avewright/proofwiki-math huggingface dataset:
Theorem: Dual of Dual Category is Isomorphic to Category

Let $\mathbf C$ be a metacategory.
Let $\mathbf C^{\text{op} }$ be the dual category of $\mathbf C$.
Let $\paren{\mathbf C^{\text{op} } }^{\text{op} }$ be the dual category of $\mathbf C^{\text{op} }$.
Then:
:$C$ is isomorphic to $\paren{\mathbf C^{\text{op} } }^{\text{op} }$
with isomorphisms:
:$F: C \to \paren{\mathbf C^{\text{op} } }^{\text{op} }$ defined by:
{{DefineFunctor
|ob =  for each object $C$ in $\mathbf C$: $FC = \paren{C^{\text{op} } }^{\text{op} }$
|mor = for each morphism $f$ in $\mathbf C$: :$Ff = \paren{f^{\text{op} } }^{\text{op} }$
}}
and:
:$G: \paren{\mathbf C^{\text{op} } }^{\text{op} } \to C$ be defined by:
{{DefineFunctor
|ob = for each object $\paren{C^{\text{op} } }^{\text{op} }$ in $\paren{\mathbf C^{\text{op} } }^{\text{op} }$: $\map G {\paren{C^{\text{op} } }^{\text{op} } }= C$
|mor = for each morphism $\paren{f^{\text{op} } }^{\text{op} }$ in $\paren{\mathbf C^{\text{op} } }^{\text{op} }$: $\map G {\paren{f^{\text{op} } }^{\text{op} } }= f$
}}

Proof:
=== $F$ is a Functor ===
Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.
Then:
{{begin-eqn}}
{{eqn | l = \map F {g \circ f}
      | r = \paren{\paren{g \circ f}^{\text{op} } }^{\text{op} }
      | c = Definition of $F$
}}
{{eqn | r = \paren{f^{\text{op} } \circ g^{\text{op} } }^{\text{op} }
      | c = {{Defof|Composition of Morphisms}} in dual category
}}
{{eqn | r = \paren{g^{\text{op} } }^{\text{op} } \circ \paren{f^{\text{op} } }^{\text{op} }
      | c = {{Defof|Composition of Morphisms}} in dual category
}}
{{eqn | r = Fg \circ Ff
      | c = Definition of $F$
}}
{{end-eqn}}
Also, for any object $C$ of $\mathbf C$:
{{begin-eqn}}
{{eqn | l = F \operatorname{id}_C
      | r = \paren{\operatorname{id}_C^{\text{op} } }^{\text{op} }
      | c = Definition of $F$
}}
{{eqn | r = \paren{\operatorname{id}_{C^{\text{op} } } }^{\text{op} }
      | c = {{Defof|Identity Morphism}} in dual category
}}
{{eqn | r = \operatorname{id}_{\paren{C^{\text{op} } }^{\text{op} } }
      | c = {{Defof|Identity Morphism}} in dual category
}}
{{eqn | r = \operatorname{id}_{FC}
      | c = Definition of $F$
}}
{{end-eqn}}
Hence $F$ is shown to be a functor.
{{qed|lemma}}
Theorem: Normalized Block Basic Sequence of Schauder Basis in Space of Zero-Limit Sequences is Isometrically Equivalent to Schauder Basis

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Let $\sequence {e_n}_{n \mathop \in \N}$ be the Schauder basis given by Space of Zero-Limit Sequences admits Schauder Basis.
Let $\sequence {y_k}_{k \mathop \in \N}$ be a block basic sequence of $\sequence {e_n}_{n \mathop \in \N}$ with:
:$\norm {y_k}_\infty = 1$ for each $k \in \N$. 
Then $\sequence {y_k}_{k \mathop \in \N}$ is isometrically equivalent to $\sequence {e_n}_{n \mathop \in \N}$.

