nyuuzyou/wikis huggingface dataset:
Negative of Triangular Matrix

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
If $\mathbf A$ is an upper triangular matrix, then so is $-\mathbf A$.
If $\mathbf A$ is a lower triangular matrix, then so is $-\mathbf A$.
== Proof ==
From the definition of negative matrix, we have:
:$\forall i, j \in \closedint 1 n: \sqbrk {-a}_{i j} = -a_{i j}$
If $\mathbf A$ is an upper triangular matrix, we have:
:$\forall i > j: a_{i j} = 0$
Hence:
:$\forall i > j: \sqbrk {-a}_{i j} = -a_{i j} = 0$
and so $-\mathbf A$ is itself upper triangular.
Similarly, if $\mathbf A$ is a lower triangular matrix, we have:
:$\forall i < j: a_{i j} = 0$
Hence:
:$\forall i < j: \sqbrk {-a}_{i j} = -a_{i j} = 0$
and so $-\mathbf A$ is itself lower triangular.
Category:Negative Matrices
Category:Triangular Matrices
Negative of Upper Bound of Set of Real Numbers is Lower Bound of Negatives

== Theorem ==
Let $S$ be a subset of the real numbers $\R$.
Let $T = \set {x \in \R: -x \in S}$ be the set of negatives of the elements of $S$.
Then:
:$U$ is an upper bound of $S$
:
:$-U$ is a lower bound of $T$.
== Proof ==
Let $V$ be the set defined as:
:$V = \set {x \in \R: -x \in T}$
From Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives:
:$B$ is a lower bound of $T$
:
:$-B$ is an upper bound of $V$
Then we have:
That is:
:$B$ is a lower bound of $T$
:
:$-B$ is an upper bound of $S$
The result follows by setting $U = -B$.
== Also see ==
* Negative of Lower Bound of Set of Real Numbers is Upper Bound of Negatives
Category:Real Analysis
Neighborhood Basis in Cartesian Product under Chebyshev Distance

== Theorem ==
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
For all $i \in \set {1, 2, \ldots, n}$, let $\BB_{a_i}$ be a basis for the system of neighborhoods for $a_i$ in $M_i$.
Let $\BB_a$ be the cartesian product of $\BB_{a_1}, \BB_{a_2}, \ldots, \BB_{a_n}$:
:$\ds \BB_a = \prod_{i \mathop = 1}^n \BB_{a_i}$
Then $\BB_a$ is a basis for the system of neighborhoods for $a$ in $M$.
== Proof ==
Let $N$ be a neighborhood of $a$ in $M$.
Then by definition of neighborhood:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} {a; d_\infty} \subseteq N$
where $\map {B_\epsilon} {a; d_\infty}$ is the open $\epsilon$-ball of $a$ under $d_\infty$.
From Open Ball in Cartesian Product under Chebyshev Distance:
:$\ds \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$
From Open Ball is Neighborhood of all Points Inside, each $\map {B_\epsilon} {a_i; d_i}$ is a neighborhood of $a_i$.
By definition of basis for system of neighborhoods:
:$\exists N_i \in \BB_{a_i}: N_i \subseteq \map {B_\epsilon} {a_i; d_i}$
From Cartesian Product of Subsets:
:$\ds \prod_{i \mathop = 1}^n N_i \subseteq \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i} \subseteq N$
But $\ds \prod_{i \mathop = 1}^n N_i \in \BB_a = \prod_{i \mathop = 1}^n \BB_{a_i}$
Hence the result by definition of basis for system of neighborhoods.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $7$
Category:Chebyshev Distance
Category:Neighborhoods
Neighborhood Basis in Real Number Line is Infinite

