avewright/proofwiki-math huggingface dataset:
Theorem: Continuous Image of Path-Connected Set is Path-Connected

Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be path-connected in $\struct {T_1, \tau_1}$.
Then $f \sqbrk S$ is path-connected in $\struct {T_2, \tau_2}$.

Proof:
Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.
Let $\mathbb I$ denote the closed unit interval:
:$\mathbb I = \closedint 0 1$
Let $p: \mathbb I \to S$ be a continuous mapping such that:
:$\map p 0 = s, \map p 1 = s'$
Such a $p$ exists since $S$ is path-connected in $\struct {T_1, \tau_1}$.
Now define $q: \mathbb I \to f \sqbrk S$ by:
:$\map q t := f \circ \map p t$
Composite of Continuous Mappings is Continuous shows that $q$ is a continuous mapping, and:
:$\map q 0 = \map f {\map p 0} = \map f s$
:$\map q 1 = \map f {\map p 1} = \map f {s'}$
Thus $\map f s$ and $\map f {s'}$ are connected by a path in $f \sqbrk S$.
Since these points were arbitrary in $f \sqbrk S$, the result follows.
{{qed}}
Theorem: Leibniz's Rule/One Variable

Let $f$ and $g$ be real functions defined on the open interval $I$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are $n$ times differentiable.
Then:
:$\ds \paren {\map f x \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \map {g^{\paren {n - k} } } x$
where $\paren n$ denotes the order of the derivative.

Proof:
Proof by induction:
Theorem: Leibniz's Rule/Real Valued Functions

Let $f, g : \R^n \to \R$ be real valued functions, $k$ times differentiable on some open set $U \subseteq \R^n$.
Let $\alpha = \tuple {\alpha_1, \ldots, \alpha_n}$ be a multiindex indexed by $\set {1, \ldots, n}$ with $\size \alpha \le k$.
For $i \in \set {1, \ldots, n}$, let $\partial_i$ denote the partial derivative:
:$\partial_i = \dfrac {\partial} {\partial {x_i} }$
Let $\partial^\alpha$ denote the partial differential operator:
:$\partial^\alpha = \partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n}$
Then as functions on $U$, we have:
:$\ds \map {\partial^\alpha} {f g} = \sum_{\beta \mathop \le \alpha} \binom \alpha \beta \paren {\partial^\beta f} \paren {\partial^{\alpha - \beta} g}$

Proof:
First, inserting the definitions, the statement of the theorem reads:
:$\ds \map {\partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} } {f g} = \sum_{\beta_1 = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_1} {\beta_1} \cdots \binom {\alpha_n} {\beta_n} \paren {\partial_1^{\beta_1} \cdots \partial_n^{\beta_n} f} \paren {\partial_1^{\alpha_1 - \beta_1} \cdots \partial_n^{\alpha_n - \beta_n} g}$
We prove this by induction over $n \ge 1$.
Theorem: Open Sets of Double Pointed Topology

Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Then $X \subseteq S \times D$ is open in $\tau$ {{iff}} for some $U \in \tau$:
:$X = U \times D$

Proof:
By definition, $\tau$ is the product topology on $X \times D$.
That is, $\tau$ has as a basis sets of the form:
:$U \times V$
with $U \in \tau$ and $V$ open in $D$.
Since $D$ is endowed with the indiscrete topology, either $V = \O$ or $V = D$.
In the former case, by Cartesian Product is Empty iff Factor is Empty, $U \times V = \O$ for all $U \in \tau$.
In particular, then, it follows that $\tau$ has the basis:
:$\BB = \set {U \times D: U \in \tau}$
since $U = \O \in \tau$, and so the basis elements with $V = \O$ are also accounted for.
It is therefore clear that all $U \times D$ with $U \in \tau$ are open in $\tau$.
That open sets of this form constitute all of $\tau$ amounts to showing that:
:$\tau = \BB$
that is, that $\BB$ is already a topology on $S \times D$.
{{finish|show this, maybe transfer/replace this result to/with general "Topology of Product with Indiscrete Space" or s.t. like that}}
{{qed}}
Category:Double Pointed Topologies
pj1sa27w606egyd5snjm2u8k9soo5cb
Theorem: Open Sets of Double Pointed Topology/Corollary

