nyuuzyou/wikis huggingface dataset:
Completeness Theorem for Propositional Tableaux and Boolean Interpretations

== Theorem ==
Tableau proofs (in terms of propositional tableaux) are a complete proof system for boolean interpretations.
That is, for every WFF $\mathbf A$:
:$\models_{\mathrm{BI} } \mathbf A$ implies $\vdash_{\mathrm{PT} } \mathbf A$
== Proof ==
This is a corollary of the Extended Completeness Theorem for Propositional Tableaux and Boolean Interpretations.
Namely, it is the special case $\mathbf H = \O$.
Hence the result.
== Also see ==
The Soundness Theorem for Propositional Tableaux and Boolean Interpretations in which it is proved that:
:If $\vdash \mathbf A$ then $\models \mathbf A$.
== Sources ==
* : $\S 1.10$: Completeness
Category:Completeness Theorem
Category:Propositional Tableaux
Category:Named Theorems
Completeness Theorem for Semantic Tableaux

== Theorem ==
Let $\mathbf A$ be a WFF of propositional logic.
Let $\mathbf A$ be unsatisfiable for boolean interpretations.
Then every completed tableau for $\mathbf A$ is closed.
== Proof ==
 some open tableau $T$ for $\mathbf A$ exists.
We will prove that $\mathbf A$ must be satisfiable.
This claim is proved by inductively establishing the following claim for all nodes $t$ of $T$:
:If a descendant leaf node of $t$ is marked open, then $\map U t$ is satisfiable.
By the Semantic Tableau Algorithm, this statement is immediate for leaf nodes.
For, sought satisfiability is the condition for a leaf to be marked open.
Inductively suppose that all children of $t$ satisfy the mentioned condition.
Now suppose that $t$ has a descendant marked open leaf.
Then obviously some child $t'$ of $t$ also has a descendant marked open leaf.
By induction hypothesis, $\map U {t'}$ is satisfiable.
First, the case that the formula $\mathbf B$ used at $t$ by the Semantic Tableau Algorithm is an $\alpha$-formula.
Then $\map U {t'}$ is obtained from $\map U t$ by replacing $\mathbf B$ by $\mathbf B_1$ and $\mathbf B_2$.
Here, $\mathbf B_1$ and $\mathbf B_2$ are the formulas such that $\mathbf B$ is semantically equivalent to $\mathbf B_1 \land \mathbf B_2$.
It follows that if:
:$v \models_{\mathrm{BI}} \map U {t'}$
for some boolean interpretation $v$, then also:
:$v \models_{\mathrm{BI}} \map U t$
by the boolean interpretation of $\land$.
Consequently, $\map U t$ is satisfiable.
Next, the case that $\mathbf B$ is a $\beta$-formula.
Then $\mathbf B$ is semantically equivalent to $\mathbf B_1 \lor \mathbf B_2$.
Then, by the Rule of Addition:
:$\mathbf B_1 \models_{\mathrm{BI}} \mathbf B$
:$\mathbf B_2 \models_{\mathrm{BI}} \mathbf B$
Consequently, by the definitions of $\map U {t'}$ and $\map U {t' '}$:
:$\map U {t'} \models_{\mathrm{BI}} \map U t$
:$\map U {t' '} \models_{\mathrm{BI}} \map U t$
By hypothesis, one of $\map U {t'}$ and $\map U {t' '}$ must be satisfiable.
Hence, so must $\map U t$ be.
This establishes our claim.
Applying the claim to the root node, that is, to the whole tableau, we obtain the result.
== Sources ==
* : $\S 2.7.2$
Category:Completeness Theorem
Category:Propositional Logic
Category:Named Theorems
Completing the Square