== Theorem ==
Let $\R$ be the real number line with the usual (Euclidean) metric.
Let $a \in \R$ be a point in $\R$.
Let $\BB_a$ be a basis for the neighborhood system of $a$.
Then $\BB_a$ is an infinite set.
== Proof ==
 $\BB_a$ be finite.
Let the elements of $\BB_a$ be enumerated as $N_1, N_2, \ldots, N_n$.
For each $N_k \in \BB_a$, let $\map {B_{\epsilon_k} } a$ be the open $\epsilon_k$-ball of $a$ for some $\epsilon_k \in \R_{>0}$.
Let $\alpha = \min \set {\epsilon_k: k \in \set {1, 2, \ldots, n} }$.
Consider the open interval:
:$I = \openint {a - \beta} {a + \beta}$
where $\beta = \dfrac \alpha 2$
By the method of construction of $I$, it follows that $\map {B_{\epsilon_k} } a \nsubseteq I$ for any $k$.
Thus $I$ is a neighborhood of $a$ which does not have an element of $\BB_a$ as a subset.
So $\BB_a$ cannot be a basis for the neighborhood system of $a$.
Hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $4$
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Neighborhood Basis of Open Subspace iff Neighborhood Basis

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $U \subseteq S$ be an open subset
Let $\tau_U$ denote the subspace topology on $U$.
Let $s \in U$.
Let $\NN \subseteq \powerset U$.
Then:
:$\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$ $\NN$ is a neighborhood basis of $s$ in $\struct{S, \tau}$
== Proof ==
Let $\map \NN s$ denote the set of neighborhoods of $s$ in $\struct{S, \tau}$
Let $\map \MM s$ denote the set of neighborhoods of $s$ in $\struct{U, \tau_U}$
From Neighborhood in Open Subspace:
:$\NN \subseteq \map \NN s$ $\NN \subseteq \map \MM s$
=== Necessary Condition ===
Let $\NN$ be a neighborhood basis of $s$ in $\struct{U, \tau_U}$.
Let $K \in \map \NN s$.
By definition of neighborhood:
:$\exists W \in \tau : s \in W \subseteq K$
By definition of subspace topology:
:$V = W \cap U \in \tau_U$
From Set is Open iff Neighborhood of all its Points:
:$V \in \map \MM s$
By definition of neighborhood basis:
:$\exists N \in \NN : x \in N \subseteq V$
From Subset Relation is Transitive:
:$N \subseteq K$
We have:
:$\exists N \in \NN : x \in N \subseteq K$
It follows that:
:$\NN$ is a neighborhood basis of $s$ in $\struct{S, \tau}$
=== Sufficient Condition ===
Let $\NN$ be a neighborhood basis of $s$ in $\struct{S, \tau}$.
Let $K \in \map \MM s$.
From Neighborhood in Open Subspace:
:$K \in \map \NN s$.
Hence:
:$\exists N \in \NN : x \in N \subseteq K$
It follows that:
:$\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$
Category:Topological Subspaces
Category:Neighborhood Bases
Neighborhood Condition for Coarser Topology

== Theorem ==
Let $S$ be a set.
Let $\tau_1$ and $\tau_2$ be two topologies on $S$.
Suppose that for all $z \in S$ and for all open neighborhoods $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$.
Then $\tau_1$ is coarser than $\tau_2$.
== Proof ==
Let $V \in \tau_1$.
For all $z \in V$, we have that $V$ is an open neighborhood of $z$ with respect to $\tau_1$.
Then for all $z \in V$, we can find $U_z \in \tau_2$ such that $z \in U_z \subseteq V$.
Then:
:$\ds V = \bigcup_{z \mathop \in V } \set z \subseteq \bigcup_{z \mathop \in V } U_z \subseteq V$
By definition of set equality, it follows that:
:$\ds V = \bigcup_{z \mathop \in V} U_z \in \tau_2$
Hence:
:$\tau_1 \subseteq \tau_2$
which by definition means that $\tau_1$ is coarser than $\tau_2$.
Category:Topology
Neighborhood Filter of Point is Filter