Proof:
Let $\sequence {r_j}_{j \mathop \in \N}$ be a strictly increasing sequence in $\N$ and $\sequence {a_j}_{j \mathop \in \N}$ be a sequence in $\GF$ such that:
:$\ds y_k = \sum_{j \mathop = r_{k - 1} + 1}^{r_k} a_j e_j$
where $r_0 = 0$.
By assumption, we have:
:$\ds 1 = \norm {y_k}_\infty = \sup_{r_{k - 1} + 1 \le j \le r_k} \cmod {a_j}$
Let $\struct {\map {c_{00} } \GF, \norm {\, \cdot \,}_\infty}$ be the space of almost-zero sequences.
By Characterization of Isometric Equivalence of Basic Sequences, it is enough to show that:
:$\ds \norm {\sum_{k \mathop = 1}^\infty \map \xi k y_k}_\infty = \norm {\sum_{k \mathop = 1}^\infty \map \xi k e_k}_\infty = \sup_{k \mathop \in \N} \cmod {\map \xi k}$ for each $\xi \in \map {c_{00} } \GF$.
We have:
:$\ds \norm {\sum_{k \mathop = 1}^\infty \map \xi k y_k}_\infty = \norm {\sum_{k \mathop = 1}^\infty \sum_{j \mathop = r_{k - 1} + 1}^{r_k} \map \xi k a_j e_j}_\infty$
Note that the coefficient of $e_j$ is $\map \xi k a_j$ for $r_{k - 1} + 1 \le j \le r_k$. 
Hence we have:
{{begin-eqn}}
{{eqn	| l = \norm {\sum_{k \mathop = 1}^\infty \sum_{j \mathop = r_{k - 1} + 1}^{r_k} \map \xi k a_j e_j}_\infty 
		| r = \sup_{k \mathop \in \N} \paren {\sup_{r_{k - 1} + 1 \le j \le r_k} \cmod {\map \xi k a_j} }
}}
{{eqn	| r = \sup_{k \mathop \in \N} \paren {\paren {\sup_{r_{k - 1} + 1 \le j \le r_k} \cmod {a_j} } \cmod {\map \xi k} }
}}
{{eqn	| r = \sup_{k \mathop \in \N} \cmod {\map \xi k}
}}
{{end-eqn}}
for each $\xi \in \map {c_{00} } \GF$.
Hence:
:$\ds \norm {\sum_{k \mathop = 1}^\infty \map \xi k y_k}_\infty = \norm {\sum_{k \mathop = 1}^\infty \map \xi k e_k}_\infty = \sup_{k \mathop \in \N} \cmod {\map \xi k}$ for each $\xi \in \map {c_{00} } \GF$.
Hence by Characterization of Isometric Equivalence of Basic Sequences, we have that $\sequence {y_k}_{k \mathop \in \N}$ is isometrically equivalent to $\sequence {e_n}_{n \mathop \in \N}$.
{{qed}}
Theorem: Space of Almost-Zero Sequences is Everywhere Dense in P-Sequence Space

Let $\GF \in \set {\R, \C}$.
Let $p \in \hointr 1 \infty$. 
Let $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$ be the $p$-sequence space.
Let $\map {c_{00} } \GF$ be the space of almost-zero sequences.
Then $\map {c_{00} } \GF$ is everywhere dense in $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$.

Proof:
Define $\sequence {e_n}_{n \mathop \in \N}$ by:
:$\ds \map {e_n} k = \begin{cases}1 & n = k \\ 0 & \text{otherwise}\end{cases} \in \map {c_{00} } \GF$
From P-Sequence Space admits Schauder Basis, $\sequence {e_n}_{n \mathop \in \N}$ is a Schauder basis for $\map {\ell_p} \GF$.
Let $\sequence {e_n^\ast}_{n \mathop \in \N}$ be the coordinate functionals associated with $\sequence {e_n}_{n \mathop \in \N}$.
From the definition of the coordinate functionals, we have:
:$\ds x = \sum_{j \mathop = 1}^\infty \map {e_j^\ast} x e_j = \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \map {e_j^\ast} x e_j$
From Space of Almost-Zero Sequences forms Vector Space, we have that:
:$\ds \sum_{j \mathop = 1}^n \map {e_j^\ast} x e_j \in \map {c_{00} } \GF$
Hence:
:$\map {\ell_p} \GF \subseteq \map \cl {\map {c_{00} } \GF}$
where $\cl$ is closure in $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$.
We therefore have:
:$\map {\ell_p} \GF = \map \cl {\map {c_{00} } \GF}$
So $\map {c_{00} } \GF$ is everywhere dense in $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$.
{{qed}}
Category:Space of Almost-Zero Sequences
Category:P-Sequence Spaces
5orpzsmeqxg83bgyck7ouqzlni7axmh
Theorem: Conditions for Constant Speed

Let $B$ be a body moving at constant speed.
Then the velocity of $B$ is not necessarily constant.