== Theorem ==
Let $a, b, c, x$ be real numbers with $a \ne 0$.
Then:
:$a x^2 + b x + c = \dfrac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}$
This process is known as completing the square.
== Proof ==
== Also presented as ==
This result can also be presented in the form:
:$a x^2 + b x + c = \dfrac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}$
== Also see ==
* Definition:Tschirnhaus Transformation
* Quadratic Formula
== Historical Note ==
== Sources ==
* 
* 
* : Historical Introduction: Polynomial Equations
* 
* 
* 
Category:Algebra
Category:Proof Techniques
Completion Theorem (Measure Space)/Lemma

== Lemma ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\NN$ and $\Sigma^*$ be defined as:
:$\NN := \set {N \subseteq X: \exists M \in \Sigma: \map \mu M = 0, N \subseteq M}$
:$\Sigma^* := \set {E \cup N: E \in \Sigma, N \in \NN}$
Next, define $\bar \mu: \Sigma^* \to \overline \R_{\ge 0}$ by:
:$\map {\bar \mu} {E \cup N} := \map \mu E$
The mapping $\bar \mu$ is well-defined, i.e.:
:$\forall E, F \in \Sigma: \forall N, M \in \NN: E \cup N = F \cup M \implies \map \mu E = \map \mu F$
== Proof ==
Let $N_0, M_0 \in \Sigma$ be null sets such that $N \subseteq N_0, M \subseteq M_0$.
Then:
:$E \subseteq E \cup N = F \cup M \subseteq F \cup M_0$
so that:
:$\map \mu E \le \map \mu {F \cup M_0} \le \map \mu F + \map \mu {M_0} = \map \mu F + 0$
Analogously:
:$F \subseteq F \cup M = E \cup N \subseteq E \cup N_0$
so that:
:$\map \mu F \le \map \mu {E \cup N_0} \le \map \mu E + \map \mu {N_0} = \map \mu E + 0$
In total:
:$\map \mu E = \map \mu F$
Category:Completion Theorem (Measure Space)
Completion Theorem (Metric Space)

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then there exists a completion $\tilde M = \struct {\tilde A, \tilde d}$ of $\struct {A, d}$.
Moreover, this completion is unique up to isometry.
That is, if $\struct {\hat A, \hat d}$ is another completion of $\struct {A, d}$, then there is a bijection $\tau: \tilde A \leftrightarrow \hat A$ such that:
:$(1): \quad \tau$ restricts to the identity on $x$:
::::$\forall x \in A: \map \tau x = x$
:$(2): \quad \tau$ preserves metrics:
::::$\forall x_1, x_2 \in A : \map {\hat d} {\map \tau {x_1}, \map \tau {x_2} } = \map {\tilde d} {x_1, x_2}$
== Proof ==
We construct the completion of a metric space as equivalence classes of the set of Cauchy sequences in the space under a suitable equivalence relation.
Let $\struct {A, d}$ be a metric space.
Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$.
Define a relation $\sim$ on $\CC \sqbrk A$ by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
By Equivalence Relation on Cauchy Sequences, $\sim$ is an equivalence relation on $\CC \sqbrk A$.
Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\eqclass {x_n} \sim$.
Denote the set of equivalence classes under $\sim$ by $\tilde A$.
By Relation Partitions Set iff Equivalence this is a partition of $\CC \sqbrk A$.
That is, each $\sequence {x_n} \in \CC \sqbrk A$ lies in one and only one equivalence class under $\sim$.
Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:
:$\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
=== Lemma 1: $\tilde d$ is Well-Defined ===
We claim that $\struct {\tilde A, \tilde d}$ is a completion of $\struct {A, d}$.
Therefore we must show that:
:$\tilde d$ is a metric on $\tilde A$
:There exists an everywhere dense inclusion $\struct {A, d} \to \struct {\tilde A, \tilde d}$ preserving $d$.
In addition the theorem claims that $\struct {\tilde A, \tilde d}$ is unique up to isometry.
=== Lemma 2: $\tilde d$ is a Metric ===
=== Lemma 3: $\tilde A$ is a Completion of $A$ ===
=== Lemma 4: Uniqueness of $\tilde A$ ===
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Completion Theorem (Metric Space)
Category:Isometries (Metric Spaces)
Category:Metric Spaces
Category:Complete Metric Spaces
Category:Completion Theorem
Completion Theorem (Metric Space)/Lemma 1