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be a point of $T$.
Let $\NN_x$ denote the neighborhood filter of $x$ in $T$.
Then $\NN_x$ is a filter on $S$.
== Proof ==
By definition, $\NN_x$ is the set of all neighborhoods of $x$ in $T$.
It is to be demonstrated that all the conditions are satisfied for $\NN_x$ to be a filter.
=== Proof of $(\text F 1)$ ===
From Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$.
Hence:
:$S \in \NN_x$
Hence $(\text F 1)$ is fulfilled by $\NN_x$.
=== Proof of $(\text F 2)$ ===
Let $N$ be an element of $\NN_x$
By definition, $x \in N$.
That is:
:$N \ne \O$
and it follows that:
:$\O \notin \NN_x$
Hence $(\text F 2)$ is fulfilled by $\NN_x$.
=== Proof of $(\text F 3)$ ===
Let $N_1, N_2 \in \NN_x$.
From Intersection of Neighborhoods in Topological Space is Neighborhood:
:$N_1 \cap N_2 \in \NN_x$
Hence $(\text F 3)$ is fulfilled by $\NN_x$.
=== Proof of $(\text F 4)$ ===
Let $N \in \NN_x$.
Let $M \subseteq S$ such that $N \subseteq M$.
From Superset of Neighborhood in Topological Space is Neighborhood:
:$M \in \NN_x$
Hence $(\text F 4)$ is fulfilled by $\NN_x$.
All criteria are fulfilled, and the result follows.
Category:Neighborhoods
Category:Filter Theory
Neighborhood Space induced by Topological Space induced by Neighborhood Space

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\NN$ on $S$.
Let $\struct {S, \NN'}$ be the neighborhood space induced by $\tau$ on $S$.
Then $\NN = \NN'$.
== Proof ==
Let $x \in S$.
Let $\NN_x$ be the set of all neighborhoods of $x$.
Let $N \in \NN_x$ be a neighborhood of $x$.
From Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is the superset of some open set $U$ in $\struct {S, \NN}$.
By Neighborhood Space is Topological Space we have that $U$ is an open set of $\struct {S, \tau}$.
Thus, by definition, $N$ is a neighborhood of $x$ in the context of the topological space $\struct {S, \tau}$.
Thus by definition of the neighborhood space induced by $\tau$ on $S$, it follows that:
:$N \in \NN'_x$
where $\NN'_x$ is the set of all neighborhoods of $x$ in $\struct {S, \NN'}$.
Thus $\NN_x \subseteq \NN'_x$.
Now suppose that $N \in \NN'_x$.
Then, by definition of neighborhood, $N$ is the superset of some open set $U$ in $\struct {S, \tau}$ such that $x \in U$.
Since $U \in \tau$ it follows that $U$ is an open setin $\struct {S, \NN}$.
So by Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is a neighborhood of $x$ in $\struct {S, \NN}$.
That is:
:$N \in \NN_x$
Thus it follows that $\NN_x \subseteq \NN'_x$
It follows by definition of set equality that $\NN_x = \NN'_x$.
As $x$ is arbitrary, it applies to all $x \in S$.
Hence the result.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces
Category:Neighborhood Spaces
Category:Topology
Neighborhood Space is Topological Space

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Let $\tau = \set {N: N \in \NN}$ be the set of all open sets of $\struct {S, \NN}$.
Then $\struct {S, \tau}$ forms a topological space.
== Proof ==
Each of the open set axioms is examined in turn:
=== ===
From Union of Open Sets of Neighborhood Space is Open, it follows that {{Open-set-axiom|1} is fulfilled.
=== ===
From Intersection of two Open Sets of Neighborhood Space is Open, it follows that is fulfilled.
=== ===
From Whole Space is Open in Neighborhood Space, it follows that is fulfilled.
All the open set axioms are fulfilled, and the result follows.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces
Category:Neighborhood Spaces
Neighborhood Sub-Basis Criterion for Filter Convergence

