avewright/proofwiki-math huggingface dataset:
Theorem: Surgery for Rings

Let $R$ and $S$ be commutative rings with unity, and $\phi: R \to S$ a ring monomorphism.
Then there is a ring $T$ isomorphic to $S$ that contains $R$ as a subring.

Proof:
Let $T$ be the disjoint union $T = R \cup \paren {S \setminus \Img \phi}$.
Define $\theta : T \to S$ as follows:
:If $x \in R$, then $\map \theta x = \map \phi x$
:If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
We claim that $\theta$ is an isomorphism.
''Injectivity'': Let $\map \theta x = \map \theta y$.
Since $\theta \sqbrk R = \phi \sqbrk R$, we have $\theta \sqbrk R \cap \paren {S \setminus \Img \phi} = \O$.
Therefore either $x, y \in R$ or $x, y \in \paren {S \setminus \Img \phi}$.
If $x, y \in R$ then $\map \phi x = \map \phi y$, so $\theta$ is injective because $\phi$ is.
If $x, y \in \paren {S \setminus \Img \phi}$ then $\theta$ acts as the identity and $x = y$.
''Surjectivity'': Under $\theta$, $R$ surjects onto $\Img \phi = \phi \sqbrk R$ by definition.
Moreover:
:$\theta \sqbrk {S \setminus \Img \phi} = I \sqbrk {S \setminus \Img \phi} = \paren {S \setminus \Img \phi}$
where $I$ denotes the identity mapping.
Therefore $\theta$ is surjective everywhere.
Now we endow $T$ with addition and multiplication: for $x, y \in T$, let
:$x + y = \map {\theta^{-1} } {\map \theta x + \map \theta y}$
:$x y = \map {\theta^{-1} } {\map \theta x \, \map \theta y}$
so trivially we have
:$\map \theta {x + y} = \map \theta x + \map \theta y$
:$\map \theta {x y} = \map \theta x \, \map \theta y$
Therefore $T$ is the isomorphic image of a ring, and an isomorphism is precisely a map that preserves the ring axioms.
Thus $T$ is a ring isomorphic to $S$ containing $R$ as a subring.
{{qed}}
{{proof wanted|I wonder whether this page might be no more than an application of the Embedding Theorem.}}
Category:Ring Isomorphisms
Category:Ring Monomorphisms
orqom74s8aylmkd8ikavss1u965boik
Theorem: Clopen Set contains Components of All its Points

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both closed and open in $T$.
Then $H$ contains the components of all of its points.

Proof:
Let $H$ be a clopen set in $T$.
By definition, $H$ is open and so $H \in \tau$.
But as $H$ is also closed, by definition $\relcomp S H \in \tau$ where $\complement_S$ denotes complement relative to $S$.
Thus $H$ and $\relcomp S H$ are both open such that:
:$H \cap \relcomp S H = \O$ from Intersection with Relative Complement is Empty
:$H \cup \relcomp S H = S$ from Union with Relative Complement
and so forming a partition of $T$.
As $H$ and $\relcomp S H$ are both closed, it follows from Closed Set Equals its Closure that:
:$H \cap \paren {\relcomp S H}^- = \O = H^- \cap \relcomp S H$
and so by definition $H$ and $\relcomp S H$ are separated.
{{AimForCont}} that $H$ does not contain the components of all of its points.
That is:
:$\exists x \in H: \map {\operatorname {Comp}_x} T \nsubseteq H$
where $\map {\operatorname {Comp}_x} T$ is the component of $x$ in $T$.
Then:
:$\exists z \in \map {\operatorname {Comp}_x} T: z \in H \cup \relcomp S H$
This means that $H \cup \relcomp S H$ is connected.
This contradicts the fact that $H$ and $\relcomp S H$ are separated.
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Connectedness Between Two Points is an Equivalence Relation

Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff \exists S \subseteq T: S$ is connected between the two points  $a$ and $b$
where $a, b \in X$.
Then $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Quasicomponent is Intersection of Clopen Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $p \in S$.
Then the quasicomponent containing $p$ equals the intersection of all sets which are both open and closed which contain $p$.