== Lemma ==
Let $M = \struct {A, d}$ be a metric space.
Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$.
Define the equivalence relation $\sim$ on $\CC \sqbrk A$ by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\eqclass {x_n} \sim$.
Denote the set of equivalence classes under $\sim$ by $\tilde A$.
Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:
:$\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
Then:
:$\tilde d$ is well-defined on $\tilde A$.
== Proof ==
Let $\sequence {x_n}$, $\sequence {\hat x_n}$, $\sequence {y_n}$, $\sequence {\hat y_n} \in \CC \sqbrk A$ be such that:
:$\sequence {x_n} \sim \sequence {\hat x_n}$
:$\sequence {y_n} \sim \sequence {\hat y_n}$
We have:
By an identical argument, we can also show that:
:$\map d {\hat x_n, \hat y_n} - \map d {x_n, y_n} \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$ 
and therefore:
:$\ds 0 \le \size {\map d {x_n, y_n} - \map d {\hat x_n, \hat y_n} } \le \map d {x_n, \hat x_n} + \map d {\hat y_n, y_n}$
Passing to the limit $n \to \infty$ and using the Combination Theorem for Sequences we have shown that:
$\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {\hat x_n, \hat y_n}$
But this precisely means that:
:$\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \map {\tilde d} {\eqclass {\hat x_n} \sim, \eqclass {\hat y_n} \sim}$
Category:Completion Theorem (Metric Space)
Completion Theorem (Metric Space)/Lemma 2

== Lemma ==
Let $M = \struct {A, d}$ be a metric space.
Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$.
Define the equivalence relation $\sim$ on $\CC \sqbrk A$ by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\eqclass {x_n} \sim$.
Denote the set of equivalence classes under $\sim$ by $\tilde A$.
Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:
:$\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
Then:
:$\tilde d$ is a metric on $\tilde A$.
== Proof ==
To prove $\tilde d$ is a metric, we verify that it satisfies the metric space axioms.
=== Proof of ===
Let $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \infty$.
Then $\sequence {x_n}$ and $\sequence {y_n}$ cannot both be Cauchy.
So $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} < \infty$ for $\eqclass {x_n} \sim, \eqclass {y_n} \sim \in \tilde A$.
By definition of $\tilde d$, for any $\eqclass {x_n} \sim, \eqclass {y_n} \sim \in \tilde A$, $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim}$ must be a limit point of $R_{\ge 0}$.
The closure of $\R_{\ge 0}$ is $\R_{\ge 0}$, so $\tilde d: \tilde A \times \tilde A \to \R_{\ge 0}$.
So holds for $\tilde d$.
=== Proof of ===
Let $\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = 0$, which means that:
:$\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
So by definition:
:$\sequence {x_n} \sim \sequence {y_n}$
and:
:$\eqclass {x_n} \sim = \eqclass {y_n} \sim$
As $d$ is a metric, we also find immediately:
:$\map {\tilde d} {\eqclass {x_n} \sim, \eqclass {x_n} \sim} = 0$
So holds for $\tilde d$.
=== Proof of ===
We have that:
So holds for $\tilde d$.
=== Proof of ===
We have that:
So holds for $\tilde d$.
Thus $\tilde d$ satisfies all the metric space axioms and so is a metric.
Category:Completion Theorem (Metric Space)
Completion Theorem (Metric Space)/Lemma 3

