avewright/proofwiki-math huggingface dataset:
Theorem: Existence of Compact Space which Satisfies No Separation Axioms

There exists at least one example of a compact space for which none of the Tychonoff separation axioms are satisfied.

Proof:
Let $T_S = \struct {S, \tau_S}$ be a finite complement topology on an infinite set $S$.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete space on an arbitrary doubleton $A = \set {a, b}$.
Let $T = T_S \times D$ be the double pointed topology on $T_S$.
From Double Pointed Finite Complement Topology is Compact, $T$ is compact.
From Double Pointed Finite Complement Topology fulfils no Separation Axioms, $T$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.
{{qed}}
Theorem: Everywhere Dense iff Interior of Complement is Empty

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subset S$.
Then $A$ is everywhere dense {{iff}}:
:$\paren {\relcomp S A}^\circ = \O$
where $A^\circ$ is the interior of $A$.

Proof:
By definition of everywhere dense, $A$ is everywhere dense {{iff}}:
:$A^- = S$
where $A^-$ is the closure of $A$.
That happens {{iff}}:
{{begin-eqn}}
{{eqn | l = \paren {\relcomp S A}^\circ
      | r = \relcomp S {A^-}
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \relcomp S S
}}
{{eqn | r = \O
      | c = Relative Complement with Self is Empty Set
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Denseness
Category:Set Interiors
c5ojyec4e826itgus8uvohb8qc6tea1
Theorem: Greatest Set is Unique

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
Then the greatest set of $\TT$, if it exists, must be unique.

Proof:
Let $A, B \in \TT$ both be greatest sets of $\TT$.
Since $A$ is the greatest set:
:$B \subseteq A$
Since $B$ is the greatest set:
:$A \subseteq B$
Hence, by definition of set equality:
:$A = B$
Therefore the greatest set of $\TT$ is unique.
{{qed}}
Category:Greatest Set
Category:Greatest Elements
Category:Set Theory
oih19nz16569xnzkh5zi342e9zh49a2
Theorem: Greatest Set may not Exist

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
The greatest set of $\TT$ may not exist.

Proof:
Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 1$ is not the greatest set of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 0$ is not the greatest set of $\TT$.
Therefore $\TT$ has no greatest set.
{{qed}}
Category:Greatest Set
Category:Greatest Elements
Category:Set Theory
fggue7zdasdfe63b6lndzc46y52uhwm
Theorem: Smallest Set is Unique

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
Then the smallest set of $\TT$, if it exists, must be unique.

Proof:
Let $A, B \in \TT$ both be smallest sets of $\TT$.
Since $A$ is the smallest set:
:$A \subseteq B$
Since $B$ is the smallest set:
:$B \subseteq A$
Hence, by definition of set equality:
:$A = B$
Therefore the smallest set of $\TT$ is unique.
{{qed}}
Category:Set Theory
Category:Smallest Elements
d0d9b0rrqd44jwrzu9n6a5t42ww69zm
Theorem: Smallest Set may not Exist

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
The smallest set of $\TT$ may not exist.

Proof:
Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 0$ is not the smallest set of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 1$ is not the smallest set of $\TT$.
Therefore $\TT$ has no smallest set.
{{qed}}
Category:Set Theory
Category:Smallest Elements
flsg5gshlaa8hw42pup33zwg70g77y2
Theorem: Mapping is Surjection iff Direct Image Mapping is Surjection

Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is a surjection.
{{iff}}
:$f: S \to T$ is also a surjection.

Proof:
=== Necessary Condition ===
Follows from Direct Image Mapping of Surjection is Surjection.
{{qed|lemma}}
Theorem: Cauchy Sequence in Metric Space is not necessarily Convergent

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a Cauchy sequence in $M$.
Then it is not necessarily the case that $M$ is a convergent sequence in $M$.

Proof:
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $M = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $\sequence {x_n}$ be a sequence in $A$ that converges to the limit $l \in A$.
From Integer Reciprocal Space contains Cauchy Sequence with no Limit, $\sequence {x_n}$ is a Cauchy sequence in $M$ which does not converge in $M$.
{{qed}}
Theorem: Nested Sequences in Complete Metric Space not Tending to Zero may be Disjoint

Let $M = \struct {A, d}$ be a complete metric space.
Let $\family {S_k}_{k \mathop \in \N}$ be a nested sequence of closed balls in $M$.
Let the radii of $\family {S_k}_{k \mathop \in \N}$ be convergent in $M$, but not to zero.
Then it is not necessarily the case that their intersection $\ds \bigcap S_k$ is non-empty.

