avewright/proofwiki-math huggingface dataset:
Theorem: Normed Vector Space is Open in Itself

Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.

Proof:
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in X$.
An open ball of $x$ in $M$ is by definition a subset of $X$.
Hence the result.
{{qed}}
Theorem: Normed Vector Space is Open in Itself

Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.

Proof:
By definition, a subset $S \subseteq X$ is open if:
:$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq S$
Let $S = X$.
{{AimForCont}} $X$ is not open.
By De Morgan's laws:
:$\exists x \in X : \forall \epsilon \in \R_{>0} : \map {B_\epsilon} x \cap \paren {X \setminus S} \ne \O$
Note that:
:$X \setminus S = X \setminus X = \O$.
By Intersection with Empty Set:
:$\map {B_\epsilon} x \cap \O = \O$
Hence:
:$\exists x \in X : \forall \epsilon \in \R_{>0} : \O \ne \O$.
Since Empty Set is Unique, we have a contradiction.
{{qed}}
Theorem: Normed Vector Space is Open in Itself/Proof 1

Let $M = \struct{X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.

Proof:
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in X$.
An open ball of $x$ in $M$ is by definition a subset of $X$.
Hence the result.
{{qed}}
Theorem: Normed Vector Space is Open in Itself/Proof 2

Let $M = \struct{X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.

Proof:
By definition, a subset $S \subseteq X$ is open if:
:$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq S$
Let $S = X$.
{{AimForCont}} $X$ is not open.
By De Morgan's laws:
:$\exists x \in X : \forall \epsilon \in \R_{>0} : \map {B_\epsilon} x \cap \paren {X \setminus S} \ne \O$
Note that:
:$X \setminus S = X \setminus X = \O$.
By Intersection with Empty Set:
:$\map {B_\epsilon} x \cap \O = \O$
Hence:
:$\exists x \in X : \forall \epsilon \in \R_{>0} : \O \ne \O$.
Since Empty Set is Unique, we have a contradiction.
{{qed}}
Theorem: Empty Set is Closed/Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then the empty set $\O$ is closed in $M$.

Proof:
From Normed Vector Space is Open in Itself, $X$ is open in $M$.
But:
:$\O = \relcomp X X$
where $\complement_X$ denotes the set complement relative to $X$.
The result follows by definition of closed set.
{{qed}}
Theorem: Empty Set is Open in Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the empty set $\O$ is an open set of $M$.

Proof:
By definition, an open set $S \subseteq X$ is one where every point inside it is an element of an open ball contained entirely within that set.
That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$.
As there are no elements in $\O$, the result follows vacuously.
{{qed}}
Theorem: Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent

Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$. 
Then $\sequence {f_n}$ is uniformly Cauchy on $S$ {{iff}} $\sequence {f_n}$ converges uniformly on $S$.

Proof:
=== Sufficient Condition ===
Let $\sequence {f_n}$ be uniformly Cauchy on $S$.
{{:Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Sufficient Condition}}
Theorem: Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Sufficient Condition

Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.
Let $\sequence {f_n}$ be uniformly Cauchy on $S$. 
Then $\sequence {f_n}$ is uniformly convergent on $S$.

Proof:
Let $\epsilon \in \R_{> 0}$ be arbitrary.
As $\sequence {f_n}$ is uniformly Cauchy on $S$, there exists $N \in \N$ such that: 
:$\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$
for all $n, m > N$ and $x \in S$.
Let $x \in S$ be fixed.
Then: 
:$\size {\map {f_n} x - \map {f_m} x} < \epsilon$
for all $n, m > N$. 
As $\epsilon$ was arbitrary, the sequence $\sequence {\map {f_n} x}$ is therefore Cauchy.
By Cauchy's Convergence Criterion on Real Numbers, it follows that $\sequence {\map {f_n} x}$ is convergent.
Define a function $f : S \to \R$ by: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for all $x \in S$. 
We aim to show that $f_n \to f$ uniformly.
For all $x \in S$, we have:
:$\ds \lim_{m \mathop \to \infty} \size {\map {f_n} x - \map {f_m} x} = \size {\map {f_n} x - \map f x}$
We established that for all $\epsilon \in \R_{> 0}$ we can find $N \in \N$ such that: 
:$\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$
for $x \in S$ and $n, m > N$. 
We therefore have: 
:$\size {\map {f_n} x - \map f x} \le \dfrac \epsilon 2 < \epsilon$
for all $x \in S$ and $n > N$.
So $\sequence {f_n}$ converges uniformly to $f$ on $S$.
{{qed|lemma}}
Theorem: Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Necessary Condition

Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.
Let $\sequence {f_n}$ be uniformly convergent on $S$.
Then $\sequence {f_n}$ is uniformly Cauchy on $S$.