== Lemma ==
Let $M = \struct {A, d}$ be a metric space.
Let $\CC \sqbrk A$ denote the set of all Cauchy sequences in $A$.
Define the equivalence relation $\sim$ on $\CC \sqbrk A$ by:
:$\ds \sequence {x_n} \sim \sequence {y_n} \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$
Denote the equivalence class of $\sequence {x_n} \in \CC \sqbrk A$ by $\eqclass {x_n} \sim$.
Denote the set of equivalence classes under $\sim$ by $\tilde A$.
Define $\tilde d: \tilde A \to \R_{\ge 0}$ by:
:$\ds \map {\tilde d} {\eqclass {x_n} \sim, \eqclass {y_n} \sim} = \lim_{n \mathop \to \infty} \map d {x_n, y_n}$
Then:
:$\tilde M = \struct {\tilde A, \tilde d}$ is a completion of $M$.
== Proof ==
We are to show that:
:$(1): \quad \tilde M$ is a complete metric space
:$(2): \quad A \subseteq \tilde A$
:$(3): \quad A$ is dense in $\tilde M$
:$(4): \quad \forall x, y \in A : \map {\tilde d} {x, y} = \map d {x, y}$
For $x \in A$, let $\hat x = \tuple {x, x, x, \ldots}$ be the constant sequence with value $x$.
Let $\phi: A \to \tilde A: x = \eqclass {\hat x} \sim$.
We first demonstrate that $(2)$ holds, by showing that $A \subseteq \tilde A$.
Thus:
:$A \subseteq \tilde A$
Henceforth we identify $A$ with its isomorphic copy in $\tilde A$ when it is convenient.
Now we demonstrate that $(4)$ holds, by showing that $\phi$ is an injection from $A$ into $\tilde A$.
For any $x, y \in A$:
That is:
:$\forall x, y \in A : \map {\tilde d} {x, y} = \map d {x, y}$
Now we demonstrate that $(3)$ holds, by showing that $A$ is dense in $\tilde A$.
Recall that the closure of $A$ is the union of $A$ and the limit points of $A$.
Let $\eqclass {x_n} \sim \in \tilde A$ and $\epsilon \in \R_{>0}$ be arbitrary.
If we can find $x \in A$ such that $\map {\tilde d} {\eqclass {\hat x} \sim, \eqclass {x_n} \sim} < \epsilon$ then we have shown that $A$ is dense in $\tilde A$.
Since $\sequence {x_n}$ is Cauchy, there exists $N \in \N$ such that:
:$\forall m, n \ge N: \map d {x_m, x_n} < \epsilon$
Then we have:
and therefore $A$ is dense in $\tilde A$.
Finally we demonstrate that $(1)$ holds, by showing that $\struct {\tilde A, \tilde d}$ is complete.
By the completeness criterion it is sufficient to show that every Cauchy sequence in $\phi \sqbrk A$ converges in $\tilde A$.
Let $\sequence {\hat w_n}$ be a Cauchy sequence in $\phi \sqbrk A$, so each $\hat w_n$ has the form $\tuple {w_n, w_n, w_n, \ldots}$.
Since $\phi$ is an isometry:
:$\forall m, n \in \N: \map {\tilde d} {\hat w_n, \hat w_m} = \map d {w_n, w_m}$
Therefore, $\tuple {w_1, w_2, w_3, \ldots}$ is Cauchy in $A$.
Let $W = \sqbrk {\tuple {w_1, w_2, w_3, \ldots} } \in \tilde A$.
We claim that $\sequence {\hat w_n}$ converges to $W$ in $\tilde A$.
Let $\epsilon \in \R_{>0}$ be arbitrary.
Since $\tuple {w_1, w_2, w_3, \ldots}$ is Cauchy in $A$, there exists $N \in \N$ such that for all $m, n \ge N$, we have $\map d {w_n, w_m} < \epsilon$.
Thus for all $n > N$:
:$\ds \map {\tilde d} {w_n, W} = \lim_{n \mathop \to \infty} \map d {w_n, W} < \epsilon$ 
Therefore, $\sequence {\hat w_n} \to W$ as $N \to \infty$, and $\tilde A$ is complete.
Category:Completion Theorem (Metric Space)
Completion Theorem (Metric Space)/Lemma 4