Proof:
Let $M = \struct {A, d}$ be Sierpiński's metric space.
{{Recall|Sierpiński's Metric Space|Sierpiński's metric space}}
{{:Definition:Sierpiński's Metric Space}}
From Sierpiński's Metric Space is Complete, $M$ is a complete metric space.
Let $S_k = \set {y \in A: \map d {y, x_k} \le 1 + \dfrac 1 {2 n} }$.
From Nested Sequence of Closed Balls in Sierpiński's Metric Space with Empty Intersection:
:$\ds \bigcap S_k = \O$
{{qed}}
Theorem: Union of Interiors is Subset of Interior of Union

Let $T$ be a topological space.
Let $\H$ be a set of subsets of $T$.
That is, let $\H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then the union of the interiors of the elements of $\H$ is a subset of the interior of the union of $\H$:
:$\ds \bigcup_{H \mathop \in \H} H^\circ \subseteq \paren {\bigcup_{H \mathop \in \H} H}^\circ $

Proof:
Let $\mathbb U$ be the set of all open subsets of $\bigcup \H$.
Then by definition of interior:
:$\ds \paren {\bigcup_{H \mathop \in \H} H}^\circ = \paren {\bigcup \H}^\circ = \bigcup \mathbb U$
As $\mathbb U$ contains all open subsets of $\bigcup \H$, and $H^\circ$ is open for any $H \in \H$:
:$\ds \set {H^\circ : H \in \H} \subseteq \mathbb U$
Thus:
:$\ds \bigcup_{H \mathop \in \H} H^\circ = \bigcup \set {H^\circ : H \in \H} \subseteq \bigcup \mathbb U$
{{qed}}
Theorem: Closed Ball is Closed/Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $x \in X$.
Let $\epsilon \in \R_{> 0}$.
Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon^-} x$ is a closed set of $M$.

Proof:
We show that the complement $X \setminus \map {B_\epsilon^-} x$ is open in $M$.
Let $y \in X \setminus \map {B_\epsilon^-} x$.
Then by definition of closed ball:
:$\norm {x - y} > \epsilon$
Put:
:$\delta := \norm {x - y} - \epsilon > 0$
Then:
:$\norm {x - y} - \delta = \epsilon$
Let $z \in \map {B_\delta} y$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {x - z}
      | o = \ge 
      | r = \norm {x - y} - \norm {y - z}
      | c = Reverse Triangle Inequality
}}
{{eqn | o = >
      | r = \norm {x - y} - \delta
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
and so:
:$z \notin \map {B_\epsilon^-} x$
Then:
:$\map {B_\delta} y \subseteq X \setminus \map {B_\epsilon^-} x$
so $X \setminus \map {B_\epsilon^-} x$ is open in $M$.
Hence, by definition of closed set:
:$\map {B_\epsilon^-} x$ is closed in $M$.
{{qed}}
Theorem: Unit Sphere is Closed/Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\Bbb S := \set {x \in X : \norm x = 1}$ be a unit sphere in $M$.
Then $\Bbb S$ is closed in $M$.

Proof:
Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an open ball.
Let $\map { {B_1}^-} 0 = \set {x \in X : \norm x \le 1}$ be a closed ball.
Then:
:$\ds X = \Bbb S \bigcup \relcomp X {\Bbb S}$
where
:$\ds \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map { {B_1}^-} 0}$
is the relative complement of $\Bbb S$ in $X$.
We have that Closed Ball is Closed in Normed Vector Space.
By definition, $X \setminus \map { {B_1}^-} 0$ is open.
Furthermore, Open Ball is Open Set in Normed Vector Space.
By Union of Open Sets of Normed Vector Space is Open, $\relcomp X {\Bbb S}$ is open.
By definition, $\Bbb S$ is closed.
{{qed}}
Theorem: Limit Points of Open Set of Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a limit point of $U$.