Proof:
Let $C$ be the quasicomponent of $p$.
Let $Q$ be the set of clopen sets containing $p$ that are not equal to $S$.
First let $x \in C$.
Let $U \in Q$.
By definition of $Q$:
:$p \in U$
Let $V = \relcomp S U$ be the complement of $U$ relative to $S$.
By Complement of Clopen Set is Clopen $V$ is also a clopen set of $T$.
By Clopen Set and Complement form Separation $U$ and $V$ separate $S$.
By the definition of quasicomponent, $U \mid V$ is a separation of $S$ which contains a single open set $U$ containing both $x$ and $p$.
Thus:
:$x \in U$
As this holds for all such $U$:
:$x \in \ds \bigcap Q$
As this holds for all such $x$:
:$C \subseteq Q$
{{qed|lemma}}
Let $x \in \ds \bigcap Q$.
Let $U \mid V$ be a separation of $S$.
Let $p \in U$.
Then $U \in Q$.
Thus $x \in U$.
Thus $x$ lies on the same side as $p$ of any separation of $X$.
So $x \in C$.
{{qed}}
Theorem: Topological Space with One Quasicomponent is Connected

Let $T = \struct {S, \tau}$ be a topological space which has one quasicomponent.
Then $T$ is connected.

Proof:
Let $x \in S$.
By hypothesis, the quasicomponent of $x$ is $S$ itself.
Thus by definition of quasicomponent:
:$\forall y \in S: y \sim x$
where $\sim$ is the relation defined on $T$ as:
:$x \sim y \iff T$ is connected between the two points $x$ and $y$
Let $K = \ds \bigcap_{x \mathop \in U} U: U$ is clopen in $T$.
By Quasicomponent is Intersection of Clopen Sets:
:$\ds \bigcap K = S$
Thus there is no non-empty clopen set of $T$ apart from $S$.
The result follows by definition of connected topological space.
{{qed}}
Theorem: Localization of Ring Exists

Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$.
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:
:$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$
Theorem: Transitivity of Integrality

Let $A \subseteq B \subseteq C$ be extensions of commutative rings with unity.
Suppose that $C$ is integral over $B$, and $B$ is integral over $A$.
Then $C$ is integral over $A$.

Proof:
First, a lemma:
Theorem: Integral Closure is Integrally Closed

Let $A \subseteq B$ be an extension of commutative rings with unity.
Let $C$ be the integral closure of $A$ in $B$.
Then $C$ is integrally closed.

Proof:
Suppose $x \in B$ is integral over $C$.
Certainly $C$ is integral over $A$, so by Transitivity of Integrality, $C \sqbrk x$ is integral over $A$.
In particular, $x$ is integral over $A$, so $x \in C$.
{{Qed}}
Category:Algebraic Number Theory
Category:Commutative Algebra
3g683hc822lfpsjlmes125b6m8fhhme
Theorem: Universal Property for Field of Quotients

Let $\struct {D, +, \circ}$ be an integral domain.
Let $\struct {F, \oplus, \cdot}$ be a field of quotients of $D$.
Then $F$ satisfies the following universal property:
There exists a (ring) homomorphism $\iota : D \to F$ such that:
::for every field $\tilde F$ and
:and:
::for every (ring) homomorphism $\phi: D \to \tilde F$
:there exists a unique field homomorphism $\psi: F \to \tilde F$ satisfying:
:::$\psi \iota = \phi$
That is, the following diagram commutes:
:<nowiki>$\xymatrix{
  D \ar[r]^{\iota} \ar[dr]_{\phi}  & F \ar[d]^{\exists!\psi} \\
  {}                               & {\tilde F}
}$</nowiki>
Namely we may take:
:$\psi: a / b \mapsto \map \phi a \map \phi b^{-1}$

Proof:
{{ProofWanted}}
Category:Fields of Quotients
Category:Integral Domains
Category:Universal Properties
ci1x76tsuac3v1qn7xlmxos3tz8cv19
Theorem: Constant Mapping is Continuous

Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $b \in B$ be any point in $B$.
Let $f_b: A \to B$ be the constant mapping defined by:
:$\forall x \in A: \map {f_b} x = b$
Then $f_b$ is continuous.

