nyuuzyou/wikis huggingface dataset:
Localization Preserves Integral Closure

== Theorem ==
Let $A \subseteq B$ be an extension of commutative rings with unity.
Let $C$ be the integral closure of $A$ in $B$.
Let $S \subseteq A$ be a multiplicatively closed subset.
Then $C_S$ is the integral closure of $A_S$ in $B_S$, where subscript $S$ indicates the localization at $S$.
== Proof ==
First we show that $C_S$ is an integral extension of $A_S$.
Let $x \in C_S$, and $\iota$ be the canonical inclusion from a ring to its localization.
There exists $s \in S$ such that $sx \in \iota(C)$, say $sx = \iota(c)$.
Since $c \in C$ is integral, there is an equation:
:$c^n + a_{n-1}c^{n-1} + \cdots + a_1 c + a_0 = 0$
with the $a_i \in A$. Since $\iota$ is a homomorphism, we have:
:$\left({s x}\right)^n + \iota \left({a_{n-1} }\right) \left({s x}\right)^{n-1} + \cdots + \iota \left({a_1}\right) \left({s x}\right) + \iota \left({a_0}\right) = 0$
Multiplying by $s^{-n}$ this gives:
:$x^n + s^{-1}\iota \left({a_{n-1} }\right)x^{n-1} + \cdots + x^{1-n} \iota \left({a_1}\right) x + s^{-n}\iota \left({a_0}\right) = 0$
This is a monic polynomial with coefficients in $A_S$, so $x$ is integral over $A_S$.
Now let $x \in B_S$ be integral over $A_S$.
So we have an equation
:$x^n + r_{n-1}x^{n-1} + \cdots + r_1 x + r_0 = 0$
Let:
:$t \in S$ be such that $t x \in \iota \left({B}\right)$
:$u_i$ be such that $u_i r_i \in \iota \left({A}\right)$ for $i = 0, \ldots, n-1$
:$s = t u_0 \cdots u_{n-1}$.
Then we have:
:$\left({s x}\right)^n + r_{n-1} s \left({s x}\right)^{n-1} + \cdots + r_1 s^{n-1} \left({s x}\right) + r_0 s^n = 0$
Now let:
: $s x = \iota \left({y}\right)$ for $y \in B$
: $r_{n-i} s^i = \iota \left({q_i}\right)$ for $i = 1, \ldots, n$
Then:
:$\iota \left({y^n + q_1 y^{n-1} + \cdots + q_{n-1} y + q_n}\right) = 0$
Now:
: $\ker \left({\iota: B \to B_S}\right) = \left\{{x \in B : s x = 0 \text{ for some } s \in S}\right\}$
So there exists $v \in S$ such that:
:$\left({v y}\right)^n + q_1 v (\left({v y}\right)^{n-1} + \cdots + q_{n-1} v^{n-1} \left({v y}\right) + q_n v^n = 0$
Therefore $v y \in B$ is integral over $A$, so $vy \in C$.
Therefore:
:$\iota \left({v y}\right) \in C_S$
and:
:$\iota \left({y}\right) \in C_S$
Thus:
:$x = s^{-1} \iota \left({y}\right) \in C_S$
Category:Algebraic Number Theory
Localization of Module Homomorphism is Exact

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset.
Let $S^{-1}A$ be the localization of $A$ at $S$.
Let:
:$M_1 \stackrel {f_1} \longrightarrow M_2 \stackrel {f_2} \longrightarrow M_3$
be an exact sequence of $A$-modules.
For $i=1,2,3$, let $\struct { S^{-1}M_i, \iota_i}$ be the localization of $M_i$ at $S$.
For $i=1,2$, let:
:$S^{-1}f_i : S^{-1}M_i \to S^{-1}M_{i+1}$
be the unique $S^{-1}A$-homomorphism such that:
:$\iota_{i+1} \circ f_i = \paren {S^{-1}f_i} \circ \iota_i$
Then:
:$S^{-1}M_1 \stackrel {S^{-1}f_1} \longrightarrow S^{-1}M_2 \stackrel {S^{-1}f_2} \longrightarrow S^{-1}M_3$
is an exact sequence of $S^{-1}A$-modules.
== Proof ==
== Also known as ==
The operation $S^{-1}$ is called exact.
== Sources ==
* : Chapter $3$: Rings and Modules of Fractions
Category:Commutative Algebra
Localization of Module Homomorphism is Multiplicative

