avewright/proofwiki-math huggingface dataset:
Theorem: Injection Induces Well-Ordering

Let $\left({T, \le}\right)$ be a well-ordered set.
Let $S$ be a set.
Let $f: S \to T$ be an injection.
Then $\le_f$, the ordering induced by $f$, is a well-ordering.

Proof:
By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering.
Let $S' \subseteq S$ be non-empty.
Then as $\le$ is a well-ordering, the set:
:$f \left({S'}\right) = \left\{{f \left({s}\right): s \in S'}\right\}$
has a minimal element, say $f \left({s_0}\right)$.
That is, for all $s \in S', f \left({s_0}\right) \le f \left({s}\right)$.
By definition of $\le_f$, this implies:
:$\forall s \in S': s_0 \le_f s$
That is, $s_0$ is a minimal element of $S'$.
Hence $\le_f$ is a well-ordering, as $S'$ was arbitrary.
{{qed}}
Category:Well-Orderings
Category:Injections
7hfxhks81racj6qp7e5oupmvgi9e4rg
Theorem: K-Connectivity Implies Lesser Connectivity

Let $G = \struct {V, E}$ be a $k$-connected graph.
Then $G$ is $l$-connected for all $l \in \Z : 0 < l < k$.

Proof:
Suppose that $G$ is $k$-connected.
Then:
:$\card V > k > l$
:$G$ is connected
:If $W$ is a vertex cut of $G$, then $\card W \ge k > l$ so $\card W \ge l$.
{{qed}}
Category:Connectedness (Graph Theory)
syax1man5j7s7nb575ekb7dgjj4g2gn
Theorem: Hartogs' Lemma (Set Theory)

Let $S$ be a set.
Then there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.

Proof:
Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$.
First of all, it is to be shown that $\alpha$ is a set.
To this end, define the set $W$ by:
:$W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}$
By the Counting Theorem, each $w \in W$ corresponds to a unique ordinal $\beta_w$.
Thus by the Axiom of Replacement, the following is a set:
:$\set {\beta: \exists w \in W: \beta = \beta_w}$
It follows from Injection Induces Well-Ordering that this set coincides with $\alpha$.
In particular, then, $\alpha$ is a set.
Next, to establish $\alpha$ is an ordinal.
Suppose $\beta \in \alpha$ and $\gamma < \beta$.
Let $i: \beta \to S$ be an injection, and let $\iota: \gamma \to \beta$ be the inclusion of $\gamma$ in $\beta$.
Then by Composite of Injections is Injection, $i \circ \iota: \gamma \to S$ is an injection.
Hence $\gamma \in \alpha$, and therefore $\alpha$ is an ordinal.
Finally, by Ordinal is not Element of Itself, $\alpha \notin \alpha$.
That is to say, there is no injection $\alpha \to S$.
{{qed}}
Theorem: Hartogs' Lemma (Set Theory)

Let $S$ be a set.
Then there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.

Proof:
Let $W$ be the set of all well-orderings on subsets of $S$.
By the Counting Theorem, there exists a mapping $F: W \to \On$ defined by letting $\map F s$ be the ordinal which is isomorphic to $s$.
{{explain|"isomorphism" between ordinals}}
By Mapping from Set to Class of All Ordinals is Bounded Above, $F \sqbrk W$ has an upper bound $\alpha_0$.
Then if $\alpha$ is any ordinal strictly greater than $\alpha_0$, then $\alpha \notin F \sqbrk W$.
{{AimForCont}} there is an injection $g: \alpha \to S$.
Then by Injection to Image is Bijection, there is a bijection from $\alpha$ onto $g \sqbrk \alpha$.
But this induces a well-ordering on $g \sqbrk \alpha \subseteq S$ which is isomorphic to $\alpha$, contradicting the fact that $\alpha \notin F \sqbrk W$.
{{qed}}
Theorem: Bourbaki-Witt Fixed Point Theorem

Let $\struct {X, \le}$ be a non-empty chain complete ordered set (that is, an ordered set in which every chain has a supremum).
Let $f: X \to X$ be an inflationary mapping, that is, so that $\map f x \ge x$.
Then for every $x \in X$ there exists $y \in X$ where $y \ge x$ such that $\map f y = y$.

