nyuuzyou/wikis huggingface dataset:
Characterization of Hausdorff Property in terms of Nets

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
We have that $\struct {X, \tau}$ is Hausdorff convergent nets have a unique limit. 
That is, whenever $\struct {\Lambda, \preceq}$ is a directed set, $x, y \in X$ and $\family {x_\lambda}_{\lambda \in \Lambda}$ is a net converging to $x$ and $y$ in $\struct {X, \tau}$, we have $x = y$. 
== Proof ==
=== Necessary Condition ===
Suppose that $\struct {X, \tau}$ is Hausdorff.
Let $\struct {\Lambda, \preceq}$ is a directed set, $x, y \in X$ and $\family {x_\lambda}_{\lambda \in \Lambda}$ is a net converging to $x$ and $y$ in $\struct {X, \tau}$.
 suppose that $x \ne y$. 
Since $\struct {X, \tau}$ is Hausdorff there exists open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $U \cap V = \O$.
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_x \in \Lambda$ such that: 
:$x_\lambda \in U$ for $\lambda_x \preceq \lambda$.
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $y$, there exists $\lambda_y \in \Lambda$ such that: 
:$x_\lambda \in V$ for $\lambda_y \preceq \lambda$.
Since $\struct {\Lambda, \preceq}$ is a directed set, there exists $\lambda_\ast \in \Lambda$ such that: 
:$\lambda_x \preceq \lambda_\ast$ and $\lambda_y \preceq \lambda_\ast$. 
Then: 
:$x_{\lambda_\ast} \in U$
and:
:$x_{\lambda_\ast} \in V$
But then $x_{\lambda_\ast} \in U \cap V = \O$, contradiction. 
So we have $x = y$.
=== Sufficient Condition ===
Suppose that $\struct {X, \tau}$ is not Hausdorff. 
Then there exists distinct $x, y \in X$ such that: 
:for all open neighborhoods $U$ and $V$ of $x$ and $y$ respectively, we have $U \cap V \ne \O$.
Let: 
:$\FF = \set {\tuple {U, V} : U \text { and } V \text { are open with } x \in U, \, y \in V}$
Define a preordering $\sqsubseteq$ on $\FF$ by: 
:$\tuple {U, V} \sqsubseteq \tuple {U', V'}$
:
:$U \supseteq U'$ and $V \supseteq V'$.
From Product of Directed Sets is Directed Set, we have that $\FF$ is directed.
For each $\tuple {U, V} \in \FF$, pick: 
:$x_{\tuple {U, V} } \in U \cap V$
We show that $\family {x_{\tuple {U, V} } }_{\tuple {U, V} \in \FF}$ converges to $x$ and $y$.
Let $U'$ be an open neighborhood of $x$. 
Then for $\tuple {U', X} \in \FF$ with $\tuple {U', X} \sqsubseteq \tuple {U, V}$, we have:
:$x_{\tuple {U, V} } \in U \cap V \subseteq U \subseteq U'$
Similarly, let $V'$ be an open neighborhood of $Y$.
Then for $\tuple {X, V'} \in \FF$ with $\tuple {X, V'} \sqsubseteq \tuple {U, V}$ we have: 
:$x_{\tuple {U, V} } \in U \cap V \subseteq V \subseteq V'$
So $\family {x_{\tuple {U, V} } }_{\tuple {U, V} \in \FF}$ converges to $x$ and $y$ with $x \ne y$.
So:
:if $\struct {X, \tau_X}$ is not Hausdorff then: 
::there exists $x, y \in X$ with $x \ne y$, a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to both $x$ and $y$.
We are done by Proof by Contraposition.
Category:Hausdorff Spaces
Category:Nets (Set Theory)
Characterization of Hausdorff Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space.
:$(1) \quad$ $\struct {X, \tau}$ is a Hausdorff topological vector space
:$(2) \quad$ the intersection of all open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$ is $\set { {\mathbf 0}_X}$
:$(3) \quad$ $\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$. 
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose that $\struct {X, \tau}$ is a Hausdorff topological vector space.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is $T_1$.
Clearly:
:$\set { {\mathbf 0}_X} \subseteq \bigcap \FF$
Now let $x \in X \setminus \set { {\mathbf 0}_X}$. 
Then there exists an open neighborhood $U \in \FF$ of ${\mathbf 0}_X$ such that $x \not \in U$. 
In particular, $x \not \in \bigcap \FF$. 
So we obtain $\bigcap \FF = \set { {\mathbf 0}_X}$.
=== $(2)$ implies $(3)$ ===
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Suppose that $\bigcap \FF = \set { {\mathbf 0}_X}$.
Let $x \in \map \cl {\set { {\mathbf 0}_X} }$.
From Condition for Point being in Closure, for each open neighborhood $U_x$ of $x$ such that ${\mathbf 0}_X \in U_x$.
From Classification of Open Neighborhoods in Topological Vector Space, we equivalently have:
:for each open neighborhood $U$ of ${\mathbf 0}_X$ we have ${\mathbf 0}_X \in U + x$.
That is: 
:$x \in -U$ for each open neighborhood $U$ of ${\mathbf 0}_X$.
From Dilation of Open Set in Topological Vector Space is Open, $U$ is an open neighborhood of ${\mathbf 0}_X$ $-U$ is.
Hence, we have $x \in U$ for each open neighborhood $U$ of ${\mathbf 0}_X$.
So $x \in \bigcap \FF = \set { {\mathbf 0}_X}$.
Hence we have $x = {\mathbf 0}_X$, so $\map \cl {\set { {\mathbf 0}_X} } \subseteq \set { {\mathbf 0}_X}$.
Since $\set { {\mathbf 0}_X} \subseteq \map \cl {\set { {\mathbf 0}_X} }$, we have:
:$\map \cl {\set { {\mathbf 0}_X} } = \set { {\mathbf 0}_X}$
From Set is Closed iff Equals Topological Closure:
:$\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$.
=== $(3)$ implies $(1)$ ===
Suppose that $\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$. 
Let $x \in X$.
We have $\set x = x + \set { {\mathbf 0}_X}$.
From Translation of Closed Set in Topological Vector Space is Closed Set, $\set x$ is closed.
So for all $x \in X$, $\set x$ is closed.
Hence from Definition 2 of a $T_1$ space, $\struct {X, \tau}$ is a $T_1$ space.
From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is Hausdorff.
Category:Hausdorff Topological Vector Spaces
Characterization of Homeomorphic Topological Spaces