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset.
Let $S^{-1}A$ be the localization of $A$ at $S$.
Let $f_1$ and $f_2$ be $A$-homomorphisms:
:$M_1 \stackrel {f_1} \longrightarrow M_2 \stackrel {f_2} \longrightarrow M_3$
For $i=1,2,3$, let $\struct { S^{-1}M_i, \iota_i}$ be the localization of $M_i$ at $S$.
Let $S^{-1}f_1$ and $S^{-1}f_i$ be the unique $S^{-1}A$-homomorphisms:
:$S^{-1}M_1 \stackrel {S^{-1}f_1} \longrightarrow S^{-1}M_2 \stackrel {S^{-1}f_2} \longrightarrow S^{-1}M_3$
such that:
:$\iota_{i+1} \circ f_i = \paren {S^{-1}f_i} \circ \iota_i$
Let:
:$S^{-1}M_1 \stackrel {S^{-1}\paren {f_2 \circ f_1}} \longrightarrow S^{-1}M_3$
be the unique $S^{-1}A$-homomorphism such that:
:$\iota_3 \circ \paren {f_2 \circ f_1} = \paren {S^{-1} \paren {f_2 \circ f_1} } \circ \iota_1$
Then we have:
:$S^{-1} \paren {f_2 \circ f_1} = \paren {S^{-1} f_2} \circ \paren {S^{-1} f_1}$
== Proof ==
Let $g : S^{-1}M_1 \to S^{-1}M_3$ be defined by:
:$g:= \paren {S^{-1} f_2} \circ \paren {S^{-1} f_1}$
Then:
Thus, by , we have:
:$g = S^{-1} \paren {f_2 \circ f_1}$
== Sources ==
* : Chapter $3$: Rings and Modules of Fractions
Category:Commutative Algebra
Localization of Ring Exists

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.
Then there exists a localization $\struct {A_S, \iota}$ of $A$ at $S$.
== Proof ==
Define a relation $\sim$ on the Cartesian product $A \times S$ by:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
=== Lemma 1 ===
Let $A_S$ denote the quotient set $\paren {A \times S} / \sim$.
Let $\dfrac a s$ denote the equivalence class of $\tuple {a, s}$ in $\paren {A \times S} / \sim$.
For $\dfrac a s, \dfrac b t \in A_S$, let the following operations be defined:
:$\dfrac a s + \dfrac b t = \dfrac {a t + b s} {s t}$
:$\dfrac a s \cdot \dfrac b t = \dfrac {a b} {s t}$
=== Lemma 2 ===
Now define $\iota: A \to A_S$ by:
:$\map \iota a := \dfrac a 1$
It is to be shown that $\struct {A_S, \iota}$ satisfy the universal property for localization.
That is, to be shown is the $(2)$ of .
Let $B$ be a ring.
Let $g: A \to B$ be a mapping such that:
:$g \sqbrk S \subseteq B^\times$
where $B^\times$ denotes the set of units of $B$.
Suppose that $h: A_S \to B$ is a ring homomorphism with $h \circ \iota = g$.
Then we must have:
:$\map h {\dfrac a 1} = \map g a$
and:
:$\map h {\dfrac 1 s} \cdot \map h {\dfrac s 1} = 1$
Therefore:
:$\map h {\dfrac 1 s} \map g s = 1$
so:
:$\map h {\dfrac 1 s} = \map g s^{-1}$
Therefore:
:$\map h {\dfrac a s} = \map h {\dfrac a 1} \cdot \map h {\dfrac 1 s} = \map g a \map g s^{-1}$.
So if such $h$ exists it must equal $\map g a \map g s^{-1}$, so is unique.
Therefore, to conclude the proof we pull out:
=== Lemma 3 ===
Hence the result.
== Notes ==
The proof shows additionally that:
:$\forall x \in A_S: \exists s \in S: s x \in \iota \sqbrk A$
:$\map \ker \iota = \set {a \in A: \exists s \in S: a s = 0}$
== Also see ==
* If $A$ is an integral domain and $S = A \setminus \set 0$ then the localization of $A$ at $S$ is precisely the field of quotients of $A$.
Category:Localization of Rings
Category:Localization of Ring Exists
Localization of Ring Exists/Lemma 1

