avewright/proofwiki-math huggingface dataset:
Theorem: Kuratowski's Closure-Complement Problem/Complement

The complement of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A'
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 1
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 2 3
      | c = {{Defof|Half-Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointl 3 4
      | c = ... adjacent to Half-Open Real Interval
}}
{{eqn | o =
      | ro= \cup
      | r = \paren {\paren {\R \setminus \Q} \cap \closedint 4 5}
      | c = Irrational Numbers from $4$ to $5$
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 5 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
For ease of analysis, let:
:$A_1 := \openint 0 1$
:$A_2 := \openint 1 2$
:$A_3 := \set 3$
:$A_4 := \Q \cap \openint 4 5$
Thus:
:$\ds A = \bigcup_{i \mathop = 1}^4 A_i$
By De Morgan's Laws:
:$\ds A' := \R \setminus A = \bigcap_{i \mathop = 1}^4 \paren {\R \setminus A_i}$
{{begin-eqn}}
{{eqn | l = \R \setminus A_1
      | r = \hointl \gets 0 \cup \hointr 1 \to
      | c = 
}}
{{eqn | l = \R \setminus A_2
      | r = \hointl \gets 1 \cup \hointr 2 \to
      | c = 
}}
{{eqn | l = \R \setminus A_3
      | r = \openint \gets 3 \cup \openint 3 \to
      | c = 
}}
{{eqn | l = \R \setminus A_4
      | r = \hointl \gets 4 \cup \paren {\R \setminus \Q} \cup \hointr 5 \to
      | c = De Morgan's Laws
}}
{{end-eqn}}
from which the result follows by inspection.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
5z8k11ykiy79ge4vz700jxkoczc7xyl
Theorem: Kuratowski's Closure-Complement Problem/Interior

The interior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^\circ
      | r = \openint 0 1 \cup \openint 1 2
      | c = Union of Adjacent Open Intervals
}}
{{end-eqn}}
:500px

Proof:
From Interior equals Complement of Closure of Complement:
:$A^\circ = A^{\prime \, - \, \prime}$
From Kuratowski's Closure-Complement Problem: Closure of Complement:
{{begin-eqn}}
{{eqn | l = A^{\prime \, -}
      | r = \hointl \gets 0
      | c = Unbounded Closed Real Interval
}}
{{eqn | o =
      | ro= \cup
      | r = \set 1
      | c = Singleton
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 2 \to
      | c = Unbounded Closed Real Interval
}}
{{end-eqn}}
It can be determined by inspection that:
:$A^{\prime \, - \, \prime} = \openint 0 1 \cup \openint 1 2$
Hence the result.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Examples of Set Interiors
qps0uotr3hdtewav5mch09naluoplo7
Theorem: Kuratowski's Closure-Complement Problem/Closure

The closure of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^-
      | r = \closedint 0 2
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 3
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 4 5
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Closure of Union of Adjacent Open Intervals:
:$\paren {\openint 0 1 \cup \openint 1 2}^- = \closedint 0 2$
From Real Number is Closed in Real Number Line:
:$\set 3$ is closed in $\R$
From Set is Closed iff Equals Topological Closure:
:$\set 3^- = \set 3$
From Closure of Rational Interval is Closed Real Interval:
:$\paren {\Q \cap \openint 4 5 }^- = \closedint 4 5$
The result follows from Closure of Finite Union equals Union of Closures.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Set Closures
l9svs3wen8l6wzv0kj3p5en5vi0iw2t
Theorem: Kuratowski's Closure-Complement Problem/Exterior

The exterior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^e
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 3 \cup \openint 3 4
      | c = {{Defof|Union of Adjacent Open Intervals}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 5 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
By definition, the exterior of $A$ in $\R$ can be defined either as:
:the complement of the closure of $A$ in $\R$: $A^{- \, \prime}$
or as:
:the interior of the complement of $A$ in $\R$: $A^{\prime \, \circ}$
From Kuratowski's Closure-Complement Problem: Closure:
{{begin-eqn}}
{{eqn | l = A^-
      | r = \closedint 0 2
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 3
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 4 5
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
It follows by inspection that:
{{begin-eqn}}
{{eqn | l = A^e = A^{- \, \prime}
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 3 \cup \openint 3 4
      | c = {{Defof|Union of Adjacent Open Intervals}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 5 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Set Exteriors
6l2oj9k014ngv3qrsqeg832zewp7oal
Theorem: Kuratowski's Closure-Complement Problem/Closure of Complement