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}$ be topological space.
Let $S_2$ be a set.
Let $\tau_2$ be a subset of the powerset $\powerset {S_2}$.
Then:
:$\struct{S_2, \tau_2}$ is a topological space homeomorphic to $T_1$
:
:there exists a mapping $f : S_1 \to S_2$:
::$(1)\quad f$ is a bijection
::$(2)\quad f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$
:where
::$f^\to \restriction_{\tau_1}$ denotes the restriction of $f^\to$ to $\tau_1$
::$f^\to$ denotes the direct image mapping of $f$
== Proof ==
=== Necessary Condition ===
Let $\struct{S_2, \tau_2}$ be a topological space homeomorphic to $T_1$.
=== Sufficient Condition ===
Let $f : S_1 \to S_2$ be a mapping such that:
:$(1)\quad f$ is a bijection
:$(2)\quad f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$
Category:Homeomorphisms
Category:Characterization of Homeomorphic Topological Spaces
Characterization of Homeomorphic Topological Spaces/Necessary Condition

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}$ be topological space.
Let $S_2$ be a set.
Let $\tau_2$ be a subset of the powerset $\powerset {S_2}$.
Let $\struct{S_2, \tau_2}$ be a topological space homeomorphic to $T_1$.
Then:
:there exists a mapping $f : S_1 \to S_2$:
::$(1)\quad f$ is a bijection
::$(2)\quad f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$
:where
::$f^\to \restriction_{\tau_1}$ denotes the restriction of $f^\to$ to $\tau_1$
::$f^\to$ denotes the direct image mapping of $f$
== Proof ==
Let $f: S_1 \to S_2$ be a homeomorphism.
By definition of a homeomorphism:
:$f$ is a bijection
:$f$ is an open mapping
:$f$ is a continuous mapping
By definition of an open mapping:
:$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$
By definition of direct image mapping:
:$\forall U \in \tau_1 : \map {f^\to} U \in \tau_2$
By definition of continuous mapping:
:$\forall V \in \tau_2 : f^{-1} \sqbrk V \in \tau_1$
We have:
It follows that $f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$.
Category:Characterization of Homeomorphic Topological Spaces
Characterization of Injective Linear Transformations with Closed Image