Proof:
Let $\mathbf v$ be the velocity of body $B$.
The speed of $B$ is given by the magnitude of the velocity:
:$v = \size {\mathbf v}$
We have that the speed is constant:
:$\ds \frac {\d v} {\d t} = 0$
As the speed is constant, then so is the square of the speed:
{{begin-eqn}}
{{eqn | l = \frac {\d v^2} {\d t}
      | r = 2 v \frac {\d v} {\d t}
      | c = Chain Rule for Derivatives and Power Rule for Derivatives
}}
{{eqn | r = 0
}}
{{end-eqn}}
We can expand the square of the speed in terms of the dot product of its velocity:
{{begin-eqn}}
{{eqn | l = 0
      | r = \frac {\d v^2} {\d t}
}}
{{eqn | r = \frac {\map \d {\mathbf v \cdot \mathbf v} } {\d t}
      | c = Dot Product of Vector with Itself
}}
{{eqn | r = 2 \mathbf v \cdot \frac {\d \mathbf v} {\d t}
      | c = Derivative of Dot Product of Vector-Valued Functions
}}
{{eqn | r = 2 \mathbf v \cdot \mathbf a
      | c = {{Defof|Acceleration}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf v \cdot \mathbf a
      | r = 0
}}
{{end-eqn}}
From this result, we can conclude that a body moving at a constant speed is subject to one of two conditions:
:$(1): \quad \map {\mathbf a} t = \mathbf 0$, from Dot Product with Zero Vector is Zero, in which case the velocity is constant.
:$(2): \quad \mathbf a$ is always perpendicular to $\mathbf v$, from Dot Product of Perpendicular Vectors
This second condition allows for a body to have a non-constant velocity while also having a constant speed.
{{qed}}
Theorem: Basic Sequence Congruent to Complemented Basic Sequence is Complemented

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a complemented basic sequence.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a basic sequence in $Y$ that is congruent to $\sequence {x_n}_{n \mathop \in \N}$.
Then $\sequence {y_n}_{n \mathop \in \N}$ is complemented in $Y$.

Proof:
Since $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ are congruent, there exists a linear isomorphism $T : X \to Y$ with $T x_n = y_n$ for each $n \in \N$.
Let $\sqbrk {x_n}_{n \mathop \in \N}$ and $\sqbrk {y_n}_{n \mathop \in \N}$ be the closed linear spans of $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ respectively.
Since $\sequence {x_n}_{n \mathop \in \N}$ is complemented in $X$, $\sqbrk {x_n}_{n \mathop \in \N}$ is topologically complemented in $X$.
From Linear Isomorphic Image of Complemented Subspace of Banach Space is Complemented, we have that $T \sqbrk {\sqbrk {x_n}_{n \mathop \in \N} }$ is topologically complemented in $Y$.
From Congruent Basic Sequences in Banach Spaces are Equivalent, we have $T \sqbrk {\sqbrk {x_n}_{n \mathop \in \N} } = \sqbrk {y_n}_{n \mathop \in \N}$. 
Hence $\sqbrk {y_n}_{n \mathop \in \N}$ is topologically complemented in $Y$.
Hence $\sequence {y_n}_{n \mathop \in \N}$ is complemented in $Y$.
{{qed}}
Category:Basic Sequences
08ssytlk07g8u828jfczoq7t7daspqa
Theorem: Closed Vector Subspace of Reflexive Banach Space is Reflexive

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a reflexive Banach space over $\GF$.
Let $Y$ be a closed vector subspace of $X$.
Then $Y$ is reflexive.