A subset $X \subseteq S \times D$ is closed in $\tau$ {{iff}} for some closed set $C$ of $\tau$:
:$X = C \times D$

Proof:
By definition, $X$ is closed {{iff}} its complement $\map \complement X$ is open.
By Open Sets of Double Pointed Topology, it follows by that for some $U \in \tau$:
:$\map \complement X = U \times D$
Then by Cartesian Product Distributes over Set Difference and Complement of Complement, we have that:
{{improve|rather, Cartesian Product Distributes over Complement (and for Relative Complement), but those don't exist}}
:$X = \paren {S \times D} \setminus \paren {U \times D} = \paren {S \setminus U} \times D$
Since $S \setminus U$ is a closed set of $\tau$, the result follows.
{{qed}}
Category:Double Pointed Topologies
54k40gvbl6yivcnruerescni5s08tly
Theorem: Open Set Less One Point is Open/Corollary

Let $S = \set {\alpha_1, \alpha_2, \ldots, \alpha_n} \subseteq U$ be a finite set of points in $U$.
Then $U \setminus S$ is open in $M$.

Proof:
Follows directly from Open Set Less One Point is Open and Finite Intersection of Open Sets of Metric Space is Open.
Let:
:$U_1 = U \setminus \set {\alpha_1}, U_2 = U \setminus \set {\alpha_2}, \ldots, U_n = U \setminus \set {\alpha_n}$
From the above, $U_1, U_2, \ldots, U_n$ are all open in $M$.
From Finite Intersection of Open Sets of Metric Space is Open $\ds \bigcap_{i \mathop = 1}^n U_i$ is open.
{{qed}}
Category:Open Set Less One Point is Open
sy43wh16zygmuzekqe5oievnhx7pesh
Theorem: Closed Real Interval is Neighborhood Except at Endpoints

Let $\R$ be the real number line considered as an Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then $\closedint a b$ is a neighborhood of all of its points except $a$ and $b$.

Proof:
Let $a < c < b$.
Let $\epsilon < \map \min {b - c, c - a}$.
From the definition of positive it follows that $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.
We have that $c + \epsilon < b$ and $a < c - \epsilon$.
Thus:
:$\map {B_\epsilon} c \subseteq \closedint a b$
{{explain|There needs to be a result linked to that proves that $\map {B_\epsilon} c$ is in fact a subset of $\closedint a b$.}}
It follows that, by definition, $\closedint a b$ is a neighborhood of $c$.
Now consider $a$.
Let $\epsilon \in \R_{>0}$.
We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\closedint a b$.
Similarly, consider $b$.
Let $\epsilon \in \R_{>0}$.
We have that $b < b + \epsilon$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entirely in $\closedint a b$.
It follows that, by definition, $\closedint a b$ is a neighborhood of neither $a$ nor $b$.
{{qed}}
Theorem: Open Real Interval is Open Set

Let $\R$ be the real number line considered as an Euclidean space.
Let $\openint a b \subset \R$ be an open interval of $\R$.
Then $\openint a b$ is an open set of $\R$.

Proof:
Let $c \in \R$ such that $a < c < b$.
Let $\epsilon = \min \set {b - c, c - a}$.
From the definition of positive it follows that $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.
Let $y \in \map {B_\epsilon} c$.
Then:
:$\size {c - y} < \epsilon $
implies:
:$-\epsilon < c - y < \epsilon$
We have that:
{{begin-eqn}}
{{eqn | l = a
      | r = c - \paren {c - a}
}}
{{eqn | o = \le
      | r = c - \epsilon
}}
{{eqn | o = <
      | r = y
}}
{{eqn | o = <
      | r = c + \epsilon
}}
{{eqn | o = \le
      | r = c + \paren {b - c}
}}
{{eqn | r = b
}}
{{end-eqn}}
In other words:
:$y \in \map {B_\epsilon} c \implies y \in \openint a b$
Thus:
:$\map {B_\epsilon} c \subseteq \openint a b$
It follows that, by definition, $\openint a b$ is a neighborhood of $c$.
The result follows by definition of open set.
{{qed}}
Theorem: Closed Real Interval is not Open Set

Let $\R$ be the real number line considered as a Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then $\closedint a b$ is not an open set of $\R$.