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces. 
Let $T : X \to Y$ be a bounded linear transformation.
Then $T$ is injective and $\Img T$ is closed :
:there exists $c > 0$ such that $\norm {T x}_Y \ge c \norm x_X$ for each $x \in X$. 
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Suppose that:
:there exists $c > 0$ such that $\norm {T x}_Y \ge c \norm x_X$ for each $x \in X$.
First, if $T x = {\mathbf 0}_Y$ for some $x \in X$, then we have:
:$c \norm x_X \le 0$
Since $c > 0$ and $\norm x_X \ge 0$, we have $\norm x_X = 0$.
Hence by , we have $x = {\mathbf 0}_X$.
So $T$ is injective.
Now let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $\Img T$ such that $y_n \to y$ with $y \in Y$. 
We show that $y \in \Img T$. 
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$ such that $y_n = T x_n$ for each $n \in \N$.
Then:
:$\norm {x_n - x_m}_X \le c^{-1} \norm {T x_n - T x_m}_Y = c^{-1} \norm {y_n - y_m}_Y$ 
from the linearity of $T$.
Since $\sequence {y_n}_{n \mathop \in \N}$ is convergent, it is Cauchy by Convergent Sequence is Cauchy Sequence.
Let $\epsilon > 0$.
Since $\sequence {y_n}_{n \mathop \in \N}$ is Cauchy, there exists $N \in \N$ such that:
:$\norm {y_n - y_m} < c \epsilon$ for $n, m \ge N$.
Then for $n, m \ge N$, we have $\norm {x_n - x_m}_X < \epsilon$.
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Since $X$ is Banach, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$.
Hence since $T$ is bounded, we have $T x_n \to T x$ from Continuous Mapping is Sequentially Continuous.
That is, $y_n \to T x$ with $T x \in \Img T$. 
So:
:whenever $\sequence {y_n}_{n \mathop \in \N}$ is a sequence in $\Img T$ converging to $y \in Y$, we have $y \in \Img T$.
Hence $\Img T$ is closed.
=== Necessary Condition ===
Suppose that $T$ is injective and $\Img T$ is closed.
Then $T$ is a linear isomorphism $X \to \Img T$.
From Closed Subspace of Banach Space forms Banach Space, $\struct {\Img T, \norm {\, \cdot \,}_Y}$ is a Banach space. 
Hence from the Banach Isomorphism Theorem, the inverse $T^{-1} : \Img T \to X$ is bounded.
That is, there exists $M > 0$ such that:
:$\norm {T^{-1} y}_X \le M \norm y_Y$ for each $y \in \Img T$. 
We have $y \in \Img T$ $y = T x$ for some $x \in X$.
So we have:
:$\norm {\map {T^{-1} } {T x} }_X \le M \norm {T x}_Y$
so:
:$\norm x_X \le M \norm {T x}_Y$
giving:
:$\norm {T x}_Y \ge M^{-1} \norm x_X$ for each $x \in X$. 
Category:Bounded Linear Transformations
Category:Banach Spaces
Category:Characterization of Injective Linear Transformations with Closed Image
Characterization of Injective Linear Transformations with Closed Image/Corollary

== Corollary ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces. 
Let $T : X \to Y$ be a bounded linear transformation.
Then $T$ is injective and $\Img T$ is closed :
:there does not exist a sequence $\sequence {x_n}_{n \mathop \in \N}$ with $\norm {x_n} = 1$ for each $n \in \N$ such that $T x_n \to {\mathbf 0}_Y$. 
== Proof ==
By Proof by Contraposition, we want to show that:
:there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ with $\norm {x_n} = 1$ for each $n \in \N$ such that $T x_n \to {\mathbf 0}_Y$ either $T$ is not injective or $\Img T$ is not closed.
=== Sufficient Condition ===
Suppose that:
:either $T$ is not injective or $\Img T$ is not closed.
Then from Characterization of Injective Linear Transformations with Closed Range:
:there does not exist $c > 0$ such that $\norm {T x}_Y \ge c \norm x_X$ for each $x \in X$.
So:
:for each $n \in \N$, there exists $u_n \in X$ such that $\norm {T u_n}_Y < n^{-1} \norm {u_n}_X$ for each $n \in \N$.
Since $\norm {u_n}_X > 0$, we have $u_n \ne {\mathbf 0}_X$ by . 
We then have:
:$\norm {T x_n} < n^{-1}$ for each $n \in \N$
where:
:$\ds x_n = \frac {u_n} {\norm {u_n} }$
We then have $T x_n \to {\mathbf 0}_Y$. 
By , we have $\norm {x_n}_X = 1$ for each $n \in \N$.
=== Necessary Condition ===
Suppose that:
:there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ with $\norm {x_n} = 1$ for each $n \in \N$ such that $T x_n \to {\mathbf 0}_Y$
 there exists $c > 0$ such that:
:$\norm {T x}_Y \ge c \norm x_X$ for each $x \in \HH$.
Then we have:
:$\norm {T x_n}_Y \ge c \norm {x_n}_X = c$ for each $n \in \N$.
We then have $T x_n \to {\mathbf 0}_Y$ as $n \to \infty$, so $\norm {T x_n} \to 0$, contradicting that $c > 0$. 
So there does not exist $c > 0$ such that:
:$\norm {T x}_Y \ge c \norm x_X$ for each $x \in \HH$.
Hence by Characterization of Injective Linear Transformations with Closed Range:
:either $T$ is not injective or $\Img T$ is not closed.
Category:Characterization of Injective Linear Transformations with Closed Image
Characterization of Integer Polynomial has Root in P-adic Integers