Proof:
Let $\epsilon \in \R_{> 0}$ be arbitrary.
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that: 
:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 2$
for all $x \in S$ and $n > N$. 
Then if $x \in S$ and $n, m > N$, we have:
{{begin-eqn}}
{{eqn	| l = \size {\map {f_n} x - \map {f_m} x}
	| r = \size {\map {f_n} x - \map f x - \paren {\map {f_m} x - \map f x} }
}}
{{eqn	| r = \size {\map {f_n} x - \map f x} + \size {\map {f_m} x - \map f x}
	| c = Triangle Inequality for Real Numbers
}}
{{eqn	| o = <
	| r = \epsilon
}}
{{end-eqn}}
Since $\epsilon$ was arbitrary, $\sequence {f_n}$ is uniformly Cauchy on $S$.
{{qed}}
Theorem: Open Ball of Point Inside Open Ball/Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {X, \norm {\, \cdot \,} }$.
Let $y \in \map {B_\epsilon} x$.
Then:
:$\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$
That is, for every point in an open $\epsilon$-ball in a normed vector space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.

Proof:
Let $\delta = \epsilon - \norm {x - y}$.
From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.
Let $z \in \map {B_\delta} y$ be arbitrary.
We have:
{{begin-eqn}}
{{eqn | l = z
      | o = \in
      | r = \map {B_\delta} y
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = \norm {y - z}
      | o = <
      | r = \delta
      | c = {{Defof|Open Ball in Normed Vector Space}}
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = \norm {x - z}
      | o = \le
      | r = \norm {x - y} + \norm {y - z}
      | c = Triangle Inequality
}}
{{eqn | o = <
      | r = \norm {x - y} + \delta
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = z
      | o = \in
      | r = \map {B_\epsilon} x
      | c = {{Defof|Open Ball in Normed Vector Space}}
}}
{{end-eqn}}
That is:
:$z \in \map {B_\delta} y \implies z \in \map {B_\epsilon} x$
Hence $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
{{qed}}
Theorem: Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function

Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$ converging uniformly to $f: S \to \R$.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Then $f$ is continuous at $x$.

Proof:
Let $\epsilon \in \R_{> 0}$. 
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that: 
:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$
for all $x \in S$ and $n \ge N$.
Since $f_N$ is continuous at $x$, there exists some $\delta > 0$ such that: 
:for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$
Then for $y$ with $\size {x - y} < \delta$ we have: 
{{begin-eqn}}
{{eqn	| l = \size {\map f x - \map f y}
	| r = \size {\map f x - \map {f_N} x + \map {f_N} x - \map {f_N} y + \map {f_N} y - \map f y}
}}
{{eqn	| r = \size {\paren {\map f x - \map {f_N} x} + \paren {\map {f_N} x - \map {f_N} y} + \paren {\map {f_N} y - \map f y} }
}}
{{eqn	| o = \le
	| r = \size {\map f x - \map {f_N} x} + \size {\map {f_N} x - \map {f_N} y} + \size {\map {f_N} y - \map f y}
	| c = Triangle Inequality for Real Numbers
}}
{{eqn	| o = <
	| r = 3 \times \frac \epsilon 3
}}
{{eqn	| r = \epsilon
}}
{{end-eqn}}
Since $\epsilon$ was arbitrary, $f$ is continuous at $x$.
{{qed}}
Theorem: Open Ball is Open Set/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 an open $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon} x$ is an open set of $M$.

Proof:
Let $y \in \map {B_\epsilon} x$.
From Open Ball of Point Inside Open Ball in Normed Vector Space, there exists $\delta \in \R_{>0}$ such that $\map {B_\delta} y \subseteq \map {B_\epsilon} x$
The result follows from the definition of open set.
{{qed}}
Theorem: Finite Intersection of Open Sets of Normed Vector Space is Open

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $U_1, U_2, \ldots, U_n$ be open in $M$.
Then $\ds \bigcap_{i \mathop = 1}^n U_i$ is open in $M$.
That is, a finite intersection of open subsets is open.