Proof:
Let $B_X^-$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,}_X}$.
Let $B_Y^-$ be the closed unit ball in $\struct {Y, \norm {\, \cdot \,}_X}$. 
From Kakutani's Theorem, $B_X^-$ is weakly compact.
From Closed Unit Ball in Normed Vector Space is Weakly Closed, $B_Y^-$ is weakly closed in $Y$.
From Closed Vector Subspace in Normed Vector Space is Weakly Closed, $Y$ is weakly closed in $X$.
From Closed Set in Topological Subspace: Corollary, $B_Y^-$ is weakly closed in $X$.
Hence $B_Y^-$ is weakly closed in $B_X^-$ from Closed Set in Topological Subspace: Corollary.
From Closed Subspace of Compact Space is Compact, $B_Y^-$ is weakly compact.
Hence $Y$ is reflexive.
{{qed}}
Category:Reflexive Spaces
tpvz1fw1ejh8qvqtwhx6iv2uljul4os
Theorem: Rational Numbers form Subfield of Constructible Numbers

The rational numbers form a subfield of the constructible numbers.

Proof:
From Constructible Numbers form Field, the constructible numbers form a field.
From Rational Numbers form Field, the rational numbers form a field.
Let $a \in \Q$ be a rational number.
From Construction of Point in Cartesian Plane with Rational Coordinates, $a$ is a constructible number.
Hence the rational numbers form a subset of the constructible numbers.
The result follows by definition of subfield.
{{qed}}
Theorem: Constructible Numbers form Subfield of Algebraic Numbers

The constructible numbers form a subfield of the algebraic numbers.

Proof:
From Constructible Numbers form Field, the constructible numbers form a field.
From Algebraic Numbers form Field, the algebraic numbers form a field.
It remains to be shown that the constructible numbers form a subset of the algebraic numbers.
{{finish}}
Theorem: Weakly Compact Set in P-Sequence Space is Weakly Metrizable

Let $\GF \in \set {\R, \C}$.
Let $p \in \closedint 1 \infty$. 
Let $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$ be the $p$-sequence space.
Let $w$ be the weak topology on $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$.
Let $K \subseteq \map {\ell_p} \GF$ be weakly compact.
Then $\struct {K, w}$ is metrizable.

Proof:
From P-Sequence Space with P-Norm forms Banach Space, $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$ is a Banach space.
Let $\struct {\map {\ell_p^\ast} \GF, \norm {\, \cdot \,}_{\ell_p^\ast} }$ be the normed dual space of $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$.
Define $\pi_j : \map {\ell_p} \GF \to \GF$ by:
:$\map {\pi_j} x = x_j$ for each $j \in \N$. 
Let $F = \set {\pi_j : j \in \N}$.
We show that $F$ separates points. 
Let $x, y \in \map {\ell_p} \GF$ be such that $x \ne y$. 
Then there exists $j \in \N$ such that $x_j \ne y_j$.
Hence $\map {\pi_j} x \ne \map {\pi_j} y$. 
Hence $F$ separates points. 
Let $\tau$ be the initial topology on $\map {\ell_p} \GF$ generated by $F$.
Since $F$ separates points and is countable, we have that $\tau$ is metrizable.
Hence, from Metric Space is Hausdorff, $\tau$ is Hausdorff.
From Coordinate Projection on P-Sequence Space is Bounded Linear Functional, we have that $\pi_j \in \map {\ell_p^\ast} \GF$ for each $j \in \N$.
Hence, by the definition of the weak topology, we have $\tau \subseteq w$.
Hence the mapping $\iota : \struct {K, w} \to \struct {K, \tau}$ defined by:
:$\iota x = x$ for each $x \in K$
is continuous.
Since $\struct {K, w}$ is compact and $\struct {K, \tau}$ is Hausdorff, we have that:
:$\iota$ is a homeomorphism
by Continuous Bijection from Compact to Hausdorff is Homeomorphism.
Hence $\struct {K, w} = \struct {K, \tau}$. 
Hence $\struct {K, w}$ is metrizable.
{{qed}}
Category:Metrizable Spaces
Category:Weakly Compact Sets
Category:P-Sequence Spaces
m8mb62kcjl7xumsfv0mnzr50ze2ymgx
Theorem: Not every Algebraic Number is Constructible

There exist algebraic numbers which are not constructible.