== Lemma for Localization of Ring Exists ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.
Let $\sim$ be the relation defined on the Cartesian product $A \times S$ by:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
The relation $\sim$ is an equivalence relation.
== Proof == 
Since by definition $1 \in S$ and $a s = a s$ for all $a \in A, s \in S$, it follows that $\sim$ is reflexive.
Also $\sim$ is clearly symmetric because if $a t u = b s u$ then $b s u = a t u$ by symmetry of $=$.
Lastly suppose that:
:$\tuple {a, s} \sim \tuple {b, t}$
and:
:$\tuple {b, t} \sim \tuple {c, u}$
Then there are $v, w \in S$ such that:
:$a t v = b s v$
:$b u w = c t w$
Therefore:
:$a u t v w = b u w s v = c t w s v$
and so:
:$a u \paren {t v w} = c s \paren {t v w}$
Since $S$ is multiplicatively closed:
:$\tuple {a, s} \sim \tuple {c, u}$
Thus $\sim$ is transitive.
Category:Localization of Ring Exists
Localization of Ring Exists/Lemma 2

== Lemma ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.
Let $\sim$ be the relation defined on the Cartesian product $A \times S$ by:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
Let $A_S$ denote the quotient set $\paren {A \times S} / \sim$.
Let $\dfrac a s$ denote the equivalence class of $\tuple {a, s}$ in $\paren {A \times S} / \sim$.
For $\dfrac a s, \dfrac b t \in A_S$, let the following operations be defined:
:$\dfrac a s + \dfrac b t = \dfrac {a t + b s} {s t}$
:$\dfrac a s \cdot \dfrac b t = \dfrac {a b} {s t}$
The operations $+$ and $\cdot$ are well defined on $A_S$.
== Proof ==
From Lemma 1 we have that $\sim$ is an equivalence relation on $A \times S$.
Let:
:$\dfrac a s = \dfrac c u$
:$\dfrac b t = \dfrac d v$
be two sets of representatives for two distinct equivalence classes in $A_S$.
We have $w, z \in S$ such that:
:$\paren {a u - c s} w = 0$
and
:$\paren {b v - d t} z = 0$
Therefore:
:$z w \paren {\paren {a u - c s} w - \paren {b v - d t} z} = 0$
and:
:$z w \paren {\paren {a t + b s} u v - \paren {c v + d u} s t}$
So:
:$\tuple {a t + b s, s t} \sim \tuple {c v + d u, u v}$
That is:
:$\dfrac {a t + b s} {s t} = \dfrac {c v + d u} {u v}$
For the operation $\cdot$, with $z, w$ as above we have:
:$\paren {a b u v - c d s t} z w = 0$
So:
:$\tuple {a b, s t} \sim \tuple {d c, u v}$
and:
:$\dfrac {a b} {s t} = \dfrac {d c} {u v}$
Category:Localization of Ring Exists
Localization of Ring Exists/Lemma 3

== Lemma for Localization of Ring Exists ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.
Let $\sim$ be the relation defined on the Cartesian product $A \times S$ by:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
Let $A_S$ denote the quotient set $\paren {A \times S} / \sim$.
Let $\dfrac a s$ denote the equivalence class of $\tuple {a, s}$ in $\paren {A \times S} / \sim$.
Let $B$ be a ring.
Let $g: A \to B$ be a mapping such that:
:$g \sqbrk S \subseteq B^\times$
where $B^\times$ denotes the set of units of $B$.
The map $\map h {\dfrac a s} = \map g a \map g s^{-1}$ is a well defined ring homomorphism $A_S \to B$.
== Proof ==
Let $\dfrac a s = \dfrac b t \in A_S$.
Then there exists $u \in S$ such that $a t u = b s u$.
Therefore, by the homomorphism property, and the fact that $g \sqbrk S \subseteq B^\times$:
:$\map g a \map g t = \map g b \map g s \to \map g a \map g s^{-1} = \map g b \map g t^{-1}$
We have that $g$ is a ring homomorphism.
As $B$ is a commutative ring it follows that $h$ is also a ring homomorphism.
Category:Localization of Ring Exists
Localization of Ring is Unique