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on $S$.
Let $p \in S$.
Then $\FF$ converges to $p$ $\FF$ contains as a subset a neighborhood sub-basis at $p$.
== Proof ==
=== Sufficient Condition ===
Let $\FF$ converges to $p$.
Then it contains every neighborhood of $p$.
The set of neighborhoods of $p$ is trivially a neighborhood sub-basis at $p$.
=== Necessary Condition ===
Let $S_p$ be a neighborhood sub-basis at $p$.
Let $S_p \subseteq \FF$.
Let $N$ be a neighborhood of $p$.
Then by the definition of neighborhood sub-basis, there is a finite $T_N \subseteq S_p$ such that:
:$\bigcap T_N \subseteq N$
Since a filter is by definition closed under finite intersections:
:$\bigcap T_N \in \FF$
Then:
:$\bigcap T_N \in \FF$
and:
:$\bigcap T_N \subseteq N$
so by the definition of a filter:
:$N \in \FF$
Because $\FF$ contains every neighborhood of $p$, $\FF$ converges to $p$. 
Category:Filter Theory
Category:Topological Bases
Neighborhood iff Contains Neighborhood

== Theorem ==
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.
== Also see ==
* Set is Open iff Neighborhood of all its Points
Category:Neighborhoods
Neighborhood in Compact Hausdorff Space Contains Compact Neighborhood

== Theorem ==
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 T4, 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 $V$.
== Also see ==
* Definition:Locally Compact Hausdorff Space
Category:Compact Spaces
Category:Hausdorff Spaces
Neighborhood in Metric Space has Subset Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$ be a point in $M$.
Let $N$ be a neighborhood of $a$ in $M$.
Then there exists a neighborhood $N'$ of $a$ such that:
:$(1): \quad N' \subseteq N$
:$(2): \quad N'$ is a neighborhood of each of its points.
== Proof ==
By definition of neighborhood:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$
By Open Ball is Neighborhood of all Points Inside, $N' = \map {B_\epsilon} a$ fulfils the conditions of the statement.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 5$
Category:Neighborhoods
Neighborhood in Open Subspace

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $U \subseteq S$ be an open subset.
Let $\tau_U$ denote the subspace topology on $U$.
Let $s \in U$
Let $N \subseteq U$ be a subset.
Then:
:$N$ is a neighborhood of $s$ in $\struct{U, \tau_U}$ 
:$N$ is a neighborhood of $s$ in $\struct{S, \tau}$
== Proof ==
=== Necessary Condition ===
Let $N$ be a neighborhood of $s$ in $\struct{U, \tau_U}$.
By definition of neighborhood:
:$\exists V \in \tau_U : x \in V \subseteq N$
From Open Set in Open Subspace:
:$V \in \tau$
Hence:
:$\exists V \in \tau : x \in V \subseteq N$
It follows that $N$ is a neighborhood of $s$ in $\struct{S, \tau}$ by definition.
=== Sufficient Condition ===
Let $N$ be a neighborhood of $s$ in $\struct{S, \tau}$.
By definition of neighborhood:
:$\exists W \in \tau : x \in W \subseteq N$
Let $V = W \cap U$.
From Subset Relation is Transitive:
:$V \subseteq N$
By definition of topological space:
:$V \in \tau$.
From Open Set in Open Subspace:
:$V \in \tau_U$
Hence:
:$\exists V \in \tau_U : x \in V \subseteq N$
It follows that $N$ is a neighborhood of $s$ in $\struct{U, \tau_U}$ by definition.
== Also see ==
* Open Set in Open Subspace
* Closed Set in Closed Subspace
Category:Neighborhoods
Category:Topological Subspaces
Neighborhood in Topological Space has Subset Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Then there exists a neighborhood $N'$ of $x$ such that:
:$(1): \quad N' \subseteq N$
:$(2): \quad N'$ is a neighborhood of each of its points.
That is:
:$\forall x \in S: \forall N \in \NN_x: \exists N' \in \NN_x, N' \subseteq N: \forall y \in N': N' \in \NN_y$
where $\NN_x$ is the neighborhood filter of $x$.
== Proof ==
By definition of neighborhood:
:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
where $U$ is an open set of $T$.
By Set is Open iff Neighborhood of all its Points, $N' = U$ fulfils the conditions of the statement.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 5$
Category:Neighborhoods
Neighborhood in Topological Subspace