Proof:
Let $a$ be a constructible number.
From Constructible Length with Compass and Straightedge, $a$ is algebraic with degree equal to a power of $2$.
So, for example, $\sqrt [3] 2$ is not constructible, although it is an algebraic number, as a root of the equation $x^3 = 2$.
{{qed}}
Theorem: Equivalence of Basic Sequences in Banach Spaces is Transitive

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$ and $\struct {Z, \norm {\, \cdot \,}_Z}$ be Banach spaces over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a basic sequence in $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a basic sequence in $\struct {Y, \norm {\, \cdot \,}_Y}$ equivalent to $\sequence {x_n}_{n \mathop \in \N}$.
Let $\sequence {z_n}_{n \mathop \in \N}$ be a basic sequence in $\struct {Z, \norm {\, \cdot \,}_Z}$ equivalent to $\sequence {y_n}_{n \mathop \in \N}$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is equivalent to $\sequence {z_n}_{n \mathop \in \N}$.

Proof:
Since $\sequence {x_n}_{n \mathop \in \N}$ is equivalent to $\sequence {y_n}_{n \mathop \in \N}$, we have that:
:for every sequence $\sequence {\alpha_n}_{n \mathop \in \N}$ in $\GF$ we have:
::$\ds \sum_{j \mathop = 1}^\infty \alpha_j x_j$ converges in $\struct {X, \norm {\, \cdot \,}_X}$ {{iff}} $\ds \sum_{j \mathop = 1}^\infty \alpha_j y_j$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$
Since $\sequence {y_n}_{n \mathop \in \N}$ is equivalent to $\sequence {z_n}_{n \mathop \in \N}$, we have that:
:for every sequence $\sequence {\alpha_n}_{n \mathop \in \N}$ in $\GF$ we have:
::$\ds \sum_{j \mathop = 1}^\infty \alpha_j y_j$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$ {{iff}} $\ds \sum_{j \mathop = 1}^\infty \alpha_j z_j$ converges in $\struct {Z, \norm {\, \cdot \,}_Z}$
Hence:
:for every sequence $\sequence {\alpha_n}_{n \mathop \in \N}$ in $\GF$ we have:
::$\ds \sum_{j \mathop = 1}^\infty \alpha_j x_j$ converges in $\struct {X, \norm {\, \cdot \,}_X}$ {{iff}} $\ds \sum_{j \mathop = 1}^\infty \alpha_j z_j$ converges in $\struct {Z, \norm {\, \cdot \,}_Z}$
{{qed}}
Category:Basic Sequences
b22yk6ans1v8rxug13jz79ktet2mi3v
Theorem: Composition of Linear Isomorphisms is Linear Isomorphism

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$ and $\struct {Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ and $S : Y \to Z$ be linear isomorphisms.
Then $S T : X \to Z$ is a linear isomorphism.

Proof:
From Composition of Linear Transformations is Linear Transformation, $S T$ is a linear transformation. 
We have that $T$, $T^{-1}$, $S$ and $S^{-1}$ are bounded.
From Composition of Bounded Linear Transformations is Bounded Linear Transformation, $S T$ and $T^{-1} S^{-1}$ are bounded.
From Inverse of Composite Bijection, we have that $\paren {S T}^{-1} = T^{-1} S^{-1}$ is bounded.. 
Hence $S T$ is a linear isomorphism.
{{qed}}
Category:Normed Vector Spaces
k2prqxy6gstwwjhok8l0o2qe88cdegy
Theorem: Complement of Prime Ideal is Prime Filter

Let $P = \struct {S, \preceq}$ be an ordered set.
Let $I \subseteq S$.
Then:
:$I$ is a prime ideal of $P$
{{iff}}:
:$S \setminus I$ is a prime filter of $P$.