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset.
Let $\struct {A_S, \iota}$ and $\struct {\tilde A_S, \tilde \iota}$ both satisfy the definition of the localization of $A$ at $S$.
Then there is a canonical isomorphism $\phi: A_S \to \tilde A_S$.
== Proof ==
By the definition of localization, there exist unique homomorphisms:
:$g : A_S \to \tilde A_S$
:$h : \tilde A_S \to A_S$
such that:
:$h \circ \iota = \tilde \iota$
and:
:$g \circ \tilde \iota = \iota$
Therefore:
:$\tilde \iota = h \circ \iota = h \circ \paren {g \circ \tilde \iota} = \paren {h \circ g} \circ \tilde \iota$
The identity mapping also satisfies this equality (that is, $\tilde \iota = \paren {h \circ g} \circ \tilde \iota$).
Therefore by the uniqueness of $h$ and $g$ we have:
:$h \circ g = I_{\tilde A_S}$
where $I_{\tilde A_S}$ denotes the identity mapping on $\tilde A_S$.
Similarly we have that:
:$g \circ h = I_{A_S}$
where $I_{A_S}$ denotes the identity mapping on $A_S$.
Therefore by Bijection iff Left and Right Inverse $g = \phi$ is a bijective homomorphisms, that is, an isomorphism.
Category:Localization of Rings
Locally Arc-Connected Space is Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally arc-connected.
Then $T$ is also locally path-connected.
== Proof ==
Let $T = \struct {S, \tau}$ be arc-connected.
Then $T$ has a basis consisting entirely of arc-connected sets.
From Arc-Connected Space is Path-Connected, this basis consisting entirely of path-connected sets.
The result follows from definition of locally path-connected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Arc-Connected Spaces
Category:Locally Path-Connected Spaces
Locally Arc-Connected Space is not necessarily Arc-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally arc-connected.
Then it is not necessarily the case that $T$ is also arc-connected.
== Proof ==
Let $T := \struct {\R, \tau_d}$ be the real number line $\R$ under the usual (Euclidean) topology $\tau_d$.
Let $a, b, c \in \R$ where $a < b < c$.
Let $A$ be the union of the two open intervals:
:$A := \openint a b \cup \openint b c$
Let $T' := \struct {A, \tau_A}$ be the subspace composed of $A$ with the subspace topology induced on $\tau_d$ by $A$.
From Union of Adjacent Open Intervals is Locally Arc-Connected, $T'$ is a locally arc-connected space.
From Union of Adjacent Open Intervals is not Arc-Connected, $T$ is not a arc-connected space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Arc-Connected Spaces
Category:Arc-Connected Spaces
Locally Bounded Hausdorff Topological Vector Space with Heine-Borel Property has Finite Dimension

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a locally bounded Hausdorff topological vector space over $\GF$ with the Heine-Borel property.
Then $X$ is a finite dimensional vector space.
== Proof ==
Since $\struct {X, \tau}$ is locally bounded, there exists a von Neumann-bounded open neighborhood $V$ of ${\mathbf 0}_X$. 
From Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded, $\map \cl V$ is von Neumann-bounded.
From Topological Closure is Closed, $\map \cl V$ is closed.
So $\map \cl V$ is von Neumann-bounded and closed.
From the Heine-Borel property, $\map \cl V$ is compact.
So $V$ is an open neighborhood of ${\mathbf 0}_X$ with compact closure.
Hence $X$ is locally compact.
Hence from Locally Compact Hausdorff Topological Vector Space has Finite Dimension, $X$ is a finite dimensional vector space.
== Sources ==
* : $1.24$: Theorem
Category:Heine-Borel Property of Topological Vector Spaces
Category:Locally Bounded Topological Vector Spaces
Category:Hausdorff Topological Vector Spaces
Locally Compact Hausdorff Topological Vector Space has Finite Dimension