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $S \subseteq X$ be a subset of $X$.
Let $\tau_S$ denote the subspace topology on $S$.
Let $x \in S$ be an arbitrary point of $S$.
Let $E \subseteq S$.
Then:
:$E$ is a neighborhood of $x$ in $\struct {S, \tau_S}$
:
:$\exists D \subseteq X$ such that:
::$D$ is a neighborhood of $x$ in $X$
::$E = D \cap S$.
== Proof ==
=== Necessary Condition ===
Let $E$ be a neighborhood of $x$ in $\struct {S, \tau_S}$. 
By the definition of neighborhood:
:$\exists U \in \tau_S : x \in U \subseteq E$
Now, by the definition of the subspace topology:
:$\exists V \in \tau: U = V \cap S$
Take $D := V \cup E$.
We have that:
:$V \subseteq D$
and:
:$V \in \tau$
Thus $D$ is a neighborhood of $x$ in $X$.
Therefore it holds true that:
:$D \cap S = \paren {V \cap S} \cup \paren {E \cap S} = U \cup E = E$
=== Sufficient Condition ===
Let $\exists D \subseteq X$ such that $D$ is a neighborhood of $x$ in $\struct {X, \tau}$.
Let $E = D \cap S$
Then, by the definition of neighborhood:
:$\exists V \in \tau: x \in V \subseteq D$
By the definition of the subspace topology:
:$V \cap S \in \tau_S$
As a consequence:
:$E = D \cap S \supseteq V \cap S \in \tau_S$
That is:
:$E$ is a neighborhood of $x$ in $\struct {S, \tau_S}$.
== Also see ==
* Closed Set in Topological Subspace
Category:Topology
Category:Neighborhoods
Neighborhood of Diagonal induces Open Cover

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $X \times X$ denote the cartesian product of $X$ with itself.
Let $\tau_{X \times X}$ denote the product topology on $X \times X$.
Let $T \times T$ denote the product space $\struct {X \times X, \tau_{X \times X} }$. 
Let $U$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in $T \times T$.
Let $\VV = \set{V \in \tau : V \times V \subseteq U}$
Then:
:$\VV$ is an open cover of $T$
== Proof ==
By definition of product topology:
:$\BB = \set {V_1 \times V_2: V_1, V_2 \in \tau}$ is a basis on $T \times T$
Let $x \in X$.
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$U$ is a neighborhood of the point $\tuple{x,x}$
From Characterization of Neighborhood by Basis:
:$\exists V_1, V_2 \in \tau : \tuple{x, x} \in V_1 \times V_2 : V_1 \times V_2 \subseteq U$
By definition of Cartesian product:
:$x \in V_1$ and $x \in V_2$
Let $V = V_1 \cap V_2$.
By definition of set intersection:
:$x \in V$
By :
:$V \in \tau$
From Intersection is Subset:
:$V \subseteq V_1$ and $V \subseteq V_2$
From Cartesian Product of Subsets:
:$V \times V \subseteq V_1 \times V2$
From Subset Relation is Transitive:
:$V \times V \subseteq U$
Hence:
:$\exists V \in \VV : x \in V$
Since $x$ was arbitrary, it follows that:
:$\forall x \in X : \exists V \in \VV : x \in V$
By definition, $\VV$ is an open cover of $T$. 
Category:Product Spaces
Category:Neighborhoods
Category:Covers
Neighborhood of Origin in Arens-Fort Space is not Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Let $N_p$ be a neighborhood of the point $p = \tuple {0, 0}$.
Then $N_p$ is not compact.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $4$
Category:Compact Spaces
Category:Arens-Fort Space
Neighborhood of Origin of Arens-Fort Space is Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Every neighborhood of $\tuple {0, 0}$ is closed in $T$.
== Proof ==
Let $H \subseteq S$ such that:
:$\exists U \in \tau: \tuple {0, 0} \in U \subseteq H \subseteq S$
that is: such that $H$ is a neighborhood of $\tuple {0, 0}$ in $T$.
As $\tuple {0, 0} \in H$ it follows that $\tuple {0, 0} \notin \relcomp S H$.
So, by definition of the Arens-Fort space, $\relcomp S H$ is open in $T$.
So by definition, we have that $H$ is closed.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $8$
Category:Closed Sets
Category:Arens-Fort Space
Neighborhood of Point in Metrizable Space contains Closed Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a metrizable topological space.
Let $x \in S$ be an arbitrary point of $T$.
Let $N$ be a neighborhood of $x$.
Then $N$ has as a subset a neighborhood $V$ of $x$ such that $V$ is closed.
== Proof ==
Since $N$ is a neighborhood of $x$, there exists an open set $U \subseteq N$ containing $x$ by definition.
As $\struct {S, \tau}$ is metrizable, the set $U$ is open with respect to some metric space $\struct {S, d}$.
Hence:
:$\exists \epsilon > 0: \map {\BB_\epsilon} x \subseteq U$
where $\map {\BB_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.
Taking $\zeta = \dfrac \epsilon 2$, we have that the closed $\zeta$-ball of $x$ is contained in $\map {\BB_\epsilon} x$, and hence is a subset of $U \subseteq N$. 
This closed ball is a closed neighborhood of $x$ in $\struct {S, \tau}$, and is a subset of $N$.
Hence the result.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $5$
Category:Metrizable Topologies
Category:Neighborhoods
Neighborhoods in Standard Discrete Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space where $d$ is the standard discrete metric.
Let $a \in A$.
Then $\set a$ is a neighborhood of $a$ which forms a basis for the system of neighborhoods of $a$.
== Proof ==
By definition of the standard discrete metric:
:$\map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$
Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.
Then:
So by definition $\set a$ is a neighborhood of $a$.
Let $\NN_a$ be the system of neighborhoods of $a$.
Let $N \in \NN_a$ be a neighborhood of $a$.
Then:
Hence the result by definition of basis of a system of neighborhoods.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $1$
Category:Standard Discrete Metric
Category:Neighborhoods
Neighbourhood of Point Contains Point of Subset iff Distance is Zero