Proof:
We have by definition that:
:$\forall x \in A: \map {f_b} x = b$
So:
$\map {f^{-1} } b = A$
For $c \in B: c \ne B$ we have that:
$\map {f^{-1} } c = \O$
Let $U \in \tau_B$ such that $b \in U$.
Then $f^{-1} \sqbrk U = A$
Let $V \in \tau_B$ such that $b \notin V$.
Then $f^{-1} \sqbrk V = \O$
From the definition of topology, $A$ is open in $T$.
From Empty Set is Element of Topology, $\O$ is also open in $T$.
So $f_b$ fulfils the conditions for it to be continuous.
{{qed}}
Theorem: Path-Connectedness is Equivalence Relation

Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is path-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Injective Path-Connectedness is Equivalence Relation

Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is injectively path-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Joining Injective Paths makes Another Injective Path

Let $T$ be a topological space.
Let $\mathbb I \subseteq \R$ be the closed unit interval $\closedint 0 1$.
Let $a, b, c$ be three distinct points of $T$.
Let $f, g: \mathbb I \to T$ be injective paths in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
Let $h: \mathbb I \to T$ be the mapping given by:
:<nowiki>$\map h x = \begin{cases}
\map f {2 x} & : x \in \closedint 0 {\dfrac 1 2} \\
\map g {2 x - 1} & : x \in \closedint {\dfrac 1 2} 1
\end{cases}$</nowiki>
Then either:
:$h$ is an injective path in $T$
or
:There exists some restriction of $h$ which, possibly after reparametrisation, is an injective path in $T$.

Proof:
From Injective Path in Topological Space is Path, $f$ and $g$ are also paths in $T$.
So by Joining Paths makes Another Path it follows that $h$ is a path in $T$.
Now if $\Img f \cap \Img g = b$ it can be seen that:
:<nowiki>$\forall x \in \Img h : x = \begin{cases}
\map f y & \text {for some $y \in \closedint 0 {\dfrac 1 2}$}, \text{ or} \\
\map g z & \text {for some $z \in \closedint {\dfrac 1 2} 1$}
\end{cases}$</nowiki>
and it follows that $h$ is an injection, and therefore an injective path.
{{mistake|that should be an XOR, really. And the intervals for $y$ and $z$ are simply $I$}}
On the other hand, suppose:
:$\exists y \in \closedint 0 {\dfrac 1 2}: \exists z \in \closedint {\dfrac 1 2} 1: \map f y = \map g z$
such that $\map f y \ne b$.
{{finish|It remains to show that you can build an injective path out of the bits up till where the injective paths cross. It's intuitively obvious but requires some analysis work.}}
{{qed}}
Theorem: Injective Path in Topological Space is Path

Let $T = \struct {X, \tau}$ be a topological space.
Let $f$ be an injective path in $T$.
Then $f$ is a path in $T$.

Proof:
By definition, an injective path from $a$ to $b$ is a path $f: I \to T$ such that $f$ is injective.
Hence the result, by definition.
{{qed}}
Theorem: Relationship between Component Types

Let $T = \struct {S, \tau}$ be a topological space.
Let $p \in S$.
Let:
:$A$ be the arc component of $p$
:$J$ be the injective path component of $p$
:$P$ be the path component of $p$
:$C$ be the component of $p$
:$Q$ be the quasicomponent of $p$.
Then:
:$A \subseteq P \subseteq C \subseteq Q$
In general, the inclusions do not hold in the other direction.