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a locally compact Hausdorff topological vector space over $\GF$. 
Then $X$ is a finite dimensional vector space.
== Proof ==
Since $X$ is locally compact, there exists a von Neumann-bounded open neighborhood $V$ of ${\mathbf 0}_X$ such that: 
:$\map \cl V$ is compact.
From Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin:
:$\BB = \set {2^{-n} V : n \in \N}$ is a local basis for ${\mathbf 0}_X$.
We have that:
:$\ds \map \cl V \subseteq \bigcup_{x \in X} \paren {x + \frac 1 2 V}$
Since $\map \cl V$ is compact, there exists $x_1, \ldots, x_m \in X$ such that:
:$\ds \map \cl V \subseteq \bigcup_{j \mathop = 1}^m \paren {x_j + \frac 1 2 V}$
Let:
:$Y = \span \set {x_j : 1 \le j \le m}$
Then $Y$ is finite dimensional.
We show that $Y = X$.
From Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Closed, $Y$ is closed.
=== Lemma ===
Recall that $\BB$ is a local basis for ${\mathbf 0}_X$.
From Expression for Closure of Set in Topological Vector Space: Corollary, we obtain: 
:$\ds \bigcap_{n \mathop = 1}^\infty \paren {Y + 2^{-n} V} = \map \cl Y$
Since $Y$ is closed, we have:
:$\map \cl Y = Y$
from Set is Closed iff Equals Topological Closure.
Hence from the Lemma, we have:
:$V \subseteq Y$
Hence for each $k \in \N$, we have:
:$k V \subseteq Y$
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have:
:$\ds X = \bigcup_{n \mathop = 1}^\infty k V$
From Set Union Preserves Subsets: General Result, we obtain:
:$\ds \bigcup_{n \mathop = 1}^\infty k V \subseteq Y$
so that:
:$X \subseteq Y$
Since we have $Y \subseteq X$, we obtain $Y = X$.
So $X$ is a finite dimensional vector space.
== Sources ==
* : $1.22$: Theorem
Category:Locally Compact Hausdorff Spaces
Category:Hausdorff Topological Vector Spaces
Category:Locally Compact Hausdorff Topological Vector Space has Finite Dimension
Locally Compact Hausdorff Topological Vector Space has Finite Dimension/Lemma

== Lemma ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a locally compact topological vector space over $\GF$. 
Let $V$ be a von Neumann-bounded open neighborhood of ${\mathbf 0}_X$ such that:
:$\map \cl V$ is compact.
Let $x_1, \ldots, x_m \in X$ be such that:
:$\ds \map \cl V \subseteq \bigcup_{j \mathop = 1}^m \paren {x_j + \frac 1 2 V}$
Let:
:$Y = \span \set {x_j : 1 \le j \le m}$
Then:
:$\ds V \subseteq \bigcap_{n \mathop = 1}^\infty \paren {Y + 2^{-n} V}$
== Proof ==
Since: 
:$\ds \map \cl V \subseteq \bigcup_{j \mathop = 1}^m \paren {x_j + \frac 1 2 V}$
we in particular have:
:$V \subseteq Y + \frac 1 2 V$
since $x_j \in Y$ for each $1 \le j \le n$.
Since $Y$ is a vector subspace, we have:
:$\lambda Y = Y$ for each $\lambda \in \GF$
and:
:$Y + Y = Y$
We now aim to prove:
:$V \subseteq Y + 2^{-n} V$ for each $n \in \N$.
We do this by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$V \subseteq Y + 2^{-n} V$
=== Basis for the Induction ===
We have already shown:
:$V \subseteq Y + \dfrac 1 2 V$
Hence $\map P 1$ is already known to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds V \subseteq Y + 2^{-k} V$
Then we need to show:
:$\ds V \subseteq Y + 2^{-\paren {k + 1} } V$
=== Induction Step ===
This is our induction step:
We have: 
:$\ds V \subseteq Y + \frac 1 2 V$
so that:
:$\ds 2^{-k} V \subseteq Y + 2^{-\paren {k + 1} } V$
We then have: 
:$\ds V \subseteq Y + 2^{-k} V \subseteq Y + Y + 2^{-\paren {k + 1} } V = Y + 2^{-\paren {k + 1} } V$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: V \subseteq Y + 2^{-n} V$
Hence from Set Intersection Preserves Subsets, we obtain:
:$\ds V \subseteq \bigcap_{n \mathop = 1}^\infty \paren {Y + 2^{-n} V}$
Category:Locally Compact Hausdorff Topological Vector Space has Finite Dimension
Locally Compact Space is Weakly Locally Compact