Proof:
Let $\gamma$ be the Hartogs number of $X$.
Suppose that $x \in X$.
Define $g : \gamma \to X$ by transfinite recursion as follows: 
:$\map g 0 = x$
:$\map g {\alpha + 1} = \map f {\map g \alpha}$
:$\map g \alpha = \sup \set {\map g \beta: \beta < \alpha}$ when $\alpha$ is a limit ordinal.
That $f$ is inflationary guarantees both that:
:$(1) \quad \set {\map g \beta: \beta < \alpha}$ is a chain for each $\alpha < \gamma$
:$(2) \quad g$ is increasing.
{{explain|I'm not sure whether further explanation is needed as to why the above should be so, or whether I'm just slow.}}
If $g$ is strictly increasing, then $g$ is strictly monotone and, by Strictly Monotone Mapping with Totally Ordered Domain is Injective, an injection.
But by definition of Hartogs number, $\gamma$ is the least ordinal such that there is no injection from $\gamma$ to $X$.
Therefore there must be some $\alpha < \beta < \gamma$ with $\map g \alpha = \map g \beta$.
Then:
:$\alpha + 1 \le \beta$
so:
:$\map g \alpha \le \map g {\alpha + 1} \le \map g \beta = \map g \alpha$
whence:
:$\map f {\map g \alpha} = \map g \alpha$
Because $x = \map g 0$:
:$\exists y \in \map g \gamma \subseteq \set {y \in X : x \le y}: \map f y = y$
{{qed}}
{{Namedfor|Nicolas Bourbaki|name2 = Ernst Witt|cat = Bourbaki|cat2 = Witt}}
Theorem: Convergent Sequence with Finite Number of Terms Deleted is Convergent

Let $\struct {X, d}$ be a metric space.
Let $\sequence {x_k}$ be a sequence in $X$.
Let $\sequence {x_k}$ be convergent.
Let a finite number of terms be deleted from $\sequence {x_k}$.
Then the resulting subsequence is convergent.

Proof:
Suppose the sequence $\sequence {x_k}$ converges to $x \in X$.
That is for every $\epsilon > 0$ there is some index $N$ such that $\map d {x_n, x} < \epsilon$ for all $n \ge N$.
The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the original sequence.
For the second part note that also adding finitely many terms to a convergent sequence will still result in a convergent sequence.
This implies that removing finitely many terms from a divergent sequence will still result in a divergent sequence (if it were convergent then the original sequence must already have been convergent).
{{handwaving|Very little more is being done here than repeat the exposition.}}
{{refactor|First fix the exposition, then determine if the structure needs amending|level = basic}}
{{MissingLinks}}
Theorem: Cauchy Condensation Test

Let $\sequence {a_n}: n \mapsto \map a n$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero.
That is, for every $n$ in the domain of $\sequence {a_n}$: $a_n > 0$, $a_n \ge a_{n + 1}$, and $a_n \to 0$ as $n \to +\infty$.
Then the series:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ 
converges {{iff}} the condensed series:
:$\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$
converges.

Proof:
=== Necessary Condition ===
We will first show that:
:if the condensed series $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges
then:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges as well.
Assume $\ds \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges.
Consider the graph of $\sequence {a_n}$ and the partial sums of $\sum 2^n \map a {2^n}$:
:700px
The dotted black line represents the sequence $\sequence {a_n}$.
The $n$th rectangle has:
:Base equal to $2^n$
:Altitude equal to $\map a {2^n}$
The series of partial sums of the areas of the rectangles are:
:$\ds \sum_{n \mathop = 1}^N 2^n \map a {2^n}$
which is precisely the defined condensed series.
The series $\sum a_n$ can be viewed as the sum of the areas of rectangles with width $1$ and height $a_n$. 
The diagram suggests that the partial sums of $\sum a_n$ are not greater than the condensed partial sums:
:$\ds \sum_{n \mathop = 1}^N a_n \le \sum_{n \mathop = 1}^N 2^n \map a {2^n}$
To formalize this claim, observe that because $\sequence {a_n}$ is decreasing:
:$a_1 + \underbrace {a_2 + a_3}_{\le 2 a_2} + \underbrace {a_4 + a_5 + a_6 + a_7}_{\le 4 a_4} + \cdots + a_N \le a_1 + 2 a_2 + 4 a_4 + \cdots + 2^N \map a {2^N}$
As $n \to +\infty$, the {{RHS}} converges by hypothesis.
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ also converges, by the Comparison Test.
{{qed|lemma}}
Theorem: Between Two Sets Exists Injection or Surjection

Let $S$ and $T$ be sets.
Then either or both of the following cases hold:
:$(1):$ There exists a mapping $f: S \to T$ such that $f$ is an injection
:$(2):$ There exists a mapping $f: S \to T$ such that $f$ is a surjection.