Proof:
Let $f \in A$.
By Arc in Topological Space is Injective Path we have that $f \in J$.
That is, $A \subseteq J$.
{{qed|lemma}}
Let $f \in J$.
By Injective Path in Topological Space is Path we have that $f \in P$.
That is, $J \subseteq P$.
{{qed|lemma}}
Let $f \in P$.
From Path-Connected Space is Connected we have directly that $P \subseteq C$.
{{qed|lemma}}
Let $f \in C$.
From Connected Space is Connected Between Two Points we have directly that $C \subseteq Q$.
{{qed|lemma}}
Then we have:
* Injective Path Component is not necessarily Arc Component
* Path Component is not necessarily Injective Path Component
* Component is not necessarily Path Component
* Quasicomponent is not necessarily Component
{{qed}}
Theorem: Increasing Sequence in Ordered Set Terminates iff Maximal Element

Let $\struct {P, \le}$ be an ordered set.
{{TFAE}}
:$(1): \quad$ Every increasing sequence $x_1 \le x_2 \le x_3 \le \cdots$ with $x_i \in P$ eventually terminates: there is $n \in \N$ such that $x_n = x_{n + 1} = \cdots$.
:$(2): \quad$ Every non-empty subset of $P$ has a maximal element.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose $(1)$ holds.
Pick $\O \ne S \subseteq P$.
Let $x_1 \in S$ be arbitrary.
Given $x_k \in S$, pick $x_{k + 1} \in S$ strictly bigger than $x_k$.
By hypothesis the process must eventually terminate, say $x_n$ is the last element.
{{explain|Why does this terminate? I think Axiom:Axiom of Choice is necessary here}}
Then by construction there are no larger elements than $x_n$, that is, $x_n$ is a maximal element of $S$.
{{qed|lemma}}
{{improve|Last three sentences can be made more explicit and less handwaving}}
Theorem: Ultraconnected Space is Path-Connected

Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is path-connected.

Proof:
Let $T = \struct {S, \tau}$ be ultraconnected.
Let $a, b \in S$.
Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the closure of $\set a$.
Such a $p$ can be chosen, as $T$ being ultraconnected guarantees that $\set a^- \cap \set b^- \ne \O$.
Consider the mapping $f: \closedint 0 1 \to X$ such that:
:<nowiki>$\map f x = \begin {cases}
a & : x \in \hointr 0 {\dfrac 1 2} \\
p & : x = \dfrac 1 2 \\
b & : x \in \hointl {\dfrac 1 2} 1 \\
\end {cases}$</nowiki>
Then $f$ is continuous.
{{explain|The above is a nontrivial claim}}
The result follows from the definition of path-connectedness.
{{qed}}
Theorem: Continuous Real-Valued Function on Irreducible Space is Constant

Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Let $f: S \to \R$ be a continuous real-valued function.
Then $f$ is constant, that is:
:$\exists a \in \R: \forall x \in S: \map f x = a$

Proof:
{{Recall|Irreducible Space|irreducible space}}
{{:Definition:Irreducible Space/Definition 3}}
For every $x \in \R$, define:
:$L_x := f^{-1} \sqbrk {\openint {-\infty} x}$
:$U_x := f^{-1} \sqbrk {\openint x \infty}$
By continuity of $f$, these are open in $T$.
They are also disjoint, because for each $s \in S$, $\map f s < x$ and $\map f s > x$ cannot both hold simultaneously.
{{AimForCont}} there exist $a, b \in \R: a < b$ such that:
:$\map f s = a$
:$\map f {s'} = b$
for some $s, s' \in S$.
Define:
:$c = \dfrac {a + b} 2$
Then $a < c < b$, whence:
:$s \in L_c$ and $s' \in U_c$
But, as noted above:
:$L_c \cap U_c = \O$
Because $T$ is irreducible, it follows that either $L_c$ or $U_c$ must be empty.
This contradicts the existence of such $a$ and $b$.
Hence $f$ must be constant.
{{qed}}
Theorem: Irreducible Space is Pseudocompact

Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is pseudocompact.