== Theorem ==
Let $T = \struct{S, \tau}$ be a locally compact topological space.
Then $T$ is weakly locally compact.
== Proof ==
We are given $T$ is locally compact, so every point of $S$ has a neighborhood basis consisting of compact sets.
Let $x \in S$ be arbitrary. 
From Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$ in $S$.
 
So, every point contained in $S$ has a compact neighborhood.
Hence, $T$ is weakly locally compact.
== Also see ==
* Strongly Locally Compact Space is Weakly Locally Compact
* Sequence of Implications of Local Compactness Properties
Category:Locally Compact Spaces
Category:Weakly Locally Compact Spaces
Locally Compact iff Open Neighborhood contains Compact Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is locally compact
:
:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
where $K^\circ$ denotes the interior of $K$.
== Proof ==
=== Sufficient Condition ===
Let $T$ be locally compact.
Let $U \in \tau, x \in U$.
Since $T$ is locally compact there exists a neighborhood basis $\BB$ of $x$ consisting of compact sets. 
From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$.
By definition of a neighborhood basis:
:$\exists K \in \BB : K \subseteq U$.
By definition of a neighborhood of $x$:
:$\exists V \in \tau : x \in V \subseteq K$
By definition of the interior of a subset then:
:$x \in V \subseteq K^\circ$
Thus
:$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
=== Necessary Condition ===
Let
:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
Let $\BB$ be the set of all compact neighborhoods of $x$.
It is shown that $\BB$ is a neighborhood basis of $x$.
Let $N$ be a neighborhood of $x$.
By definition of a neighborhood of $x$:
:$\exists U \in \tau : x \in U \subseteq N$
By assumption:
:$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
By definition of the interior of a subset $K^\circ$ is open.
By definition of a neighborhood then $K$ is a neighborhood of $x$.
Hence $K \in \BB$ and $K \subseteq U \subseteq N$.
As $N$ was arbitrary, then $B$ is a neighborhood basis of $x$ consisting of compact sets.
== Sources ==
* 
Category:Locally Compact Spaces
Locally Connected Separable Topological Space has Countably Many Components

== Theorem ==
Let $\struct {X, \tau}$ be a locally connected separable topological space. 
Let $\CC$ be the set of components of $\struct {X, \tau}$.
Then $\CC$ is countable.
== Proof ==
From Component of Locally Connected Space is Open:
:$\CC \subseteq \tau$
From Equivalence Classes are Disjoint, we have:
:the sets in $\CC$ are pairwise disjoint.
From Collection of Pairwise Disjoint Open Sets in Separable Topological Space is Countable, it follows that:
:$\CC$ is countable.
Category:Locally Connected Spaces
Category:Components
Locally Connected Space is not necessarily Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally connected.
Then it is not necessarily the case that $T$ is also a connected space.
== Proof ==
Let $T$ be a discrete topological space with more than $1$ point.
From Discrete Space is Locally Connected, $T$ is a locally connected space.
From Non-Trivial Discrete Space is not Connected, $T$ is not a connected space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Connected Spaces
Category:Connected Spaces
Locally Connected Space is not necessarily Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is locally connected.
Then it is not necessarily the case that $T$ is also an locally patj-connected space.
== Proof ==
Let $T$ be a countable finite complement Space.
From Finite Complement Space is Locally Connected, $T$ is a locally connected space.
From Countable Finite Complement Space is not Locally Path-Connected, $T$ is not a locally path-connected space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Connected Spaces
Category:Locally Path-Connected Spaces