avewright/proofwiki-math huggingface dataset:
Theorem: Kernel of Bounded Linear Transformation is Annihilator of Image of Dual Operator

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$. 
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively. 
Let $T : X \to Y$ be a Bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Then we have: 
:$\map \ker T = {}^\bot T^\ast \sqbrk {Y^\ast}$
where ${}^\bot$ denotes the annihilator operation on subspaces of $X^\ast$.

Proof:
Let $x \in \map \ker T$. 
Let $\phi \in T^\ast \sqbrk {Y^\ast}$.
Then from the definition of the dual operator, there exists $f \in Y^\ast$ such that $f \circ T = \phi$.
Then we have: 
{{begin-eqn}}
{{eqn	| l = \map \phi x
		| r = \map f {T x}
}}
{{eqn	| r = 0
		| c = since $T x = {\mathbf 0}_X$
}}
{{end-eqn}}
Since this holds for all $\phi \in T^\ast \sqbrk {Y^\ast}$, we obtain $x \in {}^\bot T^\ast \sqbrk {Y^\ast}$.
So we have:
:$\map \ker T \subseteq {}^\bot T^\ast \sqbrk {Y^\ast}$
Now let $x \in {}^\bot T^\ast \sqbrk {Y^\ast}$.
Then for all $\phi \in T^\ast \sqbrk {Y^\ast}$ we have $\map \phi x = 0$. 
That is, for all $f \in Y^\ast$, we have: 
:$\map {\paren {T^\ast f} } x = \map f {T x} = 0$
from the definition of the dual operator.
From Normed Dual Space Separates Points, we obtain $T x = {\mathbf 0}_Y$ and hence $x \in \map \ker T$. 
So we obtain:
:${}^\bot T^\ast \sqbrk {Y^\ast} \subseteq \map \ker T$
Hence we can conclude:
:$\map \ker T = {}^\bot T^\ast \sqbrk {Y^\ast}$
{{qed}}
Theorem: Annihilator of Subspace of Banach Space is Weak-* Closed

Let $X$ be a Banach space.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $M^\bot$ be the annihilator of $M$. 
Then $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$.

Proof:
From Set is Closed iff Equals Topological Closure, we aim to show:
:$M^\bot = \map {\cl_{w^\ast} } {M^\bot}$
From Set is Subset of its Topological Closure, we have: 
:$M^\bot \subseteq \map {\cl_{w^\ast} } {M^\bot}$
Now let: 
:$f \in \map {\cl_{w^\ast} } {M^\bot}$
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {f_\lambda}_{\lambda \in \Lambda}$ in $M^\bot$ converging to $f$ in $\struct {X^\ast, w^\ast}$.
That is, from Characterization of Convergent Net in Weak-* Topology: 
:for each $x \in X$ the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ in $\GF$.
Since $f_\lambda \in M^\bot$ for each $\lambda \in \Lambda$, we have: 
:$\map {f_\lambda} x = 0$ for each $x \in M$ and $\lambda \in \Lambda$.
From Constant Net is Convergent, we obtain:
:the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ and $0$ in $\GF$. 
From Metric Space is Hausdorff, $\GF$ is Hausdorff. 
Hence from Characterization of Hausdorff Property in terms of Nets, we obtain $\map f x = 0$ for each $x \in M$.
So:
:$f \in M^\bot$
So we obtain: 
:$\map {\cl_{w^\ast} } {M^\bot} \subseteq M^\bot$
and hence:
:$M^\bot = \map {\cl_{w^\ast} } {M^\bot}$
From Set is Closed iff Equals Topological Closure, we have that $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$.
{{qed}}
Theorem: Annihilator of Subspace of Banach Space is Weak-* Closed

Let $X$ be a Banach space.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $M^\bot$ be the annihilator of $M$. 
Then $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$.

Proof:
From Annihilator of Subspace of Banach Space as Intersection of Kernels, we have:
:$\ds M^\bot = \bigcap_{x \in M} \map \ker {x^\wedge}$ 
From Characterization of Continuity of Linear Functional in Weak-* Topology:
:the linear functional $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Characterization of Continuous Linear Functionals on Topological Vector Space, $\map \ker {x^\wedge}$ is closed in $\struct {X^\ast, w^\ast}$.
So $M^\bot$ is the intersection of closed sets in $\struct {X^\ast, w^\ast}$, and hence is closed itself.
{{qed}}
Theorem: Constant Net is Convergent

Let $\struct {X, \tau}$ be a topological space.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $x \in X$. 
Define a net $\family {x_\lambda}_{\lambda \in \Lambda}$ by: 
:$x_\lambda = x$ for each $\lambda \in \Lambda$. 
Then $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.