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Let $\map F X \in \Z \sqbrk X$ be a polynomial with integer coefficients.
Let $a \in \Z_p$.
Then:
:$\map F a = 0$
:there exists a sequence $\sequence{a_n}$ of integers: 
::$(1)\quad\ds\lim_{n \mathop \to \infty} {a_n} = a$
::$(2)\quad\map F {a_n} \equiv 0 \mod {p^{n+1}}$
That is, a polynomial with integer coefficients has a root it has an integer root modulo $p^k$ for every $k \in \N_{>0}$.
== Proof ==
From Characterization of Polynomial has Root in P-adic Integers:
:$\map F a = 0$
:there exists a sequence $\sequence{a_n}$ of integers: 
::$(1)\quad\ds\lim_{n \mathop \to \infty} {a_n} = a$
::$(2)\quad\map F {a_n} \equiv 0 \mod {p^{n+1}\Z_p}$
By definition of a polynomial with integer coefficients:
:$\forall n \in \N_{>0} : \map F {a_n} \in \Z$
From Congruence Modulo Equivalence for Integers in P-adic Integers:
:$\map F a = 0$
:there exists a sequence $\sequence{a_n}$ of integers: 
::$(1)\quad\ds\lim_{n \mathop \to \infty} {a_n} = a$
::$(2)\quad\map F {a_n} \equiv 0 \mod {p^{n+1}}$
== Sources ==
* : $\S 1.7$ Hensel's Lemma and Congruences: Theorem $1.42$
Category:P-adic Integers
Characterization of Integer has Square Root in P-adic Integers

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $a \in Z$ be an integer such that $p \nmid a$.
Then:
:$\exists x \in \Z_p : x^2 = a$
:$a$ is a quadratic residue of $p$.
That is, an integer $a$ not divisible by $p$ has a square root in $\Z_p$ ($p \ne 2$) $a$ is a quadratic residue of $p$.
== Proof ==
Let $F \in \Z[X]$ be the polynomial:
:$\map F X = X^2 - a$
By definition of formal derivative the formal derivative of $F$ is:
:$\map {F'} X = 2X$
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 1.7$ Hensel's Lemma and Congruences: Theorem $1.43$
Category:P-adic Integers
Category:Characterization of Integer has Square Root in P-adic Integers
Characterization of Integer has Square Root in P-adic Integers/Necessary Condition

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $a \in Z$ be an integer such that $p \nmid a$.
Let the exist $x \in \Z_p$ such that $x^2 = a$.
Then:
:$a$ is a quadratic residue of $p$.
That is:
:an integer $a$ not divisible by $p$ has a square root in $\Z_p$ ($p \ne 2$)
:
:$a$ is a quadratic residue of $p$.
== Proof ==
Let $F \in \Z \sqbrk X$ be the polynomial:
:$\map F X = X^2 - a$
By definition, the formal derivative of $F$ is:
:$\map {F'} X = 2 X$
Let there exist $x$ such that $x^2 = a$.
By definition of root of polynomial:
:$\map F X$ has a root in $\Z_p$.
From Characterization of Integer Polynomial has Root in P-adic Integers:
:there exists an integer sequence $\sequence {a_n}$ such that:
::$(1) \quad \ds \lim_{n \mathop \to \infty} {a_n} = a$
::$(2) \quad \map F {a_n} \equiv 0 \mod {p^{n + 1} }$
We have:
:$a_0^2 - a \equiv 0 \pmod p$
That is:
:$a_0^2 \equiv a \pmod p$
Hence by definition:
:$a$ is a quadratic residue of $p$.
Category:Characterization of Integer has Square Root in P-adic Integers
Characterization of Integer has Square Root in P-adic Integers/Sufficient Condition

