to be $T_5$ Space that $T^*_{\bar q}$ is not a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $10$
Category:Separation Axioms
Category:Open Extension Topology
Category:Particular Point Topology
Separation Properties of Alexandroff Extension of Rational Number Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ satisfies no Tychonoff separation axioms higher than a $T_1$ (Fréchet) space.
== Proof ==
From Alexandroff Extension of Rational Number Space is $T_1$ Space, $T^*$ is a $T_1$ space.
From Alexandroff Extension of Rational Number Space is not Hausdorff, $T^*$ is not a $T_2$ (Hausdorff) space.
From Completely Hausdorff Space is Hausdorff Space, $T^*$ is not a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $4$
Category:Alexandroff Extensions
Category:Separation Axioms
Category:Rational Number Space
Septuple Angle Formulas

== Theorem ==
=== Septuple Angle Formula for Cosine ===
where $\sin, \cos, \tan$ denote sine, cosine and tangent respectively.
== Also see ==
* Quintuple Angle Formulas
* Sextuple Angle Formulas
Category:Septuple Angle Formulas
Category:Multiple Angle Formulas
Category:Trigonometric Identities
Septuple Angle Formulas/Cosine

== Theorem ==
:$\cos 7 \theta = 64 \cos^7 \theta - 112 \cos^5 \theta + 56 \cos^3 \theta - 7 \cos \theta$
where $\cos$ denotes cosine.
== Proof ==
Hence:
Category:Cosine Function
Category:Septuple Angle Formulas
Category:Septuple Angle Formula for Cosine
Sequence Characterization of Open Sets

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $G \subseteq X$.
Then the following are equivalent:
:$(1): \quad G \subseteq X$ is an open set of $\struct {X, d}$
:$(2): \quad \forall x \in G: \forall \sequence {x_n} \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \sequence {x_n} \in G$
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $G \subseteq X$ be open in $\struct {X, d}$.
Let $x \in G$.
Let $\sequence {x_n}$ be a sequence in $X$ such that $x_n \to x$.
By definition of open set, there exists $\epsilon > 0$ such that:
: $B_\epsilon \left({x}\right) \subseteq G$
where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$.
Since $x_n \to x$, there exists $n_0 \in \N$ such that:
: $\forall n \ge n_0: d \left({x_n, x}\right) < \epsilon$
Thus:
: $\forall n \ge n_0: x_n \in B_\epsilon \left({x}\right) \subseteq G$
=== $(2)$ implies $(1)$ ===
Let $G \subseteq X$.
Let:
:$\forall x \in G: \forall \sequence {x_n} \in X: x_n \to x: \exists n_0 \in \N: \forall n \ge n_0: \sequence {x_n} \in G$
 $G$ is not open in $\struct {X, d}$.
Then:
: $\exists x \in G: \forall \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \nsubseteq G$
Therefore, for $n = 1, 2, \ldots$ we can inductively find a sequence:
: $x_n \in B_{1/n} \left({x}\right) \cap \left({X \setminus G}\right) \implies x_n \notin G$
and:
: $\forall n \in \N: d \left({x_n, x}\right) < \dfrac 1 n \implies x_n \to x \in G$ and $\forall n \in \N: x_n \notin G$
This contradicts hypothesis $(2)$.
Thus, by Proof by Contradiction, $G$ is open in $\struct {X, d}$.
== Also see ==
* Definition:Open Set (Metric Space)
Category:Open Sets
Sequence Converges to Point Relative to Metric iff it Converges Relative to Induced Topology

== Theorem ==
Let $M = \struct {S, d}$ be a metric space or a pseudometric space.
Let $T = \struct {S, \tau}$ be the topological space induced by $d$.
Let $\sequence {x_n}$ be a infinite sequence in $S$.
Let $l \in S$.
Then $\sequence {x_n}$ converges to $l$ relative to $d$ $\sequence {x_n}$ converges to $l$ relative to $\tau$.
== Proof ==
=== Necessary Condition ===
Suppose that $\sequence {x_n}$ converges to $l$ relative to $d$.
When $\map {B_\epsilon} l$ denotes the open $\epsilon$-ball of $l$, this means:
:$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies x_n \in \map {B_\epsilon} l$
Let $U \in \tau$ with $l \in U$.
By definition of induced topology, there exists $\epsilon_0 \in \R_{>0}$ such that:
:$\map {B_{\epsilon_0} } l \subseteq U$
Then there exists $N_0 \in \R$ such that for all $n > N_0$:
:$x_n \in \map {B_{\epsilon_0} } l \subseteq U$
Hence, $\sequence {x_n}$ converges to $l$ in the induced topology $\tau$.
=== Sufficient Condition ===
Suppose that $\sequence {x_n}$ converges to $l$ in $\tau$.
Let $\epsilon \in \R_{>0}$.
Then $l \in \map {B_\epsilon} l$, and $\map {B_\epsilon} l \in \tau$.
Then there exists $N \in \N$ such that for all $n > N$, we have $x_n \in \map {B_\epsilon} l$.
Hence, $\sequence {x_n}$ converges to $l$ relative to $d$.
Category:Metric Spaces
Sequence Converges to Within Half Limit

== Theorem ==
=== Sequence of Real Numbers ===
=== Sequence of Complex Numbers ===
=== Sequence in Normed Division Ring ===
== Also see ==
This is used in the Quotient Rule for Sequences.
Although this result seems a little trivial, it is often crucial to know that a sequence will be "eventually non-zero" so we know we can legitimately divide by it.
Category:Limits of Sequences
Sequence Converges to Within Half Limit/Complex Numbers

== Theorem ==
Let $\sequence {z_n}$ be a sequence in $\C$.
Let $\sequence {z_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.
Then:
:$\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$
== Proof ==
Suppose $l > 0$.
Let us choose $N$ such that:
:$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$
Then:
Category:Limits of Sequences
Sequence Converges to Within Half Limit/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero $0$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l \ne 0$
Then:
:$\exists N: \forall n > N: \norm {x_n} > \dfrac {\norm l} 2$
== Proof ==
Since $l \ne 0$, by :
:$\norm l > 0$
Let us choose $N$ such that:
:$\forall n > N: \norm {x_n - l} < \dfrac {\norm l} 2$
Then:
Category:Sequences
Category:Limits of Sequences
Category:Normed Division Rings
Sequence Converges to Within Half Limit/Real Numbers

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Suppose $l > 0$.
Then:
:$\exists N: \forall n > N: x_n > \dfrac l 2$
Similarly, suppose $l < 0$.
Then:
:$\exists N: \forall n > N: x_n < \dfrac l 2$
== Proof ==
Suppose $l > 0$.
From the definition of convergence to a limit:
:$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$
That is, $l - \epsilon < x_n < l + \epsilon$.
As this is true for all $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.
Thus:
:$\exists N: \forall n > N: x_n > \dfrac l 2$
as required.
Now suppose $l < 0$.
By a similar argument:
:$\forall \epsilon > 0: \exists N: \forall n > N: l - \epsilon < x_n < l + \epsilon$
Thus it is also true for $\epsilon = -\dfrac l 2$ for some value of $N$.
Thus:
:$\exists N: \forall n > N: x_n < \dfrac l 2$
as required.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (2)$
Category:Limits of Sequences