Proof:
Let $U$ be an open neighborhood of $x$.
Let $\lambda_0 \in \Lambda$. 
Then we have $x_\lambda \in U$ for all $\lambda \in \Lambda$, and in particular all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$. 
So $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.
{{qed}}
Category:Nets (Set Theory)
f6k99d3i4zqcjqxyo76n56adyifhsgm
Theorem: Annihilator of Subspace of Banach Space as Intersection of Kernels

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\GF$.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $M^\bot$ be the annihilator of $M$. 
Then: 
:$\ds M^\bot = \bigcap_{x \mathop \in M} \map \ker {x^\wedge}$
where $x^\wedge$ denotes the evaluation linear transformation evaluated at $x$.

Proof:
We have: 
{{begin-eqn}}
{{eqn | l = M^\bot
      | r = \set {g \in X^\ast : \map g x = 0 \text { for all } x \in M}
      | c = {{Defof|Annihilator of Subspace of Banach Space}}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \set {g \in X^\ast : \map g x = 0}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \set {g \in X^\ast : \map {x^\wedge} g = 0}
      | c = {{Defof|Evaluation Linear Transformation}}
}}
{{eqn | r = \bigcap_{x \mathop \in M} \map \ker {x^\wedge}
      | c = {{Defof|Kernel of Linear Transformation}}
}}
{{end-eqn}}
{{qed}}
Category:Annihilators of Subspaces of Banach Spaces
cj9tcz7b0pekurnlqkr4bu07oag8gxo
Theorem: Closure in Weak-* Topology in terms of Annihilators

Let $X$ be a Banach space. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $N$ be a linear subspace of $X^\ast$. 
Then: 
:$\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$
where:
:$\cl_{w^\ast}$ denotes closure in the weak-$\ast$ topology
:${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$
:$\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$.

Proof:
From Linear Subspace is Subset of Double Annihilator, we have:
:$N \subseteq \paren { {}^\bot N}^\bot$
So, from Set Closure Preserves Set Inclusion:
:$\map {\cl_{w^\ast} } N \subseteq \map {\cl_{w^\ast} } {\paren { {}^\bot N}^\bot}$
From Annihilator of Subspace of Banach Space is Weak-* Closed and Set is Closed iff Equals Topological Closure, we have:
:$\map {\cl_{w^\ast} } N \subseteq \paren { {}^\bot N}^\bot$
It remains to show:
:$\paren { {}^\bot N}^\bot \subseteq \map {\cl_{w^\ast} } N$
From Set Complement inverts Subsets, we can equivalently show:
:$X^\ast \setminus \map {\cl_{w^\ast} } N \subseteq X^\ast \setminus \paren { {}^\bot N}^\bot$
Let:
:$f \in X^\ast \setminus \map {\cl_{w^\ast} } N$
Applying Existence of Non-Zero Continuous Linear Functional vanishing on Proper Closed Subspace of Locally Convex Space to the locally convex space $\struct {X^\ast, w^\ast}$, there exists $\Phi \in \struct {X^\ast, w^\ast}^\ast$ such that $\Phi \ne {\mathbf 0}_{X^\ast}$ and:
:$\map \Phi g = 0$ for each $g \in \map {\cl_{w^\ast} } N$
and:
:$\map \Phi f \ne 0$
From Characterization of Continuity of Linear Functional in Weak-* Topology, there exists $x \in X$ such that $\Phi = x^\wedge$. 
Then: 
:$\map {x^\wedge} g = \map g x = 0$ for all $g \in N$
So $x \in {}^\bot N$.
But we also have $0 \ne \map {x^\wedge} f = \map f x$, so $f$ does not vanish at some point in ${}^\bot N$.
So $f \not \in \paren { {}^\bot N}^\bot$ by the definition of the annihilator of a subspace of $X$. 
We therefore have:
:$X^\ast \setminus \map {\cl_{w^\ast} } N \subseteq X^\ast \setminus \paren { {}^\bot N}^\bot$
hence:
:$\paren { {}^\bot N}^\bot \subseteq \map {\cl_{w^\ast} } N$
from Set Complement inverts Subsets.
We conclude:
:$\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$
{{qed}}
Category:Annihilators of Subspaces of Banach Spaces
Category:Annihilators of Subspaces of Normed Dual Spaces
Category:Weak-* Topologies
nfa5kxdsywcks89h09qo7f6oz36xtxs
Theorem: Linear Subspace is Subset of Double Annihilator

Let $X$ be a Banach space. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $N$ be a linear subspace of $X^\ast$. 
Then:
:$N \subseteq \paren { {}^\bot N}^\bot$
where:
:${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$
:$\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$.