Proof:
Since $S \setminus \paren {S \setminus I} = I$, both conditions are equivalent by definition to the conjunction of:
:$I$ is an ideal of $P$
:$S \setminus I$ is a filter of $P$
The result follows.
{{qed}}
Category:Prime Ideals (Order Theory)
m5mh9pbryrga986lpw9ufxg3rtbmzz8
Theorem: Complement of Prime Filter is Prime Ideal

Let $P = \struct {S, \preceq}$ be an ordered set.
Let $F \subseteq S$.
Then:
:$F$ is a prime filter of $P$
{{iff}}:
:$S \setminus F$ is a prime ideal of $P$.

Proof:
Let $P^{-1}$ denote the dual of $P$.
By Prime Filter is Prime Ideal in Dual Ordered Set:
:$F$ is a prime filter of $P$
{{iff}}:
:$F$ is a prime ideal of $P^{-1}$
By Prime Ideal is Prime Filter in Dual Ordered Set:
:$S \setminus F$ is a prime ideal of $P$
{{iff}}:
:$S \setminus F$ is a prime filter of $P^{-1}$
The result follows by Complement of Prime Ideal is Prime Filter.
{{qed}}
Category:Prime Ideals (Order Theory)
6mzz8ow6jywhil1b28j6q9tn3hoyn19
Theorem: Equivalence of Basic Sequences in Normed Vector Spaces is Preserved by Scalar Multiplication

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a basic sequence in $X$.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a basic sequence in $Y$ that is equivalent to $\sequence {x_n}_{n \mathop \in \N}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N}$ be a sequence in $\GF$ such that $\lambda_n \ne 0$ for each $n \in \N$.
Then $\sequence {\lambda_n x_n}_{n \mathop \in \N}$ is equivalent to $\sequence {\lambda_n y_n}_{n \mathop \in \N}$.

Proof:
From Scalar Multiple of Basic Sequence is Basic, both $\sequence {\lambda_n x_n}_{n \mathop \in \N}$ and $\sequence {\lambda_n y_n}_{n \mathop \in \N}$ are basic.
Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be a sequence in $\GF$.
Since $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ are equivalent, we have that:
:for each sequence $\sequence {\beta_j}_{j \mathop \in \N}$ in $\GF$, we have that:
::$\ds \sum_{j \mathop = 1}^\infty \beta_j x_j$ converges in $\struct {X, \norm {\, \cdot \,}_X}$ {{iff}} $\ds \sum_{j \mathop = 1}^\infty \beta_j y_j$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$.
Setting $\beta_j = \alpha_j \lambda_j$ for each $j \in \N$, we have that:
:$\ds \sum_{j \mathop = 1}^\infty \alpha_j \paren {\lambda_j x_j}$ converges in $\struct {X, \norm {\, \cdot \,}_X}$ {{iff}} $\ds \sum_{j \mathop = 1}^\infty \alpha_j \paren {\lambda_j y_j}$ converges in $\struct {Y, \norm {\, \cdot \,}_Y}$.
Hence:
:$\sequence {\lambda_n x_n}_{n \mathop \in \N}$ is equivalent to $\sequence {\lambda_n y_n}_{n \mathop \in \N}$. 
{{qed}}
Category:Basic Sequences
iozes2d09265w6663xos5fwb0vj2qk7
Theorem: Characterization of Bounded Linear Transformation from Normed Vector Space to Space of Bounded Sequences

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$ such that $X \ne \set { {\mathbf 0}_X}$.
Let $\struct {\map {\ell_\infty} \GF, \norm {\, \cdot \,}_\infty}$ be the space of bounded sequences.
Let $T : X \to \map {\ell_\infty} \GF$ be a mapping.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\struct {\map \BB {X, \map {\ell_\infty} \GF}, \norm {\, \cdot \,}_{\map \BB {X, \map {\ell_\infty} \GF} } }$ be the space of bounded linear transformations $X \to \map {\ell_\infty} \GF$.
Then $T$ is a bounded linear transformation {{iff}} there exists a sequence $\sequence {f_n}_{n \mathop \in \N}$ of bounded linear functionals $f_n : X \to \GF$ such that:
:$\ds \sup_{n \mathop \in \N} \norm {f_n}_{X^\ast} < \infty$
and:
:$T x = \sequence {\map {f_n} x}_{n \mathop \in \N}$ for each $x \in X$.
In the case that $T$ is bounded, we have that:
:$\ds \norm T_{\map \BB {X, \map {\ell_\infty} \GF} } = \sup_{n \mathop \in \N} \norm {f_n}_{X^\ast}$