Proof:
From Closed Real Interval is Neighborhood Except at Endpoints, $a$ and $b$ have no open $\epsilon$-ball lying entirely in $\closedint a b$.
The result follows by definition of open set.
{{qed}}
Theorem: Open Set may not be Open Ball

Let $M = \struct {A, d}$ be a metric space.
Let $U \subseteq M$ be an open set of $M$.
Then it is not necessarily the case that $U$ is an open ball of some $x \in A$.

Proof:
Consider the Euclidean space $\R^2$.
Let:
:$U \subseteq \R^2: U = \set {\tuple {x_1, x_2}: a < x_1 < b, c < x_2 < d}$
Let $x = \tuple {x_1, x_2} \in U$.
Then $\map {B_\epsilon} x \subseteq U$ when $\epsilon = \min \set {x_1 - a, b - x_1, x_2 - c, d - x_2}$:
:310px
So by definition, $U$ is open in $M$.
However, $U$ is not an open ball.
{{qed}}
Theorem: Quotient Mapping is Coequalizer

Let $\mathbf{Set}$ be the category of sets.
Let $S$ be a Set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $r_1, r_2: \RR \to S$ be the projections corresponding to the inclusion mapping $\RR \hookrightarrow S \times S$.
Let $q: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Then $q$ is a coequalizer of $r_1$ and $r_2$ in $\mathbf{Set}$.

Proof:
Let $f: S \to T$ be a mapping as in the following commutative diagram:
$\quad\quad \begin{xy}\xymatrix{
\RR 
\ar[r]<2pt>^*{r_1}
\ar[r]<-2pt>_*{r_2}
& 
S
\ar[r]^*{q}
\ar[rd]_*{f}
&
S / \RR
\ar@{.>}[d]^*{\bar f}
\\
& &
T
}\end{xy}$
This translates to, for $s_1, s_2 \in S$ with $s_1 \RR s_2$:
:$\map {f \circ r_1} {s_1, s_2} = \map {f \circ r_2} {s_1, s_2}$
that is:
:$\map f {s_1} = \map f {s_2}$
The commutativity of the diagram implies that we must define $\bar f: S / \RR \to T$ by:
:$\map {\bar f} {\eqclass {s_1} \RR} = \map f {s_1}$
since $\map q {s_1} = \eqclass {s_1} \RR$.
The above condition precisely states that $\bar f$ is well-defined.
In conclusion, for any $f$ with $f \circ r_1 = f \circ r_2$, there is a unique $\bar f$ making the diagram commute.
That is, $q$ is a coequalizer of $r_1$ and $r_2$.
{{qed}}
Theorem: Closed Real Interval is Closed Set

Let $\R$ be the real number line considered as an Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then $\closedint a b$ is a closed set of $\R$.

Proof:
{{begin-eqn}}
{{eqn | l = \closedint a b
      | r = \set {x \in \R: x \ge a \land x \le b}
      | c = {{Defof|Closed Real Interval}}
}}
{{eqn | ll= \leadsto
      | l = \R \setminus \closedint a b
      | r = \R \setminus \set {x \in \R: x \ge a \land x \le b}
      | c = 
}}
{{eqn | r = \set {x \in \R: x < a \lor x > b}
      | c = De Morgan's Laws: Disjunction of Negations
}}
{{eqn | r = \openint {-\infty} a \cup \openint b \infty
      | c = {{Defof|Open Real Interval}}
}}
{{end-eqn}}
From {{Corollary|Open Real Interval is Open Set}}, both $\openint {-\infty} a$ and $\openint b \infty$ are open sets in $M$.
From Union of Open Sets of Metric Space is Open it follows that $\openint {-\infty} a \cup \openint b \infty$ is open in $\R$.
But $\openint {-\infty} a \cup \openint b \infty$ is the relative complement of $\closedint a b$ in $\R$.
The result follows by definition of closed set.
{{qed}}
Theorem: Kernel of Linear Transformation is Null Space of Matrix Representation