Proof:
Let $\ds x \in \bigcap_{i \mathop = 1}^n U_i$.
For each $i \in \closedint 1 n$, we have $x \in U_i$.
Thus:
:$\exists \epsilon_i > 0: \map {B_{\epsilon_i}} x \subseteq U_i$
where $\map {B_{\epsilon_i}} x$ is the open $\epsilon_i$-ball of $x$.
Let $\ds \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$.
So:
:$\epsilon > 0$.
Let $y \in \map {B_\epsilon} x$.
Then $\norm {x - y} < \epsilon$
Hence:
:$\forall i \in \closedint 1 n : \norm {x - y} < \epsilon_i$
In other words:
:$\map {B_\epsilon} x \subseteq \map {B_{\epsilon_i}} x \subseteq U_i$
for all $i \in \closedint 1 n$.
So:
:$\ds \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$
The result follows.
{{qed}}
Theorem: Union of Open Sets of Normed Vector Space is Open

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
The union of a set of open sets of $M$ is open in $M$.

Proof:
Let $I$ be any indexing set.
Let $U_i$ be open in $M$ for all $i \in I$.
Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
Then $x \in U_k$ for some $k \in I$.
Since $U_k$ is open in $M$:
:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
The result follows.
{{qed}}
Theorem: Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions

Let $a, b \in \R$ with $a < b$. 
Let $\sequence {f_n}$ be a sequence of Riemann integrable real functions $\closedint a b \to \R$ converging uniformly to $f : \closedint a b \to \R$.
Then $f$ is integrable, and: 
:$\ds \int_a^b \map f x \rd x = \lim_{n \mathop \to \infty} \int_a^b \map {f_n} x \rd x$

Proof:
By Limit of Uniformly Convergent Sequence of Integrable Functions is Integrable, $f$ is integrable.
We have:
{{begin-eqn}}
{{eqn | l = \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x}
      | r = \size {\int_a^b \paren {\map f x - \map {f_n} x} \rd x}
}}
{{eqn | o = \le
      | r = \int_a^b \size {\map f x - \map {f_n} x} \rd x
      | c = Triangle Inequality for Integrals
}}
{{eqn | o = \le
      | r = \paren {b - a} \sup_{x \mathop \in \closedint a b} \size {\map f x - \map {f_n} x}
      | c = Darboux's Theorem
}}
{{end-eqn}}
Let $\epsilon \in \R_{> 0}$.
Since $f_n \to f$ uniformly, we can find $N \in \N$ such that for $n > N$ we have:
:$\ds \sup_{x \mathop \in \closedint a b} \size {\map f x - \map {f_n} x} < \frac \epsilon {b - a}$
by the definition of uniform convergence.
So, for $n > N$ we have: 
:$\ds \paren {b - a} \sup_{x \mathop \in \closedint a b} \size {\map f x - \map {f_n} x} < \epsilon$
and hence:
:$\ds \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x} < \epsilon$ 
Since $\epsilon$ was arbitrary it follows that: 
:$\ds \lim_{n \mathop \to \infty} \int_a^b \map {f_n} x \rd x = \int_a^b \map f x \rd x$
as required.
{{qed}}
Category:Integral Calculus
Category:Uniform Convergence
rea8661lv06qp51e40c26wnakgr45yt
Theorem: Events One of Which equals Union

Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.
Let $A$ and $B$ be such that:
:$A \cup B = A$
Then whenever $B$ occurs, it is always the case that $A$ occurs as well.

Proof:
From Union with Superset is Superset:
:$A \cup B = A \iff B \subseteq A$
Let $B$ occur.
Let $\omega$ be the outcome of $\EE$.
Let $\omega \in B$.
That is, by definition of occurrence of event, $B$ occurs.
Then by definition of subset:
:$\omega \in A$
Thus by definition of occurrence of event, $A$ occurs.
Hence the result.
{{qed}}
Theorem: De Morgan's Laws (Set Theory)/Set Complement/Complement of Union/Corollary

:$T_1 \cup T_2 = \overline {\overline T_1 \cap \overline T_2}$

Proof:
{{begin-eqn}}
{{eqn | l = T_1 \cup T_2
      | r = \overline {\overline {T_1 \cup T_2} }
      | c = Complement of Complement
}}
{{eqn | r = \overline {\overline T_1 \cap \overline T_2}
      | c = De Morgan's Laws: Complement of Union
}}
{{end-eqn}}
{{qed}}
Theorem: Intersection of Closed Sets is Closed/Normed Vector Space

Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then the intersection of an arbitrary number of closed sets of $M$ (either finitely or infinitely many) is itself closed.