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\tilde M = \struct {\tilde A, \tilde d}$ be a completion of $\struct {A, d}$.
Then:
:$\tilde M = \struct {\tilde A, \tilde d}$ is unique up to isometry.
That is, if $\struct {\hat A, \hat d}$ is another completion of $\struct {A, d}$, then there is a bijection $\tau: \tilde A \leftrightarrow \hat A$ such that:
:$(1): \quad \tau$ restricts to the identity on $x$:
:::: $\forall x \in A : \map \tau x = x$
:$(2): \quad \tau$ preserves metrics:
:::: $\forall x_1, x_2 \in A : \map {\hat d} {\map \tau {x_1}, \map \tau {x_2} } = \map {\tilde d} {x_1, x_2}$
== Proof ==
Let $M_1 = \struct {\tilde{A_1}, \tilde{d_1}, \phi_1}$ and $M_2 = \struct {\tilde{A_2}, \tilde{d_2}, \phi_2}$ be two completions of $\struct {A, d}$.
Here, $\phi_1: A \to A_1$ and $\phi_2: A \to A_2$ are isometries
By Composite of Isometries is Isometry, $\psi = \phi_1^{-1} \circ \phi_2$ gives an isometry from $\phi_1 \sqbrk A$ to $\phi_2 \sqbrk A$.
The sets $\phi_1 \sqbrk A$ and $\phi_2 \sqbrk A$ are dense in $A_1$ and $A_2$ respectively.
Thus we can extend $\psi$ continuously to a map $\psi: A_1 \to A_2$.
That is, for $x \in A_1$, we can find a Cauchy sequence $\sequence {w_n}$ in $A_1$ with limit $x$.
Then we define:
:$\ds \map \psi x := \lim_{n \mathop \to \infty} \map \psi {w_n}$
which converges as $A_2$ is complete.
By Metric Space is Hausdorff, $A$ is a Hausdorff space.
By Hausdorff Condition is Preserved under Homeomorphism, $A_1$ and $A_2$ are also Hausdorff.
Therefore, by Convergent Sequence in Hausdorff Space has Unique Limit, $\psi$ is well-defined.
It is to be shown that $\psi$ is surjective:
For $y \in \tilde {A_2}$, let $\sequence {w_n'}$ be a Cauchy sequence in $\phi_2 \sqbrk A$ with limit $y$ in $\tilde {A_2}$.
Let $z_n$ be the preimage of the $w_n'$ under $\psi$.
Then, as $A_2$ is Hausdorff:
:$\ds \lim_{n \mathop \to \infty} \map \psi {z_n} = y$
as required.
injectivity of $\psi$ holds because $A_1$ is Hausdorff, as follows:
Suppose that:
:$\ds \lim_{n \mathop \to \infty} \map \psi {w_n} = \lim_{n \mathop \to \infty} \map \psi {w_n'}$
:$\ds \lim_{n \mathop \to \infty} w_n = w$
and:
:$\ds \lim_{n \mathop \to \infty} w_n' = w'$
For $\epsilon > 0$, pick $M \in \N$ such that $\map \psi {w_n}$, $\map \psi {w_n'}$ lie in the open $\epsilon$-ball $\map {B_{\epsilon / 2} } {\map \psi w}$ for all $n \ge M$.
Then we have:
:$\map {\tilde {d_1} } {w_n, w_n'} = \map {\tilde {d_2} } {\map \psi {w_n}, \map \psi {w_n'} } \le \epsilon$
As $A_1$ is Hausdorff, we conclude $w = w'$, and the task is complete.
Finally, from Distance Function of Metric Space is Continuous, it follows that $\psi$ is an isometry on all of $A_1$, and the proof is complete.
Category:Completion Theorem (Metric Space)
Completion Theorem (Normed Vector Space)