Proof:
Every open set of $T = \struct {S, \tau_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a set, $x$ is a limit point of $U$.
{{qed}}
Theorem: Limit Points of Open Set of Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a limit point of $U$.

Proof:
Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Limit Points in Subset of Closed Extension Space
{{qed}}
Theorem: Limit Points in Closed Extension Space/Subset

Let $U \subseteq S^*_p$ such that $p \in U$.
Let $x \in S$.
Then $x$ is a limit point of $U$.

Proof:
Every open set of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau^*_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a set, $x$ is a limit point of $U$.
{{qed}}
Theorem: Convergent Sequence in Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $\sequence {a_i}$ be a convergent sequence in $T$.
Except for a finite number of indices, the terms of $\sequence {a_i}$ for which $a_i \ne p$ are all equal.

Proof:
{{Recall|Convergent Sequence (Topology)|convergent sequence}}
{{:Definition:Convergent Sequence/Topology/Definition 2}}
Let $\sequence {a_i}$ be a convergent sequence in $T$ whose limit is $\alpha$.
Then by definition every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {a_i}$.
This includes the open set $\set {\alpha, p}$.
Thus $\sequence {a_i}$ is such that, except for a finite number of terms, all are equal either to $\alpha$ or $p$.
Thus $\sequence {a_i}$ is such that, except for a finite number of terms, all such that $a_i \ne p$, are equal to $\alpha$.
The result follows.
{{qed}}
Theorem: Accumulation Points for Sequence in Particular Point Space

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $\sequence {a_i}$ be an infinite sequence in $T$.
Let $\beta$ be an accumulation point of $\sequence {a_i}$.
Then $\beta$ is such that an infinite number of terms of $\sequence {a_i}$ are equal either to $\beta$ or to $p$.

Proof:
{{Recall|Accumulation Point of Sequence|accumulation point}}
{{:Definition:Accumulation Point of Sequence}}
Let $\beta$ be an accumulation point of $\sequence {a_i}$.
Then by definition:
:all open sets of $T$ which contain $\beta$ also contain an infinite number of terms of $\sequence {a_i}$.
This condition applies to the open set $\set {\beta, p}$.
So $\set {\beta, p}$ contains an infinite number of terms of $\sequence {a_i}$.
The result follows.
{{qed}}
Theorem: Absolute Value of Absolutely Continuous Function is Absolutely Continuous

Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an absolutely continuous function.
Then $\size f$ is absolutely continuous.

Proof:
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
By the Reverse Triangle Inequality, we have: 
:$\size {\map f {b_i} - \map f {a_i} } \ge \size {\size {\map f {b_i} } - \size {\map f {a_i} } }$
Therefore: 
:$\ds \sum_{i \mathop = 1}^n \size {\size {\map f {b_i} } - \size {\map f {a_i} } } \le \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary:
:$\size f$ is absolutely continuous.
{{qed}}
Category:Absolute Value Function
Category:Absolutely Continuous Real Functions
13ukm8resgulwehz7r33cylqi92nvru
Theorem: Constant Real Function is Absolutely Continuous

Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an constant real function.
Then $f$ is absolutely continuous.

Proof:
Let $\delta, \epsilon$ be positive real numbers.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of disjoint closed real intervals with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $f$ is constant, for all $i \in \set {1, 2, \ldots, n}$ we have:
:$\size {\map f {b_i} - \map f {a_i} } = 0$
so:
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } = 0 < \epsilon$
Since $\epsilon$ was arbitrary: 
:$f$ is absolutely continuous.
{{qed}}
Category:Constant Mappings
Category:Absolutely Continuous Real Functions
ehlo9zemjk467fzmnthc84sk71iqfy4
Theorem: Sum of Absolutely Continuous Functions is Absolutely Continuous

Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous real functions.
Then $f + g$ is absolutely continuous.