The closure of the complement of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\prime \, -}
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 1
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 2 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Complement:
{{begin-eqn}}
{{eqn | l = A'
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 1
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 2 3
      | c = {{Defof|Half-Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointl 3 4
      | c = adjacent to Half-Open Real Interval
}}
{{eqn | o =
      | ro= \cup
      | r = \paren {\R \setminus \Q \cap \closedint 4 5}
      | c = Irrational Numbers from $4$ to $5$
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 5 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
From Real Number is Closed in Real Number Line:
:$\set 1$ is closed in $\R$
and:
:$\hointl \gets 0$ is closed in $\R$
and:
:$\hointr 5 \to$ is closed in $\R$
Then from Set is Closed iff Equals Topological Closure:
:$\set 1^- = \set 1$
:$\hointl \gets 0^- = \hointl \gets 0$
:$\hointr 5 \to^- = \hointr 5 \to$
From Closure of Half-Open Real Interval is Closed Real Interval:
:$\hointr 2 3 = \closedint 2 3$
and:
:$\hointl 3 4 = \closedint 3 4$
From Closure of Irrational Interval is Closed Real Interval:
:$\paren {\R \setminus \Q \cap \closedint 4 5}^- = \closedint 4 5$
From Closure of Finite Union equals Union of Closures:
{{begin-eqn}}
{{eqn | l = A^{\prime \, -}
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 1
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 2 3
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 3 4
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 4 5
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 5 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
15gmqwzwejx3ktmxj34fs5pb5j9cl89
Theorem: Kuratowski's Closure-Complement Problem/Closure of Interior

The closure of the interior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\circ \, -}
      | r = \closedint 0 2
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Interior:
:$A^\circ = \openint 0 1 \cup \openint 1 2$
From Closure of Union of Adjacent Open Intervals:
:$A^{\circ \, -} = \closedint 0 2$
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Examples of Set Interiors
Category:Set Closures
4gwoy8l3ag4cni02ik7jxpsazra2kzv
Theorem: Kuratowski's Closure-Complement Problem/Interior of Closure

The interior of the closure of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ}
      | r = \openint 0 2
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 4 5
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Closure:
{{begin-eqn}}
{{eqn | l = A^-
      | r = \closedint 0 2
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 3
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 4 5
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint 0 2^\circ = \openint 0 2$
and:
:$\closedint 4 5^\circ = \openint 4 5$
From Interior of Singleton in Real Number Line is Empty:
:$\set 3^\circ = \O$
Hence the interior of the closure of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ}
      | r = \openint 0 2
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 4 5
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Examples of Set Interiors
Category:Set Closures
e19mnu8vw292wekb7r318pzh5i4ssfq
Theorem: Kuratowski's Closure-Complement Problem/Interior of Closure of Interior

The interior of the closure of the interior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\circ \, - \, \circ}
      | r = \openint 0 2
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Closure of Interior:
{{begin-eqn}}
{{eqn | l = A^{\circ \, -}
      | r = \closedint 0 2
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint 0 2^\circ = \openint 0 2$
Hence the result.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Examples of Set Interiors
Category:Set Closures
a28srut15eovh1jqy4g4tmlmwgqliw1
Theorem: Kuratowski's Closure-Complement Problem/Interior of Complement of Interior

The interior of the complement of the interior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\circ \, \prime \, \circ}
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Complement of Interior equals Closure of Complement:
:$A^{\circ \, \prime} = A^{\prime \, -}$
From Kuratowski's Closure-Complement Problem: Closure of Complement:
{{begin-eqn}}
{{eqn | l = A^{\prime \, -}
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \set 1
      | c = {{Defof|Singleton}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 2 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
From Interior of Closed Real Interval is Open Real Interval:
:$\hointl \gets 0^\circ = \openint \gets 0$
and:
:$\hointr 2 \to^\circ = \openint 2 \to$
From Interior of Singleton in Real Number Line is Empty:
:$\set 1^\circ = \O$
Hence the interior of the complement of the interior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\circ \, \prime \, \circ}
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Set Interiors
Category:Set Complement
23xx3v9f0ods9pf50gk7shf1xj0kyf1
Theorem: Kuratowski's Closure-Complement Problem/Closure of Interior of Complement

The closure of the interior of the complement of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\prime \, \circ \, -}
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 2 4
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 5 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
By definition, the exterior is the interior of the complement.
Hence from Kuratowski's Closure-Complement Problem: Exterior:
{{begin-eqn}}
{{eqn | l = A^e
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 3 \cup \openint 3 4
      | c = {{Defof|Union of Adjacent Open Intervals}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 5 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
From Closure of Open Real Interval is Closed Real Interval:
:$\openint \gets 0^- = \hointl \gets 0$
and:
:$\openint 5 \to^- = \hointr 5 \to$
From Closure of Union of Adjacent Open Intervals:
:$\paren {\openint 2 3 \cup \openint 3 4}^- = \closedint 2 4$
The result follows from Closure of Finite Union equals Union of Closures.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Examples of Set Interiors
Category:Set Closures
Category:Set Complement
qpa4tdc306r34fpg92j0iodnltr6mo7
Theorem: Kuratowski's Closure-Complement Problem/Closure of Interior of Closure