Proof:
We have that Continuous Real-Valued Function on Irreducible Space is Constant.
A constant mapping is trivially bounded.
Hence the result by definition of pseudocompact.
{{qed}}
Theorem: Sequence of Implications of Connectedness Properties

Let $P_1$ and $P_2$ be connectedness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | ||
| align="center" | ||
| align="center" | Ultraconnected ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Injectively Path-Connected ||
| align="center" | $\implies$ ||
| align="center" | Path-Connected ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Irreducible ||
| align="center" | $\implies$ ||
| align="center" | Connected ||
|-
|}
If there is no implication indicated, then there exists a counterexample demonstrating that no such implication exists.

Proof:
The relevant justifications are listed as follows:
:Ultraconnected Space is Connected
:Ultraconnected Space is Path-Connected
:Injectively Path-Connected Space is Path-Connected
:Path-Connected Space is Connected
:Irreducible Space is Connected
{{qed|lemma}}
Then we have:
:Connected Space is not necessarily Locally Connected
:Connected Space is not necessarily Path-Connected
:Connected Space is not necessarily Irreducible
:Path-Connected Space is not necessarily Injectively Path-Connected
:Path-Connected Space is not necessarily Irreducible
:Path-Connected Space is not necessarily Ultraconnected
:Injectively Path-Connected Space is not necessarily Irreducible
:Injectively Path-Connected Space is not necessarily Ultraconnected
:Irreducible Space is not necessarily Ultraconnected
:Ultraconnected Space is not necessarily Irreducible
:Ultraconnected Space is not necessarily Injectively Path-Connected
:Irreducible Space is not necessarily Injectively Path-Connected
:Irreducible Space is not necessarily Path-Connected
although we do have:
:Finite Irreducible Space is Path-Connected
:Irreducible Space with Finitely Many Open Sets is Path-Connected
{{qed}}
Theorem: Injectively Path-Connected Space is Path-Connected

Let $T = \struct {S, \tau}$ be a topological space which is injectively path-connected.
Then $T$ is path-connected.

Proof:
Let $T = \struct {S, \tau}$ be injectively path-connected.
Then $\forall x, y \in S$, there exists a continuous injection $f: \closedint 0 1 \to S$, such that $\map f 0 = x$ and $\map f 1 = y$.
As $f$ is a continuous injection, it is also simply a continuous mapping.
The result follows from the definition of path-connectedness.
{{qed}}
Theorem: Irreducible Space is Connected

Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is connected.

Proof:
Let $T = \struct {S, \tau}$ be irreducible.
Then:
:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
So trivially there are no two open sets that can form a separation of $T$.
The result follows from definition of connected.
{{qed}}
Theorem: Non-Trivial Ultraconnected Space is not T1

Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
If $S$ has more than one element, then $T$ is not a $T_1$ space.
That is, if $T$ is a $T_1$ space with more than one element, it is not ultraconnected.

Proof:
Let $T = \struct {S, \tau}$ be ultraconnected.
{{Recall|Ultraconnected Space|ultraconnected space|index = 2}}
{{:Definition:Ultraconnected Space/Definition 2}}
Let $a, b \in S$ such that $a \ne b$.
{{AimForCont}} $T$ is a $T_1$ space.
{{Recall|T1 Space|$T_1$ space|index = 3}}
{{:Definition:T1 Space/Definition 3}}
Hence $\set a$ and $\set b$ are both closed.
From Closed Set Equals its Closure we have that $\set a^- = \set a$ and $\set b^- = \set b$.
It immediately follows that:
:$\set a^- \cap \set b^- = \O$
But that contradicts the definition of ultraconnected space above. 
The result follows by Proof by Contradiction.
{{qed}}
Theorem: Ultraconnected Space is T4

Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
Then $T$ is a $T_4$ space.