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $a \in Z$ be an integer such that $p \nmid a$.
Let $a$ be a quadratic residue of $p$.
Then:
:$\exists x \in \Z_p : x^2 = a$
== Proof ==
Let $F \in \Z[X]$ be the polynomial:
:$\map F X = X^2 - a$
By definition of formal derivative the formal derivative of $F$ is:
:$\map {F'} X = 2X$
Let $a$ be a quadratic residue of $p$.
By definition of quadratic residue of $p$:
:$\exists b \in \Z : a \equiv b^2 \pmod p$
Then:
:$\map F b = b^2 - a \equiv 0 \pmod p$
and
:$\map {F'} b = 2b$
By hypothesis:
:$p \nmid 2$
and
:$p \nmid b^2$
From the contrapositive statement of Divisor Divides Multiple:
:$p \nmid b$
From the contrapositive statement of Euclid's Lemma for Prime Divisors:
:$p \nmid 2b$
Hence:
:$\map {F'} b = 2b \not\equiv 0 \pmod p$
From Congruence Modulo Equivalence for Integers in P-adic Integers:
:$\map F b \equiv 0 \pmod {p\Z}$
and
:$\map {F'} b \not\equiv 0 \pmod {p\Z}$
From Hensel's Lemma for P-adic Integers:
:$\exists x \in \Z_p : \map F x = 0$
That is:
:$\exists x \in \Z_p : x^2 = a$
Category:Characterization of Integer has Square Root in P-adic Integers
Characterization of Integrable Functions

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R, f \in \MM_{\overline \R}$ be a $\Sigma$-measurable function.
Then the following are equivalent:
:$(1): \quad f \in \map {\LL_{\overline \R} } \mu$, that is, $f$ is $\mu$-integrable.
:$(2): \quad$ The positive and negative parts $f^+$ and $f^-$ are $\mu$-integrable.
:$(3): \quad$ The absolute value $\size f$ of $f$ is $\mu$-integrable.
:$(4): \quad$ There exists an $\mu$-integrable function $g: X \to \overline \R$ such that $\size f \le g$ pointwise.
== Proof ==
We prove the whole cycle of implications:
:$(1) \implies (2) \quad$ by definition of $(1)$
:$(2) \implies (3)\quad$ because $\size f = f^+ + f^-$ and Integral of Positive Measurable Function is Additive 
:$(3) \implies (4)\quad$ because $g:= \size f$ exists
It remains to demonstrate $(4) \implies (1)$.
Let $f \in \MM_{\overline \R}$ and $g$ according to $(4)$.
Then:
:$f = f^+ - f^-$
where $f^+$ is the positive and $f^-$ is the negative part of $f$.
We have that $f^+$ and $f^-$ are positive and measurable.
Let $f^0$ stand for either $f^+$ or $f^-$.
We have that:
:$\size f = f^+ + f^-$
Therefore:
:$f^0 \le \size f \le g$
It is to be shown that the Integral of Positive Measurable Function of $f^0$ exists and is finite.
Let $\EE^+$ and $\map {I_\mu} h$ be defined as in Integral of Positive Measurable Function.
Then:
:$\forall h \in \EE^+$: $h \le f^0 \implies h \le g$
Hence:
:$\set {h: h \le f^0, h \in \EE^+} \subseteq \set {h: h \le g, h \in \EE^+}$
:$\ds \int f^0 \rd \mu := \sup \set {\map {I_\mu} h: h \le f^0, h \in \EE^+} \le \sup \set {\map {I_\mu} h: h \le g, h \in \EE^+} < \infty$
We have that the integrals for $f^+$ and $f^-$ both are finite.
Therefore $f$ is $\mu$-integrable according to definition.
.
== Sources ==
* : $10.3$
Category:Measure Theory
Characterization of Invariant Subspaces