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then there exists a Banach space $\struct {\widetilde X, \widetilde {\norm {\, \cdot \,} } }$ and a linear isometry $\phi : X \to \widetilde X$ such that $\phi \sqbrk X$ is dense in $\widetilde X$.
Further, the Banach space $\struct {\widetilde X, \widetilde {\norm {\, \cdot \,} } }$ is unique up to isometric isomorphism.
== Proof ==
=== Proof of Existence ===
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm {\, \cdot \,} }$.
From Normed Dual Space is Banach Space, $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ is a Banach space. 
Let $\phi : X \to X^{\ast \ast}$ be the evaluation transformation.
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, $\phi$ is a linear isometry. 
Let $\widetilde X = \map \cl {\phi \sqbrk X}$ where $\cl$ denotes the closure in $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$.
Let $\widetilde {\norm {\, \cdot \,} }$ be the restriction of $\norm {\, \cdot \,}_{X^{\ast \ast} }$ to $\map \cl {\phi \sqbrk X}$.
From Subspace of Complete Metric Space is Closed iff Complete, $\struct {\widetilde X, \widetilde {\norm {\, \cdot \,} } }$ is a Banach space.
Further, $\phi \sqbrk X$ is dense in $\map \cl {\phi \sqbrk X}$.
So $\struct {\widetilde X, \widetilde {\norm {\, \cdot \,} } }$ satisfies the conditions of the theorem.
=== Proof of Uniqueness ===
Let:
:$\struct {X_1, \norm {\, \cdot \,}_1}$ and $\struct {X_2, \norm {\, \cdot \,}_2}$ be Banach spaces
:$\phi_1 : X \to X_1$, $\phi_2 : X \to X_2$ be linear isometries
such that $\phi_1 \sqbrk X$ and $\phi_2 \sqbrk X$ are everywhere dense in $X_1$ and $X_2$ respectively.
In particular, $\phi_1 : X \to \phi_1 \sqbrk X$ and $\phi_2 : X \to \phi_2 \sqbrk X$ are isometric isomorphisms. 
From Inverse of Linear Isometry is Linear Isometry, $\phi_1^{-1} : \phi_1 \sqbrk X \to X$ is an isometric isomorphism.
From Composition of Linear Isometries is Linear Isometry and Composite of Bijections is Bijection, $\phi_2 \circ \phi_1^{-1} : \phi_1 \sqbrk X \to \phi_2 \sqbrk X$ is an isometric isomorphism.
Since $\phi_1 \sqbrk X$ is dense in $X_1$ and $\struct {X_2, \norm {\, \cdot \,}_2}$ is a Banach space, we obtain:
:there exists a unique bounded linear transformation $\psi : X_1 \to X_2$ extending $\phi_2 \circ \phi_1^{-1}$
from Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain.
As shown in , $\psi$ is an isometric isomorphism.
Hence the completion is unique up to isometric isomorphism.
Category:Completion Theorem (Normed Vector Space)
Category:Banach Spaces
Category:Completion Theorem
Completion of Nondegenerate Bases