Proof:
From Zermelo's Theorem, exactly one of the following cases holds:
:$(1): \quad S \le T$, that is, $T$ dominates $S$
:$(2): \quad S \equiv T$, that is, $T$ is equivalent to $S$
:$(3): \quad S \ge T$, that is, $S$ dominates $T$.
Suppose $(1)$ holds, that is, $T$ dominates $S$.
By definition, there exists an injection from $S$ to $T$.
Hence in this case the result holds.
{{qed|lemma}}
Suppose $(2)$ holds, that is, $T$ is equivalent to $S$.
By definition, there then exists a bijection from $S$ to $T$.
A bijection is, by definition, a mapping which is both an injection and a surjection.
Hence in this case the result holds.
{{qed|lemma}}
Suppose $(3)$ holds, that is, $S$ dominates $T$.
By definition, there exists an injection from $S$ to $T$.
By Injection has Surjective Left Inverse Mapping, $f$ has a left inverse $g: T \to S$ which is surjective.
Hence in this case the result holds.
{{qed|lemma}}
All cases have been shown to lead to the same conclusion, that is, that there exists either an injection or a surjection or both between $S$ and $T$.
Hence the result, by Proof by Cases.
{{qed}}
{{AoC|Zermelo's Theorem (Set Theory)}}
Category:Injections
Category:Surjections
7i5scj9fe0as9hfjtfpi0sdn2csd23b
Theorem: Injection from Set to Power Set

For every set $S$, there exists an injection from $S$ to its power set $\powerset S$.

Proof:
If $S = \O$, the empty mapping suffices, as it is vacuously an injection.
Let $f: S \to \powerset S$ be the mapping defined as:
:$\forall s \in S: \map f s = \set s$
Let $s, t \in S$ such that $\map f s = \map f t$.
Then $\set s = \set t$.
By definition of set equality it follows directly that $s = t$.
Hence $f$ is the required injection, by definition.
{{qed}}
Category:Injections
Category:Power Set
hzgqv1clittk5b4hlrxn3gj96gfklbg
Theorem: Minimum Degree is at Least Connectivity

Let $G = \struct {V, E}$ be a simple graph.
Then:
:$\map \delta G \ge \map \kappa G$
That is, the minimum degree of $G$ is at least its connectivity.

Proof:
Pick a vertex $v \in G$ with $\map {\deg_G} v = \map \delta G$, that is, a vertex with minimum degree.
Recall that $\map {\Gamma_G} v$ is the neighborhood of $v$ in $G$.
Suppose that $V = \map {\Gamma_G} v \cup \set v$.
That is, that $v$ is adjacent to all other vertices of $G$.
Then:
:$\card V = \card {\map {\Gamma_G} v} + 1 = \map \delta G + 1$
By definition, $\map \kappa G < \card V$, so $\map \kappa G \le \map \delta G$.
Otherwise, there is at least one vertex of $G$ not adjacent to $v$.
So, deleting $\map {\Gamma_G} v$ from $G$ leaves $v$ isolated.
Thus $\map {\Gamma_G} v$ is a vertex cut of size $\map \delta G$. 
Since $G$ has a vertex cut of size $\map \delta G$, it follows that $\map \kappa G \le \map \delta G$.
{{qed}}
Theorem: Homeomorphism Relation is Equivalence

Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \sim T_2$ denote that $T_1$ and $T_2$ are homeomorphic.
The relation $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Identity Mapping is Homeomorphism

Let $T$ be a topological space.
The identity mapping $I_T: T \to T$ defined as:
:$\forall x \in T: \map {I_T} x = x$
is a homeomorphism.

Proof:
We have Identity Mapping is Bijection.
We also have Identity Mapping is Continuous.
Hence, by definition, $I_T$ is a homeomorphism.
{{qed}}
Category:Homeomorphisms (Topological Spaces)
Category:Identity Mappings
onel9o918gu70i9ncm8rbow9peprcdu
Theorem: Composite of Homeomorphisms is Homeomorphism

Let $T_1, T_2, T_3$ be topological spaces. 
Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be homeomorphisms.
Then $g \circ f: T_1 \to T_3$ is also a homeomorphism.