Let $V$ and $W$ be finite dimensional vector spaces.
Let $\phi: V \to W$ be a linear transformation from $V$ to $W$.
Let $\tuple {e_1, \ldots, e_n}$ and $\tuple {f_1, \ldots, f_m}$ be ordered bases of $V$ and $W$ respectively.
Let $A$ be the matrix of $\phi$ in these bases.
Define $f: V \to \R^n$ by:
:$\ds \sum_{i \mathop = 1}^n a_i e_i \mapsto \tuple {a_1, \ldots, a_n}$
and $g : W \to \R^m$ by:
:$\ds \sum_{i \mathop = 1}^m b_i f_i \mapsto \tuple {b_1, \ldots, b_m}$
Let $\map N A = \set {x \in \R^n: A x = 0}$ be the null space of $A$.
Let $\map \ker \phi = \set {x \in V: \phi x = 0}$ be the kernel of $\phi$.
Then:
:$f \sqbrk {\map \ker \phi} = \map N A$
and
:$f^{-1} \sqbrk {\map N A} = \map \ker \phi$
where $f \sqbrk X$ denotes the image set of a subset $X$ of the domain of $f$.

Proof:
{{MissingLinks}}
{{proofread|Use of injectivity of $g$ here is implicit; but I ''think'' it's still rigorous without details: second opinion welcome}}
By the definition of the matrix $A$:
:$A \circ f = g \circ \phi$
Therefore if $x \in \map \ker \phi$ we have:
:$A \map f x = \map g {\map \phi x} = \map g 0 = 0$
This shows that:
:$f \sqbrk {\map \ker \phi} \subseteq \map N A$
Now let $x = \tuple {x_1, \ldots, x_n} \in \map N A$.
Let $y = x_1 e_1 + \cdots + x_n e_n \in V$.
We have:
:$\map {g \circ \phi} y = \map {A \circ f} y = \map A {x_1, \ldots, x_n}^T = 0$
so:
:$\map \phi y = 0 f_1 + \cdots + 0 f_m = 0$
This shows that $y \in \map \ker \phi$.
Since $x = \map f y$, we have shown that:
:$\map N A \subseteq \map f {\map \ker \phi}$
Therefore:
:$\map f {\map \ker \phi} = \map N A$
We deduce immediately from the definitions that:
:$\map \ker \phi \subseteq f^{-1} \sqbrk {\map N A}$
Now suppose that $x \in f^{-1} \sqbrk {\map N A}$.
Then:
:$\map f x \in \map N A$
Therefore:
:$\map {g \circ \phi} x = \map {A \circ f} x = 0$
so:
:$\map \phi x = 0 f_1 + \cdots + 0 f_m = 0$
This shows that:
:$f^{-1} \sqbrk {\map N A} \subseteq \map \ker \phi$
as required.
{{Qed}}
Category:Kernels of Linear Transformations
Category:Null Spaces
amlv6yurdxinwi8dm4vpdza1kc84xhb
Theorem: Half-Open Real Interval is neither Open nor Closed

Let $\R$ be the real number line considered as an Euclidean space.
Let $\hointr a b \subset \R$ be a right half-open interval of $\R$.
Then $\hointr a b$ is neither an open set nor a closed set of $\R$.
Similarly, the left half-open interval $\hointl a b \subset \R$ is neither an open set nor a closed set of $\R$.

Proof:
From Half-Open Real Interval is not Open Set we have that neither $\hointr a b$ nor $\hointl a b$ is an open set of $\R$.
From Half-Open Real Interval is not Closed in Real Number Line we have that neither $\hointr a b$ nor $\hointl a b$ is a closed set of $\R$.
{{qed}}
Theorem: Compact Sets of Double Pointed Topology

Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Then $X \subseteq S \times D$ is compact in $\tau$ {{iff}} for some compact set $C$ of $\tau_S$:
:$\map {\pr_1} X = C$
where $\pr_1$ denotes the first projection on $S \times D$.

Proof:
=== Necessary Condition ===
Suppose that $X \subseteq S \times D$ is a compact set in $\vartheta$.
It is to be shown that $C = \map {\pr_1} X$ is compact in $\tau$.
This follows from Compactness Properties Preserved under Projection Mapping.
{{qed|lemma}}
Theorem: Closure of Subset of Double Pointed Topological Space

Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Let $X \subseteq S \times D$ be a subset of $S \times D$.
Then the closure of $X$ in $\tau$ is:
:$\map \cl X = \map \cl {\map {\pr_1} X} \times D$
where $\pr_1$ denotes the first projection on $S \times D$.