Proof:
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
Similarly, since $g$ is absolutely continuous, there exists real $\delta_2 > 0$ such that whenever: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
Let: 
:$\delta = \map \min {\delta_1, \delta_2}$
Then, for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
and:
:$\ds \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
We then have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {b_i} - \map {\paren {f + g} } {a_i} }
      | r = \sum_{i \mathop = 1}^n \size {\paren {\map f {b_i} - \map f {a_i} } + \paren {\map g {b_i} - \map g {a_i} } }
}}
{{eqn | o = \le
      | r = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | o = <
      | r = \frac \epsilon 2 + \frac \epsilon 2
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
whenever: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary:
:$f + g$ is absolutely continuous.
{{qed}}
Category:Absolutely Continuous Real Functions
gk12g8hky7t78no2uwrikv07oyvddp9
Theorem: Multiple of Absolutely Continuous Function is Absolutely Continuous

Let $k$ be a real number.
Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an absolutely continuous real function.
Then $k f$ is absolutely continuous.

Proof:
Note that if $k = 0$, then $k f$ is constant. 
Hence, by Constant Real Function is Absolutely Continuous: 
:$k f$ is absolutely continuous if $k = 0$.
Take now $k \ne 0$.
Let $\epsilon$ be a positive real number.
Since $f$ is  absolutely continuous, there exists real $\delta > 0$ such that for every set of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {\size k}$
Then:
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {k f} } {b_i} - \map {\paren {k f} } {a_i} }
	| r = \size k \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} }
}}
{{eqn	| o = <
	| r = \size k \times \frac \epsilon {\size k}
}}
{{eqn	| r = \epsilon
}}
{{end-eqn}}
whenever:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary: 
:$k f$ is absolutely continuous if $k \ne 0$.
Therefore: 
:$k f$ is absolutely continuous for all $k \in \R$.
{{qed}}
Category:Absolutely Continuous Real Functions
rv4n85ls8km0ilpbolvjhpbh8ah5fcf
Theorem: Norms on Finite-Dimensional Real Vector Space are Equivalent

Norms on finite-dimensional real vector space are equivalent.

Proof:
We will prove that all norms are equivalent to $\norm {\, \cdot \,}_2$.
By definition, two norms are equivalent on $\R^d$ {{iff}}:
:$ \exists m, M \in \R_{> 0} : \forall \mathbf x \in \R^d : m \norm {\mathbf x}_a \le \norm {\mathbf x}_b \le M \norm {\mathbf x}_a$
Theorem: Linear Combination of Absolutely Continuous Function is Absolutely Continuous

Let $\alpha, \beta$ be real numbers.
Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous real functions.
Then $\alpha f + \beta g$ is absolutely continuous.

Proof:
From Multiple of Absolutely Continuous Function is Absolutely Continuous, we have: 
:$\alpha f$ and $\beta g$ are absolutely continuous.
We therefore have, by Sum of Absolutely Continuous Functions is Absolutely Continuous:
:$\alpha f + \beta g$ is absolutely continuous.
{{qed}}
Category:Absolutely Continuous Real Functions
18e2jgiydfjndn920s46gegk2q4e0l3
Theorem: Integral of Distribution Function

Let $\struct {X, \Sigma, \mu}$ be a measure space and $f$ be a $\mu$-measurable function. Let $p > 0, r \geq 0$.
For $\lambda > 0$, let $E_\lambda = \set {x \in X: \size {\map f x} > \lambda}$, so that $\map m \lambda = \map \mu {E_\lambda}$ is the distribution function of $f$.
Then:
:$\ds \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda = \int_X \size f^{p + r} \rd \mu$
and in particular:
:$\ds \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda = \int_X \size f^p \rd \mu$

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda
      | r = \int_0^\infty \int_{E_\lambda} p \lambda^{p - 1} \size f^r \rd \mu \rd \lambda
}}
{{eqn | r = \int_X \int_0^{\size {\map f x} } p \lambda^{p - 1} \size f^r \rd \lambda \rd \mu
      | c = by Tonelli's Theorem
}}
{{eqn | r = \int_X \size f^r \int_0^{\size {\map f x} } p \lambda^{p - 1} \rd \lambda \rd \mu
}}
{{eqn | r = \int_X \size f^r \size f^p \rd \mu
      | c = by Integral of Power
}}
{{eqn | r = \int_X \size f^{p + r} \rd \mu
}}
{{end-eqn}}
We have that for any measurable $A \in \Sigma$:
:$\map \mu A = \ds \int_A 1 \rd \mu$
Therefore, for $\lambda > 0$:
:$\map \mu {E_\lambda} = \ds \int_{E_\lambda} 1 \rd \mu$
which can also be written:
:$\map \mu {E_\lambda} = \ds \int_{E_\lambda} \size f^0 \rd \mu$
Therefore, taking $r = 0$ in the above, we obtain:
{{begin-eqn}}
{{eqn | l = \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda
      | r = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^0 \rd \mu \rd \lambda
      | c = 
}}
{{eqn | r = \int_X \size f^p \rd \mu
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Measure Theory
nofifbl2i67qb3oufufm1d66v3o7vet
Theorem: Limit Points in Open Extension Space/Subset

Let $U \subseteq S^*_p$.
Then $p$ is a limit point of $U$.