Proof:
By definition of homeomorphism, $f$ and $g$ are both bijections.
From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection.
By definition of homeomorphism, $f$ and $g$ are both continuous mappings.
From Composite of Continuous Mappings is Continuous it follows that $g \circ f$ is also a continuous mapping.
From Inverse of Bijection is Bijection, $f^{-1}: T_2 \to T_1$ and $g^{-1}: T_3 \to T_2$ are also both bijections.
From Composite of Bijections is Bijection it follows that $f^{-1} \circ g^{-1}$ is also a bijection.
By definition of homeomorphism, $f^{-1}$ and $g^{-1}$ are both continuous mappings.
Again from Composite of Continuous Mappings is Continuous it follows that $f^{-1} \circ g^{-1}$ is also a continuous mapping.
Hence the result, from definition of homeomorphism.
{{qed}}
Category:Homeomorphisms (Topological Spaces)
Category:Composite Mappings
7jvzxhi26k7nnghjcj75ksqn3f85z3k
Theorem: Identity Mapping is Continuous

Let $T = \struct {S, \tau}$ be a topological space.
The identity mapping $I_S: S \to S$ defined as:
:$\forall x \in S: \map {I_S} x = x$
is a continuous mapping.

Proof:
Let $U \in \tau$.
We have Identity Mapping is Bijection.
So $I_S^{-1}$ is well-defined and:
:$\forall x \in U: \map {I_S^{-1} } x = x$
Thus $I_S^{-1} \sqbrk U = U \in \tau$.
Hence, by definition of continuous mapping, $I_S$ is continuous.
{{qed}}
Theorem: Clopen Sets in Finite Complement Topology

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then the only clopen sets of $T$ are $S$ and $\O$.

Proof:
Let $U \in \tau$ be open in $T$.
Then by definition of finite complement topology, $S \setminus U$ is finite.
By definition of open set, $S \setminus U$ is closed.
As $S$ is infinite, it follows that $U$ must also be infinite.
Thus unless $U = S$, $S \setminus U$ can not be open.
Hence the result.
{{qed}}
Theorem: Included Set Topology is Topology

Let $T = \struct {S, \tau_H}$ be an included set space.
Then $\tau_H$ is a topology on $S$, and $T$ is a topological space.

Proof:
=== {{Open-set-axiom|1|nolink}} ===
Let $U_1, U_2 \in \tau_H$.
By definition of included set topology:
:$H \subseteq U_1$ and $H \subseteq U_2$
From Intersection is Largest Subset:
:$H \subseteq U_1 \cap U_2$
By definition of included set topology:
:$U_1 \cap U_2 \in \tau_H$
{{qed|lemma}}
Theorem: Compact Linear Transformation is Bounded

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $\map {B_0} {\HH, \KK}$ be the space of compact linear transformations $\HH \to \KK$. 
Let $\map B {\HH, \KK}$ be the space of bounded linear transformations $\HH \to \KK$. 
Let $T \in \map {B_0} {\HH, \KK}$.
Then $T \in \map B {\HH, \KK}$.
That is:
:$\map {B_0} {\HH, \KK} \subseteq \map B {\HH, \KK}$

Proof:
Let $T : \HH \to \KK$ be a linear transformation.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$. 
Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.
We show that:
:if $T$ is not bounded, then it is not compact.
That is: 
:if $T \not \in \map B {\HH, \KK}$, then $T \not \in \map {B_0} {\HH, \KK}$.
Then, we will obtain:
:$\map {B_0} {\HH, \KK} \subseteq \map B {\HH, \KK}$
Suppose that $T$ is not bounded.
That is: 
:there does not exist a real number $c > 0$ such that $\norm {T h}_\KK \le c \norm h_\HH$ for all $h \in \HH$. 
Then:
:for each $n \in \N$, there exists some $x_n \in \HH$ such that $\norm {T x_n}_\KK > n \norm {x_n}_\HH$.
Now define sequence $\sequence {h_n}$ by: 
:$\ds h_n = \frac {x_n} {\norm {x_n} }$
for each $n \in \N$. 
We have: 
:$\norm {h_n} = 1$
for each $n \in \N$, so $\sequence {h_n}$ is bounded. 
We also have: 
:$\norm {T h_n} > n$
for each $n \in \N$. 
So, for any increasing sequence of positive integers $\sequence {n_j}$ we have:
:$\norm {T h_{n_j} } > n_j$
Note that $n_j \to \infty$.  
So, for every subsequence $\sequence {h_{n_j} }$ of $\sequence {h_n}$ we have that: 
:$\sequence {T h_{n_j} }$ is unbounded. 
So from Convergent Sequence in Normed Vector Space is Bounded, we have that: 
:$\sequence {T h_{n_j} }$ does not converge. 
So:
:the bounded sequence $\sequence {h_n}$ does not have a subsequence $\sequence {h_{n_j} }$ such that $\sequence {T h_{n_j} }$ converges.
So:
:$T$ is not compact.
Hence the demand.
{{qed}}
Theorem: Compact Linear Transformations Composed with Bounded Linear Operator

Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_0} {H, K}$ be a compact linear transformation.
Let $A \in \map B H, B \in \map B K$ be bounded linear operators.
Then the compositions $T A$ and $B T$ are also compact linear transformations.

Proof:
Let $\sequence {h_n}_{n \in \N}$ is a bounded sequence in $H$.
That is, there exists a $M > 0$ such that:
:$\forall n \in \N : \norm {h_n}_H \le M$
Then $\sequence {A h_n}_{n \in \N}$ is also bounded, since:
:$\norm {A h_n}_H \le \norm {A}_{\map B H} \norm {h_n}_H \le \norm {A}_{\map B H} M$
As $T$ is compact,
the sequence $\sequence {T A h_n}_{n \in \N}$ has a subsequence convergent in $K$.
Hence $T A$ is compact.
{{qed|lemma}}
Let $\sequence {k_n}_{n \in \N}$ is a bounded sequence in $K$.
As $T$ is compact,
the sequence $\sequence {T k_n}_{n \in \N}$ has a subsequence convergent in $H$.
Say $\sequence {T k_{n_r} }_{r \in \N}$ is convergent.
Recall that $B$ is continuous due to Continuity of Linear Transformation between Normed Vector Spaces.
Thus $\sequence {B T k_{n_r} }_{r \in \N}$ is convergent.
Hence $B T$ is compact.
{{qed}}
Theorem: Expansion of Included Set Topology

Let $S$ be a set.
Let $A_1 \subseteq S$ and $A_2 \subseteq S$.
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$.
Then:
:$(1): \quad T_1 \ge T_2 \iff A_1 \subseteq A_2$
:$(2): \quad T_1 > T_2 \iff A_1 \subsetneq A_2$
where:
:$T_1 \ge T_2$ denotes that $T_1$ is finer than $T_2$
:$T_1 > T_2$ denotes that $T_1$ is strictly finer than $T_2$.

Proof:
Let $T_1 = \struct {S, \tau_{A_1} }$ and $T_2 = \struct {S, \tau_{A_2} }$ be included set spaces on $S$.
Theorem: Linear Transformation Compact iff Adjoint Compact

Let $H, K$ be Hilbert spaces.
Let $T: H \to K$ be a linear transformation.
Then $T$ is compact {{iff}} its adjoint $T^*$ is.

Proof:
First suppose that $T$ is compact.
From the construction of the adjoint, $T^\ast$ is bounded.
From Right Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, $T T^\ast$ is compact.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $K$.
Then there exists a convergent subsequence of $\sequence {T T^\ast x_n}_{n \mathop \in \N}$, say $\sequence {T T^\ast x_{n_j} }_{j \mathop \in \N}$. 
We show that $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ converges.
Pick $M > 0$ such that: 
:$\norm {x_n}_K \le M$
for each $n \in \N$. 
We have: 
{{begin-eqn}}
{{eqn	| l = \norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }^2_H
		| r = \innerprod {\map {T^\ast} {x_{n_j} - x_{n_k} } } {\map {T^\ast} {x_{n_j} - x_{n_k} } }_H
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \innerprod {\map {T T^\ast} {x_{n_j} - x_{n_k} } } {x_{n_j} - x_{n_k} }_K
		| c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn	| o = \le
		| r = \norm {\map {T T^\ast} {x_{n_j} - x_{n_k} } } \norm {x_{n_j} - x_{n_k} }_K
		| c = Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces
}}
{{eqn	| o = \le
		| r = 2 M \norm {\map {T T^\ast} {x_{n_j} - x_{n_k} } }_K
		| c = {{NormAxiomVector|3}}
}}
{{end-eqn}}
Since $\sequence {\map {T T^\ast} {x_{n_j} } }_{j \mathop \in \N}$ is convergent, it is Cauchy from Convergent Sequence is Cauchy Sequence. 
Let $\epsilon > 0$. 
Pick $N \in \N$ so that: 
:$\ds \norm {\map {T T^\ast} {x_{n_j} - x_{n_k} } }_K < \frac {\epsilon^2} {2 M}$
for $n, m \ge N$. 
Then for $n, m \ge N$ we have: 
:$\norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }_K^2 < \epsilon^2$
so:
:$\norm {\map {T^\ast} {x_{n_j} - x_{n_k} } }_K < \epsilon$
So $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ is Cauchy.
Since $H$ is Hilbert, $\sequence {T^\ast x_{n_j} }_{j \mathop \in \N}$ is convergent.
So for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ in $K$, $\sequence {T^\ast x_n}_{n \mathop \in \N}$ has a convergent subsequence. 
So $T^\ast$ is compact.
So for a bounded linear transformation $T : H \to K$, if $T$ is compact then $T^\ast$ is compact.
So for a bounded linear transformation $T : H \to K$, if $T^\ast$ is compact then $T^{\ast \ast}$ is compact.
From Adjoint is Involutive, we therefore have that that if $T^\ast$ is compact then $T$ is compact.
So $T$ is compact {{iff}} its adjoint $T^*$ is.
{{qed}}
Theorem: Finite Rank Operators Dense in Compact Linear Transformations