Proof:
By Closed Sets of Double Pointed Topology, $\map \cl {\map {\pr_1} X} \times D$ is closed in $\tau$.
Furthermore, for $\tuple {s, d} \in X$, one has:
:$s \in \map {\pr_1} X \subseteq \map \cl {\map {\pr_1} X}$
by definition of closure.
Since also $d \in D$, we conclude that:
:$X \subseteq \map \cl {\map {\pr_1} X} \times D$
By $(3)$ of Equivalence of Definitions of Topological Closure, it now suffices to prove that:
{{explain|Check the link above -- probably better to link directly to the specific definition of Closure, and also below}}
:$\map \cl {\map {\pr_1} X} \times D \subseteq C$
for any closed set $C$ with $X \subseteq C$.
By Closed Sets of Double Pointed Topology, for some $C'$ closed in $\tau$, we have:
:$C = C' \times D$
so that $C' = \map {\pr_1} C$.
By Image Preserves Subsets, it follows that:
:$\map {\pr_1} X \subseteq C'$
and by $(3)$ of Equivalence of Definitions of Topological Closure, this means:
:$\map \cl {\map {\pr_1} X} \subseteq C'$
whence by Cartesian Product of Subsets:
:$\map \cl {\map {\pr_1} X} \times D \subseteq C' \times D = C$
Hence the result.
{{qed}}
Theorem: Interior of Subset of Double Pointed Topological Space

Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Let $X \subseteq S \times D$ be a subset of $S \times D$.
Define $A \subseteq S$ by:
:$A := \set {s \in S: \paren {\forall d \in D: \tuple {s, d} \in X} }$
Then the interior of $X$ in $\tau$ is:
:$X^\circ = A^\circ \times D$

Proof:
By Open Sets of Double Pointed Topology, $X^\circ$ must be of the form:
:$X^\circ = U \times D$
with $U$ open in $\tau_S$.
By Set Interior is Largest Open Set, we have for any open set $U' \times D$ of $\tau$ that:
:$U' \times D \subseteq X \iff U' \times D \subseteq X^\circ = U \times D$
By Cartesian Product of Subsets, we have for open sets $V, V'$:
:$V \subseteq V' \iff V \times D \subseteq V' \times D$
since $D$ is non-empty.
Combining these two equivalences, we have:
:$U' \times D \subseteq X \iff U' \subseteq U$
The condition that $U' \times D \subseteq X$ can be expressed by:
:$\forall s \in U': \forall d \in D: \tuple {s, d} \in X$
By definition of the set $A$, this is equivalent to:
:$U' \subseteq A$
Combining this with the above, it follows that:
:$U' \subseteq A \iff U' \subseteq U$
Recall that $U$ is open in $\tau_S$.
By Set Interior is Largest Open Set, we conclude:
:$U = A^\circ$
and so:
:$X^\circ = A^\circ \times D$
as desired.
{{qed}}
Theorem: Cartesian Product with Complement

Let $S$ and $T$ be sets.
Let $A \subseteq S$ and $B \subseteq T$ be subsets of $S$ and $T$, respectively.
Let $\relcomp S A$ denote the relative complement of $A$ in $S$.
Then:
{{begin-eqn}}
{{eqn | l = \relcomp S A \times T 
      | r = \relcomp {S \times T} {A \times T}
}}
{{eqn | l = S \times \relcomp T B
      | r = \relcomp {S \times T} {S \times B}
}}
{{end-eqn}}

Proof:
By definition of relative complement we have:
:$\relcomp S A = S \setminus A$
where $S \setminus A$ denotes set difference.
By Cartesian Product Distributes over Set Difference, we have:
:$\paren {S \setminus A} \times T = \paren {S \times T} \setminus \paren {A \times T}$
and the latter equals $\relcomp {S \times T} {A \times T}$.
In conclusion, we obtain:
:$\relcomp S A \times T = \relcomp {S \times T} {A \times T}$
as desired.
And {{mutatis}}, the other statement follows from this argument as well.
{{qed}}
Category:Cartesian Product
Category:Set Complement
fv2u7rxt67wdhczh48hy9qvhsen5mpg
Theorem: Projection of Complement Contains Complement of Projection