Proof:
Let $I$ be an indexing set (either finite or infinite).
Let $\ds \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $M$ indexed by $I$.
By definition of closed set, each of $X \setminus V_i$ are by definition open in $M$.
From De Morgan's laws: Difference with Intersection:
:$\ds X \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$
We have that $\ds \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$ is the union of a indexed family of open sets of $M$ indexed by $I$.
By Union of Open Sets of Normed Vector Space is Open, $\ds \bigcup_{i \mathop \in I} \paren {X \setminus V_i} = X \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $M$.
Then by definition of closed set, $\ds \bigcap_{i \mathop \in I} V_i$ is closed in $M$.
{{qed}}
Theorem: Normed Vector Space is Closed in Itself

Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then $X$ is closed in $M$.

Proof:
From Empty Set is Open in Normed Vector Space, $\O$ is open in $M$.
But:
:$X = \relcomp X \O$
where $\complement_X$ denotes the set complement relative to $X$.
The result follows by definition of closed set.
{{qed}}
Theorem: Infinite Series of Functions is Uniformly Convergent iff Sequence of Partial Sums is Uniformly Cauchy

Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.
Then the infinite series: 
:$\ds \sum_{n \mathop = 1}^\infty f_n$
converges uniformly on $S$ {{iff}} for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that:
:$\ds \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$
for all $x \in S$ and $n > m > N$.

Proof:
Let $\sequence {s_n}$ be a sequence of real functions $S \to \R$ with: 
:$\ds \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_k} x$
for each $n \in \N$ and $x \in S$.
By definition of uniform convergence of an infinite series:
:$\ds \sum_{n \mathop = 1}^\infty f_n$ is uniformly convergent {{iff}} $\sequence {s_n}$ is uniformly convergent.
By Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent:
:$\sequence {s_n}$ is uniformly convergent {{iff}} it is uniformly Cauchy.
That is:
:for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that for all $m, n > N$ we have:
::$\size {\map {s_n} x - \map {s_m} x} < \varepsilon$ for all $x \in S$.
Note that if $n = m$: 
:$\size {\map {s_n} x - \map {s_m} x} = 0 < \varepsilon$
{{WLOG}}, we can therefore take $n > m$.
If $n > m$, then:
{{begin-eqn}}
{{eqn	| l = \size {\map {s_n} x - \map {s_m} x}
	| r = \size {\sum_{k \mathop = 1}^n \map {f_k} x - \sum_{k \mathop = 1}^m \map {f_k} x}
}}
{{eqn	| r = \size {\sum_{k \mathop = m + 1}^n \map {f_k} x}
}}
{{end-eqn}}
So $\sequence {s_n}$ is uniformly Cauchy {{iff}} for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that:
:$\ds \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$
for all $n > m > N$ and $x \in S$.
{{qed}}
Theorem: De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection/Corollary

:$T_1 \cap T_2 = \overline {\overline T_1 \cup \overline T_2}$

Proof:
{{begin-eqn}}
{{eqn | l = T_1 \cap T_2
      | r = \overline {\overline {T_1 \cap T_2} }
      | c = Complement of Complement
}}
{{eqn | r = \overline {\overline T_1 \cup \overline T_2}
      | c = De Morgan's Laws: Complement of Intersection
}}
{{end-eqn}}
{{qed}}
Theorem: LCM of 3 Integers in terms of GCDs of Pairs of those Integers/Lemma

Let $a, b, c \in \Z_{>0}$ be strictly positive integers.
Then:
:$\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = \gcd \set {a, b, c}$

Proof:
Let $\gcd \set {a, b, c} = d_1$.
From definition:
:$d_1 \divides a$, $d_1 \divides b$ and $d_1 \divides c$.
By Common Divisor Divides GCD:
:$d_1 \divides \gcd \set {a, b}$ and $d_1 \divides \gcd \set {a, c}$.
By Common Divisor Divides GCD again:
:$d_1 \divides \gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$.
On the other hand, let $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = d_2$.
From definition:
:$d_2 \divides \gcd \set {a, b}$ and $d_2 \divides \gcd \set {a, c}$.
From definition again:
:$d_2 \divides a$, $d_2 \divides b$ and $d_2 \divides c$.
Hence $d_2 \divides \gcd \set {a, b, c}$.
Since $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$ and $\gcd \set {a, b, c}$ divide each other, by Absolute Value of Integer is not less than Divisors they must be equal.
{{qed}}
Category:Greatest Common Divisor
Category:LCM of 3 Integers in terms of GCDs of Pairs of those Integers
1breqfkqm4awy95mtgjcqzl0krs2783
Theorem: Events One of Which equals Intersection

Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.
Let $A$ and $B$ be such that:
:$A \cap B = A$
Then whenever $A$ occurs, it is always the case that $B$ occurs as well.