Proof:
From Set Complement inverts Subsets, we can equivalently show:
:$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$
Let $f \in X^\ast \setminus \paren { {}^\bot N}^\bot$, then:
:$\map f x \ne 0$ for some $x \in {}^\bot N$.
By the definition of the annihilator of a subspace of $X$, we have: 
:$\map g x = 0$ for all $g \in N$. 
In particular, $g \in X^\ast \setminus N$.
So we obtain:
:$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$
and hence:
:$N \subseteq \paren { {}^\bot N}^\bot$
from Set Complement inverts Subsets.
{{qed}}
Category:Annihilators of Subspaces of Banach Spaces
Category:Annihilators of Subspaces of Normed Dual Spaces
1pwsvzhl98fkrb082tunxlnf2m526xl
Theorem: Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Let $T \sqbrk X^\bot$ be the annihilator of $T \sqbrk X$.
Then:
:$T \sqbrk X^\bot = \map \ker {T^\ast}$

Proof:
We have: 
:$f \in \map \ker {T^\ast}$
{{iff}}:
:$\map {\paren {T^\ast f} } x = \map f {T x} = 0$ for each $x \in X$. 
That is, {{iff}}:
:$\map f y = 0$ for all $y \in T \sqbrk X$
To conclude, we have $f \in \map \ker {T^\ast}$ {{iff}}:
:$f \in T \sqbrk X^\bot$
{{qed}}
Theorem: Existence of Complementary Subspace

Let $X$ be a vector space.
Let $N \subseteq X$ be a subspace.
Then $N$ has a complementary subspace.
That is, there exists a subspace $Y \subseteq X$ such that:
:$X = N \oplus Y$

Proof:
Let $S$ be the set of all subspaces $V \subseteq X$ such that:
:$N \cap V = \set 0$
By Zorn's Lemma, $\struct {S, \subseteq}$ has a maximal element $Y$.
We claim:
:$X = N \oplus Y$
Indeed, if there would be an $x \in X \setminus \paren {N \oplus Y}$, then:
:$Y \subsetneq \map \span {Y \cup \set x} \in S$
where $\span$ denotes the linear span.
This contradicts the maximality of $Y$.
{{finish|Details}}
Theorem: Cauchy-Riemann Equations/Polar Form

The Cauchy-Riemann equations can be expressed in polar form as:
{{begin-eqn}}
{{eqn | n = 1
      | l = \dfrac {\partial u} {\partial r}
      | r = \dfrac 1 r \dfrac {\partial v} {\partial \theta}
}}
{{eqn | n = 2
      | l = \dfrac 1 r \dfrac {\partial u} {\partial \theta}
      | r = -\dfrac {\partial v} {\partial r}
}}
{{end-eqn}}
where $z$ is expressed in exponential form as:
:$z = r e^{i \theta}$

Proof:
{{ProofWanted}}
{{Namedfor|Augustin Louis Cauchy|name2 = Georg Friedrich Bernhard Riemann}}
Theorem: Degenerate Linear Operator Plus Identity is Fredholm Operator

Let $U$ be a vector space.
Let $T : U \to U$ be a degenerate linear operator.
Let $I_U : U \to U$ be the identity operator.
Then:
:$T + I_U$
is a Fredholm operator.

Proof:
We need to show that both:
:$\map \ker {T + I_U}$
and:
:$U / {\Img {T + I_U} }$
are finite-dimensional.
Theorem: Recursively Enumerable Set is Image of Primitive Recursive Function

Let $S \subseteq \N$ be recursively enumerable.
Suppose $S$ is non-empty.
Then there exists a primitive recursive function $f : \N^k \to \N$ such that:
:$\Img f = S$