== Theorem ==
Let $\struct {V, q}$ be an $n$-dimensional scalar product space.
Let $\tuple {v_1, \ldots, v_k}$ be a nondegenerate $k$-tuple in $V$ with $0 \le k < n$. 
Then there exist vectors $v_{k+1}, \ldots, v_n \in V$ such that $\tuple {v_1, \ldots, v_n}$ is a nondegenerate basis for $V$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Scalar Product Spaces
Completion of Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,}_R }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
Let $Q = \CC / \NN$ where $\CC / \NN$ denotes a quotient ring.
Let $\norm {\, \cdot \,}_Q: Q \to \R_{\ge 0}$ be the norm on the quotient ring $Q$ defined by:
:$\ds \forall \sequence {x_n} + \NN: \norm {\sequence {x_n} + \NN }_Q = \lim_{n \mathop \to \infty} \norm{x_n}_R$
Then:
:$\struct {Q, \norm {\, \cdot \,}_Q}$ is a normed division ring completion of $\struct {R, \norm {\, \cdot \,}_R }$
== Proof ==
From Quotient Ring of Cauchy Sequences is Normed Division Ring:
:$\struct {Q, \norm {\, \cdot \,}_Q}$ is a normed division ring.
Let $d_R$ be the metric induced by $\struct {R, \norm {\, \cdot \,}_R }$.
Let $d_Q$ be the metric induced by $\struct {Q, \norm {\, \cdot \,}_Q}$.
From Quotient of Cauchy Sequences is Metric Completion, $\struct {Q, d_Q}$ is the metric completion of $\struct {R, d_R}$.
Let $\phi: R \to Q$ be the mapping from $R$ to the quotient ring $Q$ defined by:
:$\forall a \in R: \map \phi a = \tuple {a, a, a, \ldots} + \NN$
where $\sequence {a, a, a, \ldots} + \NN$ is the left coset in $Q$ that contains the constant sequence $\sequence {a, a, a, \ldots} $.
From Quotient of Cauchy Sequences is Metric Completion, $\map \phi R$ is a dense subset of $\struct {Q, d_Q }$.
From Embedding Division Ring into Quotient Ring of Cauchy Sequences, $\phi$ is a distance-preserving ring monomorphism.
By the definition of a normed division ring completion, $\struct {Q, \norm {\, \cdot \,}_Q }$ is a normed division ring completion of $\struct {R, \norm {\, \cdot \,}_R }$.
== Sources ==
* : $\S 3.2$: Completions
Category:Normed Division Rings
Category:Complete Metric Spaces
Category:Completion of Normed Division Ring
Completion of Valued Field

== Theorem ==
Let $\struct {k, \norm {\,\cdot\,} }$ be a valued field.
Then there exists a completion $\struct {k', \norm {\,\cdot\,}'}$ of $\struct {k, \norm {\,\cdot\,} }$ such that $\struct {k', \norm {\,\cdot\,}'}$ is a valued field.
Furthermore, every completion of $\struct{k, \norm {\,\cdot\,} }$ is isometrically isomorphic to $\struct {k', \norm {\,\cdot\,}'}$.
== Proof ==
By Completion of Normed Division Ring then $\struct {k, \norm {\, \cdot \,} }$ has a normed division ring completion $\struct {k', \norm {\, \cdot \,}'}$
By Normed Division Ring is Field iff Completion is Field then $\struct {k', \norm {\, \cdot \,}'}$ is a field.
By Normed Division Ring Completions are Isometric and Isomorphic then every completion of $\struct {k, \norm {\,\cdot\,} }$ is isometrically isomorphic to $\struct {k', \norm {\,\cdot\,}'}$.
== Examples ==
* The completion of $\Q$ with respect to the usual absolute value is $\R$
* The completion of $\Q$ with respect to the $p$-adic norm is known as the field of $p$-adic numbers and denoted $\Q_p$
* By Ostrowski's Theorem there are no other completions of $\Q$ as a valued field.
Category:Valued Fields
Complex-Differentiable Function is Continuous

== Theorem ==
Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set.
Let $f$ be complex-differentiable at $a \in D$.
Then $f$ is continuous at $a$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 7.1$
Category:Complex-Differentiable Function is Continuous
Category:Complex Differential Calculus
Category:Continuous Mappings
Complex-Differentiable Function is Continuous/Proof 1

== Theorem ==
== Proof ==
Let $\map {N_r} 0$ denote the $r$-neighborhood of $0$ in $\C$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {N_r} 0 \setminus \set 0$:
:$(1): \quad \map f {a + h} = \map f a + h \paren {\map {f'} a + \map \epsilon h}$
where $\epsilon: \map {N_r} 0 \setminus \set 0 \to \C$ is a complex function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
We rewrite the of $(1)$ to get:
By definition of continuous complex function, it follows that $f$ is continuous at $a$.
Category:Complex-Differentiable Function is Continuous