Proof:
From Intersection with Subset is Subset:
:$A \cap B = A \iff A \subseteq B$
Let $A$ occur.
Let $\omega$ be the outcome of $\EE$.
Let $\omega \in A$.
That is, by definition of occurrence of event, $A$ occurs.
Then by definition of subset:
:$\omega \in B$
Thus by definition of occurrence of event, $B$ occurs.
Hence the result.
{{qed}}
Theorem: Union of Event with Complement is Certainty

Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A \in \Sigma$ be an events of $\EE$, so that $A \subseteq \Omega$.
Then the union of $A$ with its complementary event can be evaluated as:
:$A \cup \overline A = \Omega$
where $\overline A$ denotes the complementary event to $A$.
That is:
:$A \cup \overline A$ is a certainty
or:
:$\map \Pr {A \cup \overline A} = 1$

Proof:
By definition:
:$A \subseteq \Omega$
and:
:$\overline A = \relcomp \Omega A$
From Union with Relative Complement:
:$A \cup \overline A = \Omega$
We then have from Kolmogorov axiom $(2)$ that:
:$\map \Pr \Omega = 1$
The result follows by definition of certainty.
{{qed}}
{{LEM|Union with Relative Complement}}
Theorem: Equivalence of Definitions of Adherent Point of Set/Definition by Open Neighborhood iff Definition by Neighborhood

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
{{TFAE|def = Adherent Point of Set|view = adherent point of $H$}}

Proof:
=== Necessary Condition ===
Let every open neighborhood $U$ of $x$ satisfy:
:$H \cap U \ne \O$
Let $N$ be an arbitrary neighborhood of $x$.
By definition of a neighborhood:
:$\exists V \in \tau: x \in V \subseteq N$
From Set is Open iff Neighborhood of all its Points, $V$ is an open neighborhood of $x$.
Thus:
:$H \cap V \ne \O$
By the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
:$H \cap N \ne \O$
Because $N$ was arbitrary, it follows that every neighborhood $N$ of $x$ satisfies:
:$H \cap N \ne \O$
{{qed|lemma}}
=== Sufficient Condition ===
Let every neighborhood $N$ of $x$ satisfy:
:$H \cap N \ne \O$
By definition, every open neighborhood $U$ of $x$ is a neighborhood of $x$.
So every open neighborhood $U$ of $x$ satisfies:
:$H \cap U \ne \O$
Theorem: Compact Sets in Fortissimo Space

A subset of a Fortissimo space is compact {{iff}} it is finite.

Proof:
Let $T = \struct {S, \tau}$ be a Fortissimo space.
Theorem: Equivalence of Definitions of Boundary

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
{{TFAE|def = Boundary (Topology)|view = boundary of $H$}}

Proof:
=== Definition $1$ is equivalent to Definition $3$ ===
This is demonstrated in Boundary is Intersection of Closure with Closure of Complement.
{{qed|lemma}}
Theorem: Uniformly Convergent Series of Continuous Functions Converges to Continuous Function

Let $S \subseteq \R$.
Let $x \in S$.
Let $\sequence {f_n}$ be a sequence of real functions.
Let $f_n$ be continuous at $x$ for all $n \in \N$.
Let the infinite series: 
:$\ds \sum_{n \mathop = 1}^\infty f_n$
be uniformly convergent to a real function $f : S \to \R$.
Then $f$ is continuous at $x$.

Proof:
Let $\sequence {s_n}$ be sequence of real functions $S \to \R$ such that: 
:$\ds \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_n} x$
for each $n \in \N$ and $x \in S$.
By Sum Rule for Continuous Real Functions:
:$s_n$ is continuous at $x$ for all $n \in \N$.
Since additionally $s_n \to f$ uniformly, we have by Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function: 
:$f$ is continuous at $x$.
{{qed}}
Theorem: Uniformly Convergent Series of Continuous Functions Converges to Continuous Function/Corollary

Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions.
Let $f_n$ be continuous for all $n \in \N$.
Let the infinite series:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
be uniformly convergent to a real function $f : S \to \R$.
Then $f$ is continuous.

Proof:
Let $x \in S$.
Then $f_n$ is continuous at $x$ for all $n \in \N$.
Since:
:$\ds \sum_{n \mathop = 1}^\infty f_n$
converges uniformly to $f$, we have by Uniformly Convergent Series of Continuous Functions Converges to Continuous Function:
:$f$ is continuous at $x$.
As $x \in S$ was arbitrary, we have that: 
:$f$ is continuous.
{{qed}}
Category:Uniform Convergence
Category:Continuous Functions
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