Proof:
{{Proofread}}
{{tidy|minor stuff, I'll be on the case in due course}}
By definition of recursively enumerable, there exists a recursive function $g : \N^\ell \to \N$ such that:
:$\Img g = S$
By definition of non-empty, there exists some $x \in S$.
By Kleene's Normal Form Theorem, there exist:
* A primitive recursive relation $T \subset \N^{\ell + 2}$
* A primitive recursive function $U : \N \to \N$
such that a partial function $h : \N^\ell \to \N$ is recursive {{iff}} there exists $e \in \N$ such that:
:$\map h {x_1, \dotsc, x_\ell} \approx \map U {\mu z \map T {e, x_1, \dotsc, x_\ell, z}}$
for every $\tuple {x_1, \dotsc, x_\ell} \in \N^\ell$.
Thus, as $g$ is recursive, there exists such an $e$.
Define $V : \N^{\ell + 1} \to \N$ as:
:$\map V {x_1, \dotsc, x_\ell, t} = \map {\mu z \le t} {\map T {e, x_1, \dotsc, x_\ell, z}}$
$V$ is primitive recursive by Bounded Minimization is Primitive Recursive.
Define $f : \N^{\ell + 1} \to \N$ as:
:$\map f {x_1, \dotsc, x_\ell, t} = \begin{cases}
\map U {\map V {x_1, \dotsc, x_\ell, t}} & : \map V {x_1, \dotsc, x_\ell, t} \le t \\
x & : \map V {x_1, \dotsc, x_\ell, t} > t
\end{cases}$
By Definition by Cases is Primitive Recursive and Ordering Relations are Primitive Recursive, $f$ is primitive recursive.
It remains to show that $\Img f = \Img g$.
Suppose $y \in \Img g$.
Then, for some $\tuple {x_1, \dotsc, x_\ell} \in \N^\ell$:
:$\map g {x_1, \dotsc, x_\ell} = y$
Thus, by definition of $T$ and $U$:
:$\map U {\map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}} = y$
Let $t = \map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}$.
Hence:
:$\map V {x_1, \dotsc, x_\ell, t} = \map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}} = t$.
Therefore, as $t \le t$:
:$\map f {x_1, \dotsc, x_\ell, t} = \map U t = y$.
Thus, $y \in \Img f$.
{{qed|lemma}}
Now, suppose $y \in \Img f$.
Then there exists $\tuple {x_1, \dotsc, x_\ell, t} \in \N^{\ell + 1}$ such that:
:$\map f {x_1, \dotsc, x_\ell, t} = y$
By the definition of $f$, either:
:$\map U {\map V {x_1, \dotsc, x_\ell, t}} = y$ and $\map V {x_1, \dotsc, x_\ell, t} \le t$, or
:$x = y$ and $\map V {x_1, \dotsc, x_\ell, t} > t$.
In the second case, $x \in S = \Img g$ by definition.
In the first, let $r = \map V {x_1, \dotsc, x_\ell, t}$.
As $r \le t$, by definition of $V$ and bounded minimization:
:$r = \map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}$
But then, $y = \map U r = \map U {\map {\mu z} {\map T {e, x_1, \dotsc, x_\ell, z}}} = \map g {x_1, \dotsc, x_\ell}$.
Therefore, $y \in \Img g$.
{{qed}}
Category:Recursion Theory
ea5waelny7cvnakitfn5v2v7vjzg4xf
Theorem: Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary

Let $S \subseteq \N$ be recursively enumerable.
Suppose $S$ is non-empty.
Then there exists a primitive recursive function $f : \N \to \N$ such that:
:$\Img f = S$

Proof:
By Recursively Enumerable Set is Image of Primitive Recursive Function, there is some primitive recursive $g : \N^k \to \N$ such that:
:$\Img g = S$
As $S$ is non-empty, let $n \in S$.
Define:
:$\map f x = \begin{cases}
\map g {\map {\operatorname{pred}} {\paren{x}_1}, \dotsc, \map {\operatorname{pred}} {\paren{x}_k}} & : x \in \operatorname{Seq} \land \map {\operatorname{len}} x = k \\
n & : \text{otherwise}
\end{cases}$
That $f$ is primitive recursive follows from:
* Definition by Cases is Primitive Recursive/Corollary
* Predecessor Function is Primitive Recursive
* Set of Sequence Codes is Primitive Recursive
* Length Function is Primitive Recursive
* Equality Relation is Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Prime Exponent Function is Primitive Recursive
Suppose $y \in \Img g$.
Then, there exist $\tuple {x_1, \dotsc, x_k} \in \N^k$ such that:
:$\map g {x_1, \dotsc, x_k} = y$
Let $X$ be the sequence coding for $\sequence {x_1 + 1, \dotsc, x_k + 1}$.
Then, $X \in \operatorname{Seq}$ and $\map {\operatorname{len}} X = k$.
Therefore, $\map f X = y$.
{{qed|lemma}}
Now, suppose $y \in \Img f$.
Then, there exists $x \in \N$ such that:
:$\map f x = y$
If $x \in \operatorname{Seq} \land \map {\operatorname{len}} x = k$ holds, then:
:$y = \map g {\paren{x}_1 - 1, \dotsc, \paren{x}_k - 1}$
and thus $y \in \Img g$.
Otherwise, $y = n \in S$.
But as $S = \Img g$, it follows that $y \in \Img g$.
{{qed}}
Category:Recursion Theory
its7swb32hgqrgztpg85yopfwjzcty6
Theorem: Set is Recursively Enumerable iff Domain of Recursive Function