Proof:
Let $\struct {\map {\ell_\infty^\ast} \GF, \norm {\, \cdot \,}_{\ell_\infty^\ast} }$ be the normed dual space of $\struct {\map {\ell_\infty} \GF, \norm {\, \cdot \,}_\infty}$.
For each $j \in \N$, define $\pi_j : \map {\ell_\infty} \GF \to \GF$ by:
:$\map {\pi_j} x = x_j$ for each $x \in \map {\ell_\infty} \GF$.
From Coordinate Projection on P-Sequence Space is Bounded Linear Functional, we have that $\norm {\pi_j}_{\map {\ell_\infty^\ast} \GF} = 1$ for each $J \in \N$.
Theorem: Contact Forces are Equal and Opposite

Let $A$ and $B$ be bodies in contact.
The contact force of $A$ on $B$ is equal in magnitude and in the opposite direction to the contact force of $B$ on $A$.

Proof:
This is a direct consequence of Newton's Third Law of Motion.
Theorem: Composite of Continuous Real Functions at Point is Continuous

Let $\R$ denote the real numbers.
Let $f$ and $g$ be real functions.
Let $g$ be continuous at a point $a \in \R$.
Let $f$ be continuous at the point $\map g a \in \R$.
Then their composite $f \circ g$ is continuous at $a$.

Proof:
This follows from Limit of Composite Function.
Indeed, we have, from continuity hypotheses:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \rightarrow a} \map g x
      | r = \map g a
      | c = 
}}
{{eqn | l = \lim_{y \mathop \rightarrow \map g a} \map f y
      | r = \map f {\map g a}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \lim_{x \mathop \rightarrow a} {\map f {\map g x} }
      | r = \map f {\map g a}
      | c = 
}}
{{end-eqn}}
which is the result we are looking for.
{{qed}}
Theorem: Space of Bounded Sequences Supported on Subset of Natural Numbers is Closed Vector Subspace of Space of Bounded Sequences

Let $\GF \in \set {\R, \C}$.
Let $A \subseteq \N$. 
Let $\struct {\map {\ell_\infty} {\N, \GF}, \norm {\, \cdot \,}_\infty}$ be the space of bounded sequences.
Let:
:$\map {\ell_\infty} {A, \GF} = \set {\xi \in \map {\ell_\infty} \N : \map \xi n = 0 \text { for } n \in \N \setminus A}$
Equip $\map {\ell_\infty} {A, \GF}$ with pointwise addition and pointwise scalar multiplication.
Then $\struct {\map {\ell_\infty} {A, \GF}, \norm {\, \cdot \,}_\infty}$ is a closed vector subspace of $\struct {\map {\ell_\infty} {\N, \GF}, \norm {\, \cdot \,}_\infty}$.

Proof:
=== $\map {\ell_\infty} A$ is closed ===
Let $\sequence {\xi_n}_{n \mathop \in \N}$ be a sequence of elements of $\map {\ell_\infty} {A, \GF}$ converging to $\xi$. 
Then:
:$\map {\xi_n} j = 0$ for $j \in \N \setminus A$
We have:
:$\cmod {\map \xi j} = \cmod {\map {\xi_n} j - \map \xi j} \le \norm {\xi_n - \xi}_\infty$ for each $j \in \N \setminus A$.
Taking $n \to \infty$, we have:
:$\norm {\xi_n - \xi}_\infty \to 0$
and hence we have:
:$\map \xi j = 0$ for each $j \in \N \setminus A$.
Hence $\xi \in \map {\ell_\infty} {A, \GF}$.
Hence every convergent sequence in $\map {\ell_\infty} {A, \GF}$ has its limit in $\map {\ell_\infty} {A, \GF}$. 
Hence $\map {\ell_\infty} {A, \GF}$ is closed.
{{qed|lemma}}