== Theorem ==
Let $H$ be a Hilbert space.
Let $A \in B \left({H}\right)$ be a bounded linear operator.
Let $\begin{pmatrix}
 W & X \\
 Y & Z
\end{pmatrix}$ be the matrix notation for $A$ with respect to $M$.
Let $M$ be a closed linear subspace of $H$; denote by $P$ the orthogonal projection on $M$.
Then the following three statements are equivalent:
:$(1): \qquad M$ is an invariant subspace for $A$
:$(2): \qquad PAP = AP$
:$(3): \qquad Y = 0$
== Proof ==
== Sources ==
* $II.3.7$
Category:Hilbert Spaces
Category:Linear Transformations on Hilbert Spaces
Characterization of Invertible Bounded Linear Transformations

== Theorem ==
<onlyinclude>
Let $\struct {U, \norm \cdot_U}$ and $\struct {V, \norm \cdot_V}$ be normed vector spaces.
Let $A : V \to U$ be a linear transformation with inverse $A^{-1} : U \to V$.
Then $A^{-1}$ is a bounded linear transformation :
:there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.
That is, $A$ is invertible as a bounded linear transformation :
:there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.
<onlyinclude>
== Proof ==
From Inverse of Linear Transformation is Linear Transformation, we have: 
:$A^{-1}$ is a linear transformation.
So we are interested in determining when $A^{-1}$ is bounded.
=== Necessary Condition ===
Suppose that $A^{-1}$ was bounded.
Then:
:there exists a real number $M > 0$ such that $\norm {A^{-1} y}_V \le M \norm y_U$ for all $y \in U$.
Let $x \in V$ and set $y = A x \in U$.
Then: 
:$\norm {A^{-1} y}_V \le M \norm y_U$
We have: 
and:
:$\norm y_U = \norm {A x}_U$
giving:
:$\norm x_V \le M \norm {A x}_U$
So, we have: 
:$\norm {A x}_U \ge \dfrac 1 M \norm x_V$
for all $x \in V$.
Then, setting $c = 1/M$, we have that: 
:there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.
=== Sufficient Condition ===
Suppose that: 
:there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.
Let $y \in U$ and let $x = A^{-1} y \in V$.
Then:
:$\norm {A x}_U \ge c \norm x_V$
We have: 
and:
:$\norm x_V = \norm {A^{-1} y}_V$ 
so:
:$\norm y_U \ge c \norm {A^{-1} y}_V$
So, we have: 
:$\norm {A^{-1} y}_V \le \dfrac 1 c \norm y_U$
for all $y \in U$. 
So:
:$A^{-1}$ is bounded.
Category:Bounded Linear Transformations
Characterization of Join Irreducible Element

== Theorem ==
Let $\struct{S, \vee, \preceq}$ be a join semilattice.
Let $z \in S$.
Then:
:$z$ is join-irreducible
:$\forall x, y \in S : \leftparen{x \prec z}$ and $\rightparen{y \prec z} \implies \paren{x \vee y \prec z}$
where $x \prec z$ denotes that $x \preceq z$ and $x \neq z$. 
== Proof ==
By Dual Pairs (Order Theory):
* join semilattice is dual to meet semilattice.
* join irreducible element is dual to meet irreducible element.
* join is dual to meet.
* succeeds is dual to precedes.
Thus the theorem holds by the duality principle applied to Characterization of Meet Irreducible Element.
Category:Join Irreducible Elements
Characterization of Left Null Space

== Definition ==
Let $\mathbf A_{m \times n}$ be a matrix in the matrix space $\map {\MM_{m, n} } \R$.
Let $\map {\operatorname {N^\gets} } {\mathbf A}$ be used to denote the left null space of $\mathbf A$.
Then:
:$\map {\operatorname {N^\gets} } {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf x^\intercal \mathbf A = \mathbf 0^\intercal}$
where $\mathbf X^\intercal$ is the transpose of $\mathbf X$.
== Proof ==
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.
We have that $\mathbf A^\intercal \mathbf x = \mathbf 0$ is equivalent to $\mathbf x^\intercal \mathbf A = \mathbf 0^\intercal$.
This implies that $\mathbf x \in \map {\operatorname N} {\mathbf A^\intercal} \iff \mathbf x^\intercal \mathbf A = \mathbf 0^\intercal$.
Recall that:
:$\mathbf x \in \map {\operatorname N} {\mathbf A^\intercal} \iff \mathbf x \in \map {\operatorname {N^\gets} } {\mathbf A}$
Hence the result, by definition of set equality.
== Sources ==
* 
Category:Linear Algebra
Category:Null Spaces