Proof:
{{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
As no two closed sets of an ultraconnected space are actually disjoint, it follows that $T_4$-ness follows vacuously.
{{qed}}
Theorem: Dot Product of Vector with Itself

:$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$

Proof:
Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf u
      | r = u_1 u_1 + u_2 u_2 + \cdots + u_n u_n
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = {u_1}^2 + {u_2}^2 + \cdots + {u_n}^2
      | c = 
}}
{{eqn | r = \paren {\sqrt {\sum_{i \mathop = 1}^n {u_i}^2} }^2
      | c = 
}}
{{eqn | r = \norm {\mathbf u}^2
      | c = {{Defof|Vector Length|subdef = Real Vector Space|Vector Length in $\R^n$}}
}}
{{end-eqn}}
{{qed}}
Theorem: Dot Product of Vector with Itself

:$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf u \cdot \mathbf u
      | r = \norm {\mathbf u} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf u
      | c = Cosine Formula for Dot Product
}}
{{eqn | r = \norm {\mathbf u}^2 \cos 0
      | c = since the angle between a vector and itself is $0$
}}
{{eqn | r = \norm {\mathbf u}^2
      | c = Cosine of Zero is One
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine Formula for Dot Product

Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^n$.
The dot product of $\mathbf v$ and $\mathbf w$ can be calculated by:
:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
:$\norm {\, \cdot \,}$ denotes vector length and
:$\theta$ is the angle between $\mathbf v$ and $\mathbf w$.

Proof:
Let $\mathbf v$ and $\mathbf w$ be considered to be embedded in a Cartesian plane $\CC$.
By Dot Product is Invariant under Coordinate Rotation, we may rotate $\CC$ arbitrarily, and $\mathbf v \cdot \mathbf w$ will not change.
So, let us rotate $\CC$ to $\CC'$ such that the $x$-axis is parallel to $\mathbf v$.
Hence $\mathbf v$ can be expressed as:
{{begin-eqn}}
{{eqn | l = \mathbf v
      | r = \norm {\mathbf v} \mathbf i + 0 \mathbf j + 0 \mathbf k
}}
{{eqn | l = \mathbf w
      | r = \norm {\mathbf w} \cos \theta \mathbf i + \norm {\mathbf w} \sin \theta \mathbf j + 0 \mathbf k
}}
{{end-eqn}}
Hence: by definition of dot product:
{{begin-eqn}}
{{eqn | l = \mathbf v \cdot \mathbf w
      | r = \norm {\mathbf v} \times \norm {\mathbf w} \cos \theta + 0 \times \norm {\mathbf w} \sin \theta + 0 \times 0
      | c = {{Defof|Dot Product|subdef = General Context}}
}}
{{eqn | r = \norm {\mathbf v} \norm {\mathbf w} \cos \theta
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Irreducible Space is Locally Connected

Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is locally connected.

Proof:
{{Recall|Locally Connected Space|locally connected space}}
{{:Definition:Locally Connected Space/Definition 3}}
Let $T = \struct {S, \tau}$ be irreducible.
{{Recall|Irreducible Space|irreducible space}}
{{:Definition:Irreducible Space/Definition 3}}
So trivially there are no two open sets that can form a separation of $T$.
As a basis consists of open sets, this applies to all sets in a basis for $T$.
The result follows from definition of locally connected.
{{qed}}
Theorem: Locally Injectively Path-Connected Space is Locally Path-Connected

Let $T = \struct {S, \tau}$ be a topological space which is locally injectively path-connected.
Then $T$ is also locally path-connected.

Proof:
Let $T = \struct {S, \tau}$ be locally injectively path-connected.
Then $T$ has a basis consisting entirely of injectively path-connected sets.
From Injectively Path-Connected Space is Path-Connected, this basis consisting entirely of path-connected sets.
The result follows from definition of locally path-connected.
{{qed}}
Theorem: Locally Path-Connected Space is Locally Connected

Let $T = \struct {S, \tau}$ be a topological space which is locally path-connected.
Then $T$ is also locally connected.

Proof:
Let $x \in S$ be any point of $T$.
Let $\BB$ be a local basis of path-connected sets for $x$.
From Path-Connected Space is Connected, $\BB$ is a local basis of connected sets.
Thus, $T$ is locally connected by definition.
{{qed}}