Let $S \subseteq \N$ be a set of natural numbers.
Then:
:$S$ is recursively enumerable
{{iff}}:
:There exists a recursive function $f : \N \to \N$ such that $S = \Dom f$
where $\Dom f$ is the domain of $f$.

Proof:
=== Necessary Condition ===
Suppose $S$ is the empty set.
Define $\map f x = \map {\mu z} {z \ne z}$.
Then, $f$ is undefined for every $x \in \N$, and thus is defined on the empty set.
Additionally, $f$ is recursive by:
* Equality Relation is Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Primitive Recursive Function is Total Recursive Function
Otherwise, let $S$ be non-empty.
Then, by Recursively Enumerable Set is Image of Primitive Recursive Function/Corollary, there exists a primitive recursive function $g : \N \to \N$ such that:
:$S = \Img g$
Define $\map f x = \map {\mu z} {\map g z = x}$.
It remains to be shown that $\Dom f = \Img f$.
Let $y \in \Dom f$.
Then, there exists $z \in \N$ such that $\map g z = y$.
But then, $y \in \Img g$.
Now, let $y \in \Img g$.
Then, there exists $z \in \N$ such that $\map g z = y$.
Let $z'$ be the smallest such $z$.
Then, $\map f y = z'$ be definition.
Thus, $y \in \Dom f$.
{{qed|lemma}}
Theorem: Set is Recursive iff Set and Complement are Recursively Enumerable

Let $S \subseteq \N$ be a set of natural numbers.
Then:
:$S$ is a recursive set
{{iff}}:
:$S$ and $\N \setminus S$ are recursively enumerable sets.

Proof:
=== Necessary Condition ===
{{ExtractTheorem|Complement of Recursive Set. More generally, we should show that constructions with primitive recursive ''things'' work on recursive ones as well, such as Definition by Cases is Primitive Recursive, etc.}}
By Complement of Primitive Recursive Set, $\N \setminus S$ is a recursive set as well.
Thus, it suffices to show that every recursive $S$ is recursively enumerable.
By definition of recursive set, the characteristic function $\chi_S$ is recursive.
Define:
:$\map f x = \map {\mu z} {\map {\overline \sgn} {\map {\chi_S} x} }$
As $f$ is obtained by substitution and minimization, and:
* Signum Complement Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
it follows that $f$ is a recursive function.
We now show that $\Dom f = S$.
Suppose $x \in S$.
Then, by definition of characteristic function:
:$\map {\chi_S} x = 1$
Thus, by definition of the signum-bar function:
:$\map {\overline \sgn} {\map {\chi_S} x} = 0$
Therefore, by the definition of minimization:
:$\map {\mu z} {\map {\overline \sgn} {\map {\chi_S} x}} = 0$
as the previous equation is true for every value of $z$, and $0$ is the smallest natural number.
Hence, $f$ is defined at $x$, so $x \in \Dom f$.
Now, suppose $x \notin S$.
By definition of characteristic function:
:$\map {\chi_S} x = 0$
By definition of the signum complement function:
:$\map {\overline \sgn} {\map {\chi_S} x} = 1$
Therefore, by the definition of minimization:
:$\map {\mu z} {\map {\overline \sgn} {\map {\chi_S} x} }$ is undefined
as there is no $z$ that causes the function to be equal to $0$.
Hence, $x \notin \Dom f$.
As we have shown $x \in S \iff x \in \Dom f$, it follows that $S = \Dom f$.
By Set is Recursively Enumerable iff Domain of Recursive Function, $S$ is then recursively enumerable.
{{qed|lemma}}
Theorem: Union of Recursively Enumerable Sets

Let $S, T \subseteq \N$ be recursively enumerable sets.
Then $S \cup T$ is recursively enumerable.