The closure of the interior of the closure of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ \, -}
      | r = \closedint 0 2
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 4 5
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Interior of Closure:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ}
      | r = \openint 0 2
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 4 5
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
From Closure of Open Real Interval is Closed Real Interval:
:$\openint 0 2^- = \closedint 0 2$
and:
:$\openint 4 5^- = \closedint 4 5$
The result follows from Closure of Finite Union equals Union of Closures.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
Category:Examples of Set Interiors
Category:Set Closures
kw8oqespawq2p8fpss58pnsl93uv8je
Theorem: Kuratowski's Closure-Complement Problem/Interior of Complement of Interior of Closure

The interior of the complement of the interior of the closure of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ \, \prime \, \circ}
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 4
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 5 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Interior of Closure:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ}
      | r = \openint 0 2
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 4 5
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
By inspection:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ \, '}
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \closedint 2 4
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 5 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
From Interior of Closed Real Interval is Open Real Interval:
:$\hointl \gets 0^\circ = \openint \gets 0$
and:
:$\closedint 2 4^\circ = \openint 2 4$
and:
:$\hointr 5 \to^\circ = \openint 5 \to$
Hence the interior of the complement of the interior of the closure of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{- \, \circ \, \prime \, \circ}
      | r = \openint \gets 0
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 2 4
      | c = {{Defof|Open Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \openint 5 \to
      | c = {{Defof|Unbounded Open Real Interval}}
}}
{{end-eqn}}
{{qed}}
Category:Kuratowski's Closure-Complement Problem
hcqjudn11ubzosxyx1ydvxttptzknfj
Theorem: Kuratowski's Closure-Complement Problem/Complement of Interior of Closure of Interior

The complement of the interior of the closure of the interior of $A$ in $\R$ is given by:
{{begin-eqn}}
{{eqn | l = A^{\circ \, - \, \circ \, \prime}
      | r = \hointl \gets 0
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{eqn | o =
      | ro= \cup
      | r = \hointr 2 \to
      | c = {{Defof|Unbounded Closed Real Interval}}
}}
{{end-eqn}}
:500px

Proof:
From Kuratowski's Closure-Complement Problem: Interior of Closure of Interior:
{{begin-eqn}}
{{eqn | l = A^{\circ \, - \, \circ}
      | r = \openint 0 2
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
The result follows by inspection.
{{qed}}
Category:Kuratowski's Closure-Complement Problem
26ctbparebwc0llvkfjhoz4ersprnij
Theorem: Sum of Integrals on Adjacent Intervals for Integrable Functions

Let $f$ be a real function which is Darboux integrable on any closed interval $\mathbb I$.
Let $a, b, c \in \mathbb I$.
Then:
:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$

Proof:
This is an instance of Lower Sum of Refinement.
{{qed}}
Theorem: Sum of Integrals on Adjacent Intervals for Integrable Functions

Let $f$ be a real function which is Darboux integrable on any closed interval $\mathbb I$.
Let $a, b, c \in \mathbb I$.
Then:
:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$