Let $H, K$ be Hilbert spaces.
Then:
:$\map {B_{00} } {H, K}$ is everywhere dense in $\map {B_0} {H, K}$
where:
:$\map {B_{00} } {H, K}$ is the space of continuous finite rank operators from $H$ to $K$
:$\map {B_0} {H, K}$ is the space of compact linear transformations from $H$ to $K$.
That is, for every $T \in \map {B_0} {H, K}$, there is a sequence $\sequence {T_n}_{n \mathop \in \N}$ in $\map {B_{00} } {H, K}$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {T_n - T} = 0$
where $\norm {\, \cdot \,}$ denotes the norm on bounded linear transformations.

Proof:
{{proof wanted|Proceeds by sequence definition of limit point}}
Theorem: Inverse of Matrix Product

Let $\mathbf {A, B}$ be square matrices of order $n$
Let $\mathbf I$ be the $n \times n$ unit matrix.
Let $\mathbf A$ and $\mathbf B$ be nonsingular.
Then the matrix product $\mathbf {AB}$ is also nonsingular, and:
:$\paren {\mathbf A \mathbf B}^{-1} = \mathbf B^{-1} \mathbf A^{-1}$

Proof:
We are given that $\mathbf A$ and $\mathbf B$ are nonsingular.
From Product of Matrices is Nonsingular iff Matrices are Nonsingular, $\mathbf A \mathbf B$ is also nonsingular.
By the definition of inverse matrix:
:$\mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$
and 
:$\mathbf B \mathbf B^{-1} = \mathbf B^{-1} \mathbf B = \mathbf I$
Now, observe that:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A \mathbf B} \paren {\mathbf B^{-1} \mathbf A^{-1} }
      | r = \paren {\mathbf A \paren {\mathbf B \mathbf B^{-1} } } \mathbf A^{-1}
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \paren {\mathbf A \mathbf I} \mathbf A^{-1}
}}
{{eqn | r = \mathbf A \mathbf A^{-1}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \mathbf I
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf B^{-1} \mathbf A^{-1} } \paren {\mathbf A \mathbf B}
      | r = \paren {\mathbf B^{-1} \paren {\mathbf A^{-1} \mathbf A} } \mathbf B
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \paren {\mathbf B^{-1} \mathbf I} \mathbf B
}}
{{eqn | r = \mathbf B^{-1} \mathbf B
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \mathbf I
}}
{{end-eqn}}
The result follows from the definition of inverse.
{{qed}}
Theorem: Elementary Row Operations as Matrix Multiplications

Let $e$ be an elementary row operation.
Let $\mathbf E$ be the elementary row matrix of order $m$ defined as:
:$\mathbf E = \map e {\mathbf I}$
where $\mathbf I$ is the unit matrix.
Then for every $m \times n$ matrix $\mathbf A$:
:$\map e {\mathbf A} = \mathbf {E A}$
where $\mathbf {E A}$ denotes the conventional matrix product.

Proof:
Let $s, t \in \closedint 1 m$ such that $s \ne t$.