Proof:
Suppose $S = \O$.
Then by Union with Empty Set:
:$S \cup T = T$
which is recursively enumerable by assumption.
{{qed|lemma}}
Suppose $T = \O$.
For the same reason:
:$S \cup T = S$
which is also recursively enumerable by assumption.
{{qed|lemma}}
Now, suppose neither $S$ nor $T$ is empty.
By {{Corollary|Recursively Enumerable Set is Image of Primitive Recursive Function}}, there exist primitive recursive $f, g : \N \to \N$ such that:
:$S = \Img f$
:$T = \Img g$
Define:
:<nowiki>$\map h {x, c} = \begin{cases}
\map f x & : c = 0 \\
\map g x & : \text{otherwise}
\end{cases}$</nowiki>
By:
* Equality Relation is Primitive Recursive
* {{Corollary|Definition by Cases is Primitive Recursive}}
$h$ is a primitive recursive function.
By Primitive Recursive Function is Total Recursive Function, it is thus a recursive function.
It remains to be shown that $S \cup T = \Img h$.
Let $x \in S \cup T$.
Then $x \in \Img f$ or $x \in \Img g$.
If $x \in \Img f$, then there is some $z \in \N$ such that $\map f z = x$.
Therefore, $\map h {z, 0} = x$.
Likewise, if $x \in \Img g$, there is some $z \in \N$ such that $\map g z = x$.
Therefore, $\map h {z, 1} = x$.
Thus, $S \cup T \subseteq \Img h$.
Now, $x \in \Img h$.
Then, there are $z, c \in \N$ such that:
:$\map h {z, c} = x$
Depending on $c$, either:
:$x = \map f z$
or:
:$x = \map g z$
In either case, $x \in \Img f \cup \Img g$.
Therefore, $\Img h \subseteq S \cup T$.
Hence, by definition of set equality:
:$S \cup T = \Img h$
Therefore, by definition, $S \cup T$ is recursively enumerable.
{{qed}}
Category:Recursion Theory
679l26uwj06rq0me55gwiurd8aptkng
Theorem: Intersection of Recursively Enumerable Sets

Let $S, T \subseteq \N$ be recursively enumerable.
Then $S \cap T$ is recursively enumerable.

Proof:
By Set is Recursively Enumerable iff Domain of Recursive Function, there exist recursive $f, g : \N \to \N$ such that:
:$S = \Dom f$
:$T = \Dom g$
Define:
:$\map h x = \map f x + \map g x$
By:
* Addition is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
it follows that $h$ is recursive.
Let $x \in S \cap T$.
Then $x \in \Dom f$ and $x \in \Dom g$.
Thus, there exist $y, z \in \N$ such that:
:$y = \map f x$
:$z = \map g x$
Therefore:
:$\map h x = y + z$
Hence, $x \in \Dom h$.
Then, by definition of subset, $S \cap T \subseteq \Dom h$.
Let $x \in \Dom h$.
By definition of substitution, there must exist $y, z \in \N$ such that:
:$y = \map f x$
:$z = \map g x$
Therefore, $x \in \Dom f$ and $x \in \Dom g$.
It follows that $x \in S \cap T$.
Therefore, by definition of subset, $\Dom h \subseteq S \cap T$.
By definition of set equality, $S \cap T = \Dom h$.
Thus, by Set is Recursively Enumerable iff Domain of Recursive Function, $S \cap T$ is recursively enumerable.
{{qed}}
Category:Recursion Theory
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Theorem: Generalized Sum with Finite Non-zero Summands

Let $G$ be a commutative topological semigroup with identity $0_G$.
Let $\family{g }_{i \in I}$ be an indexed family of elements of $G$.
Let $\set{i \in I : g_i \ne 0_G}$ be finite.
Let $\set{i_1, i_2, \cdots, i_n}$ be a finite enumeration of $\set{i \in I : g_i \ne 0_G}$.
Then the generalized sum $\ds \sum_{i \mathop \in I} g_i$ converges to the summation $\ds \sum_{k \mathop = 1}^n g_{i_k}$

Proof:
Let $J = \set{i \in I : g_i \ne 0_G}$.
From Finite Generalized Sum Converges to Summation:
:the generalized sum $\ds \sum_{j \mathop \in J} g_j$ converges to the summation  $\ds \sum_{k \mathop = 1}^n g_{i_k}$
From Generalized Sum Restricted to Non-zero Summands:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ converges to the summation  $\ds \sum_{k \mathop = 1}^n g_{i_k}$
{{qed}}
Category:Generalized Sums
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