Proof:
Let $P = \set {x_0, x_1, \ldots, x_n}$.
Suppose that:
:$c \in P$
Then:
:$Q = P$
We have:
:$\map L P \ge \map L P$
:$\leadsto \map L Q \ge \map L P$ as $Q = P$
This finishes the proof for this case.
The only other possibility for $c$ is:
:$x_{j-1} < c < x_j$
where $1 \le j \le n$.
Let $m_i$ be the infimum of $f$ on the interval $\closedint {x_{i - 1}} {x_i}$ for all $i \in \set {1, 2, \ldots, n}$.
We have:
{{begin-eqn}}
{{eqn | l = \map L P
      | r = \sum_{i \mathop = 1}^n m_i \paren {x_i - x_{i - 1} }
      | c = {{Defof|Lower Darboux Sum}}
}}
{{eqn | r = \sum_{i \mathop = 1}^{j - 1} m_i \paren {x_i - x_{i - 1} } + m_j \paren {x_j - x_{j - 1} }  + \sum_{i \mathop = j + 1}^n m_i \paren {x_i - x_{i - 1} }
}}
{{eqn | r = \sum_{i \mathop = 1}^{j - 1} m_i \paren {x_i - x_{i - 1} } + m_j \paren {x_j − c + c - x_{j - 1} } + \sum_{i \mathop = j + 1}^n m_i \paren {x_i - x_{i - 1} }
}}
{{eqn | r = \sum_{i \mathop = 1}^{j - 1} m_i \paren {x_i - x_{i - 1} } + m_j \paren {x_j − c} + m_j \paren {c - x_{j - 1} } + \sum_{i \mathop = j + 1}^n m_i \paren {x_i - x_{i - 1} }
}}
{{eqn | r = \sum_{i \mathop = 1}^{j - 1} m_i \paren {x_i - x_{i - 1} } + m_j \paren {c - x_{j - 1} } + m_j \paren {x_j − c} + \sum_{i \mathop = j + 1}^n m_i \paren {x_i - x_{i - 1} }
}}
{{end-eqn}}
Let $r$ be the infimum of $f$ on the interval $\closedint {x_{j - 1}} c$.
Let $s$ be the infimum of $f$ on the interval $\closedint c {x_j}$.
The interval $\closedint {x_{j - 1}} c$ is a subset of $\closedint {x_{j - 1}} {x_j}$.
Therefore, a lower bound of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound of $f$ on the interval $\closedint {x_{j - 1}} c$ as well.
The infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound for $f$ on this interval.
Accordingly, the infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound of $f$ on the interval $\closedint {x_{j - 1}} c$ as well.
Therefore, the infimum of $f$ on the interval $\closedint {x_{j - 1}} c$ is greater than or equal to the infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$.
In other words:
:$r \ge m_j$
A similar analysis of $\closedint c {x_j}$ gives:
:$s \ge m_j$
We find:
{{begin-eqn}}
{{eqn | l = \map L P
      | r = \sum_{i \mathop = 1}^{j - 1} m_i \paren {x_i - x_{i - 1} } + m_j \paren {c - x_{j - 1} } + m_j \paren {x_j − c} + \sum_{i \mathop = j + 1}^n m_i \paren {x_i - x_{i - 1} }
}}
{{eqn | o = \le
      | r = \sum_{i \mathop = 1}^{j - 1} m_i \paren {x_i - x_{i - 1} } + r \paren {c - x_{j - 1} } + s \paren {x_j − c} + \sum_{i \mathop = j + 1}^n m_i \paren {x_i - x_{i - 1} }
}}
{{eqn | r = \map L Q
      | c = {{Defof|Lower Darboux Sum}}
}}
{{end-eqn}}
This finishes the proof.
{{qed}}
Theorem: Closure of Irrational Numbers is Real Numbers

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\struct {\R \setminus \Q, \tau_d}$ be the irrational number space under the same topology.
Then:
:$\paren {\R \setminus \Q}^- = \R$
where $\paren {\R \setminus \Q}^-$ denotes the closure of $\R \setminus \Q$.

Proof:
From Irrationals are Everywhere Dense in Topological Space of Reals, $\R \setminus \Q$ is everywhere dense in $\R$.
It follows by definition of everywhere dense that $\paren {\R \setminus \Q}^- = \R$.
{{qed}}
Category:Irrational Number Space
Category:Real Number Line with Euclidean Topology
Category:Set Closures
2gq8yknqyjbxr1x3ixgvxqorlkv30f5
Theorem: Closure of Union of Adjacent Open Intervals

Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the two adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^- = \closedint a c$
where:
:$A^-$ is the closure of $A$.

Proof:
{{begin-eqn}}
{{eqn | l = A^-
      | r = \paren {\openint a b \cup \openint b c}^-
      | c = 
}}
{{eqn | r = \openint a b^- \cup \openint b c^-
      | c = Closure of Finite Union equals Union of Closures
}}
{{eqn | r = \closedint a b \cup \closedint b c
      | c = Closure of Open Ball in Metric Space
}}
{{eqn | r = \closedint a c
      | c = {{Defof|Closed Real Interval}}
}}
{{end-eqn}}
{{qed}}
Category:Union of Adjacent Open Intervals
Category:Examples of Set Closures
7gwi6avlpl5ldqm8lrgwf7tteg124oq
Theorem: Interior equals Complement of Closure of Complement

Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $H^\prime$ denote the complement of $H$ in $T$:
:$H^\prime = T \setminus H$
Then:
:$H^\circ = H^{\prime \, - \, \prime}$

Proof:
{{begin-eqn}}
{{eqn | l = H^{\circ \, \prime}
      | r = H^{\prime \, -}
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | ll= \leadsto
      | l = \paren {H^{\circ \, \prime} }^\prime
      | r = \paren {H^{\prime \, -} }^\prime
      | c = 
}}
{{eqn | ll= \leadsto
      | l = H^{\circ \, \prime \, \prime}
      | r = H^{\prime \, - \, \prime}
      | c = Composition of Mappings is Associative
}}
{{eqn | ll= \leadsto
      | l = H^\circ
      | r = H^{\prime \, - \, \prime}
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}