== Theorem ==
Let $M = \struct {X, d}$ be a metric space.
Let $A \subseteq X$ be a non-empty subset of $X$.
Let $x \in X$.
Then every neighborhood of $x$ contains a point of $A$ :
:$\map d {x, A} = 0$
where $\map d {x, A}$ denotes the distance from $x$ to $A$.
== Proof ==
=== Sufficient Condition ===
Let $x \in X$.
Let every neighborhood of $x$ contain a point in $A$.
Every open ball $\map {B_\epsilon} x$ centered at $x$ is seen to be a neighborhood of $x$ in $M$.
But then this implies that for every $\epsilon \in \R_{\gt 0}$ there must exists a $y \in A$ such that:
:$\map d {x, y} < \epsilon$
From this it is seen that $\map d {x, A} = 0$.
=== Necessary Condition ===
Let $\map d {x, A} = 0$ for $x \in X$.
Let $S$ be a neighborhood of $x$ in $M$.
From the definition of a neighborhood, there exists an open ball $\map {B_\epsilon} x$ centered at $x$ such that:
:$\map {B_\epsilon} x \subseteq S$
We have that:
:$\epsilon < 0$
and
$\map d {x, A} = 0$
Hence there must exists a $y \in A$ such that:
:$\map d {x, y} < \epsilon \implies \map {B_\epsilon} x \cap A \ne \O$
If otherwise then $\epsilon$ would be a lower bound for $\set {\map d {x, y} : y \in A}$ that is greater than $0$, which is a contradiction.
Then from Set Intersection Preserves Subsets we have:
:$\map {B_\epsilon} x \cap A \subseteq S \cap A$
and so:
:$\map {B_\epsilon} x \cap A \ne \O \implies S \cap A \ne \O$
Hence the result.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $8$
Category:Neighborhoods
Category:Metric Spaces
Nesbitt's Inequality

== Theorem ==
Let $a$, $b$ and $c$ be (strictly) positive real numbers.
Then:
:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
Category:Nesbitt's Inequality
Category:Algebra
Category:Inequalities
Nesbitt's Inequality/Proof 1

== Theorem ==
== Proof ==
These are the arithmetic mean and the harmonic mean of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$.
From AM-HM Inequality the last inequality is true.
Thus Nesbitt's Inequality holds.
Category:Nesbitt's Inequality