Proof:
Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition.
So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$.
So by definition of the limit point of a set, $p$ is a limit point of $U$.
{{qed}}
Category:Limit Points in Open Extension Space
llb70cgx6g8g14m8cjv8omro4q7e0qo
Theorem: Differentiable Function with Bounded Derivative is Absolutely Continuous

Let $a, b$ be real numbers with $a < b$.
Let $f: \closedint a b \to \R$ be a continuous function.
Let $f$ be differentiable on $\openint a b$, with bounded derivative.
Then $f$ is absolutely continuous.

Proof:
Since the derivative of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that: 
:$\size {\map {f'} x} \le M$
for all $x \in \openint a b$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of disjoint closed real intervals.
Note that for each $i \in \set {1, 2, \ldots, n}$:
:$f$ is continuous on $\closedint {a_i} {b_i}$ and differentiable on $\openint {a_i} {b_i}$.
So, by the Mean Value Theorem, for each $i$ there exists some $\xi_i \in \openint {a_i} {b_i}$ such that:
:$\map {f'} {\xi_i} = \dfrac {\map f {b_i} - \map f {a_i} } {b_i - a_i}$
We then have:
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} }
	| r = \sum_{i \mathop = 1}^n \size {\map {f'} {\xi_i} } \paren {b_i - a_i}
}}
{{eqn	| o = \le
	| r = M \sum_{i \mathop = 1}^n \paren {b_i - a_i}
	| c = since $\xi_i \in \openint a b$, we have $\size {\map {f'} {\xi_i} } \le M$
}}
{{end-eqn}}
Let $\epsilon$ be a positive real number. 
Then for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \epsilon M$
we have:
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Since $\epsilon$ was arbitrary:
:$f$ is absolutely continuous.
{{qed}}
Category:Differentiable Real Functions
Category:Absolutely Continuous Real Functions
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Theorem: Lipschitz Continuous Real Function is Absolutely Continuous

Let $I \subseteq \R$ be a real interval.
Let $f$ be Lipschitz continuous on $I$.
Then $f$ is absolutely continuous on $I$.

Proof:
By the definition of Lipschitz continuity, there exists $K \in \R$ such that: 
:$\size {\map f x - \map f y} \le K \size {x - y}$
for all $x, y \in I$.
If $K = 0$, then: 
:$\size {\map f x - \map f y} = 0$
for all $x, y \in I$. 
In this case, $f$ is constant.
Hence, by Constant Real Function is Absolutely Continuous:
:$f$ is absolutely continuous in the case $K = 0$.
Now, take $K > 0$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of disjoint closed real intervals.
We have: 
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} }
      | o = \le
      | r = \sum_{i \mathop = 1}^n K \size {b_i - a_i}
}}
{{eqn | r = K \sum_{i \mathop = 1}^n \paren {b_i - a_i}
      | c = since $b_i > a_i$
}}
{{end-eqn}}
Let $\epsilon$ be a positive real number. 
Then for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \epsilon K$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Since $\epsilon$ was arbitrary:
:$f$ is absolutely continuous in the case $K > 0$.
Hence the result.
{{qed}}
Theorem: Either-Or Topology is Compact

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a compact space.

Proof:
Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$.
So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets.
So $\CC$ has a subcover containing at most three sets.
Hence $T$ is a compact space by definition.
{{qed}}
Theorem: Compact Space is Lindelöf

Every '''compact space''' is also a '''Lindelöf space'''.

Proof:
We have:
:Compact Space is $\sigma$-Compact Space
:$\sigma$-Compact Space is Lindelöf Space
Hence the result.
{{qed}}
Category:Compact Topological Spaces
Category:Lindelöf Spaces
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