Let $S$ and $T$ be non-empty sets.
Let $X \subseteq S \times T$ be a subset of the Cartesian product $S \times T$.
Denote with $\pr_1, \pr_2$ and $\complement$ the first and second projections, and the complement operation, respectively.
Then:
{{begin-eqn}}
{{eqn|l = \map \complement {\map {\pr_1} X}
     |o = \subseteq
     |r = \map {\pr_1} {\map \complement X}
}}
{{eqn|l = \map \complement {\map {\pr_2} X}
     |o = \subseteq
     |r = \map {\pr_2} {\map \complement X}
}}
{{end-eqn}}

Proof:
Let $s \in S$.
Then:
{{begin-eqn}}
{{eqn|l = s
     |o = \in
     |r = \map \complement {\map {\pr_1} X}
}}
{{eqn|ll= \leadstoandfrom
     |l = s
     |o = \notin
     |r = \map {\pr_1} X
     |c = {{Defof|Set Complement}}
}}
{{eqn|ll= \leadstoandfrom
     |q = \forall t \in T
     |l = \tuple {s, t}
     |o = \notin
     |r = X
     |c = {{Defof|First Projection}}
}}
{{eqn|ll= \leadsto
     |q = \exists t \in T
     |l = \tuple {s, t}
     |o = \notin
     |r = X
     |c = Universal Instantiation, Existential Generalisation
}}
{{eqn|ll= \leadstoandfrom
     |q = \exists t \in T
     |l = \tuple {s, t}
     |o = \in
     |r = \map \complement X
     |c = {{Defof|Set Complement}}
}}
{{eqn|ll= \leadstoandfrom
     |l = s
     |o = \in
     |r = \map {\pr_1} {\map \complement X}
     |c = {{Defof|First Projection}}
}}
{{end-eqn}}
In conclusion:
:$s \in \map \complement {\map {\pr_1} X} \implies s \in \map {\pr_1} {\map \complement X}$
and by definition of subset, the first relation follows.
And {{mutatis}}, the other relation can be established.
{{qed}}
Category:Projections
Category:Set Complement
t2rb9lgv316e0h8eixv9f8epjtsc2bi
Theorem: Interior of Subset

Let $\left({S, \tau}\right)$ be a topological space.
Let $X$ and $Y$ be subsets of $S$, and suppose that $X \subseteq Y$.
Then:
:$X^\circ \subseteq Y^\circ$
where $X^\circ$ denotes the interior of $X$.

Proof:
By definition of interior, $X^\circ$ is open in $\tau$, and:
:$Y^\circ \subseteq Y$
Hence, by Subset Relation is Transitive:
:$X^\circ \subseteq Y$
is an open set contained in $Y$.
The result follows by Set Interior is Largest Open Set.
{{qed}}
Category:Subsets
Category:Set Interiors
rh05xi8umelkm4d6icp3t0sg58yibbu
Theorem: Real Interval is Bounded in Real Numbers

Let $\R$ be the real number line considered as an Euclidean space.
Let $a, b \in \R$.
Let $\II$ be one of the following real intervals:
{{begin-eqn}}
{{eqn | l = \openint a b
      | o = :=
      | r = \set {x \in \R: a < x < b}
      | c = Open Real Interval
}}
{{eqn | l = \hointr a b
      | o = :=
      | r = \set {x \in \R: a \le x < b}
      | c = Half-Open (to the right) Real Interval
}}
{{eqn | l = \hointl a b
      | o = :=
      | r = \set {x \in \R: a < x \le b}
      | c = Half-Open (to the left) Real Interval
}}
{{eqn | l = \closedint a b
      | o = :=
      | r = \set {x \in \R: a \le x \le b}
      | c = Closed Real Interval
}}
{{end-eqn}}
where $b \ge a$.
Then $\II$ is bounded in $\R$.

Proof:
Consider the open $\epsilon$-ball $\map {B_\epsilon} a$ where $\epsilon = b + 1 - a$.
As $b \ge a$ we have that $b + 1 > a$ and so $\epsilon > 0$.
Let $x \in \II$.
Then, whatever type of real interval $\II$ actually is, $z \ge a$ and $x \le b$.
As $\epsilon > 0$ it follows that $x > a - \epsilon$.
Also:
{{begin-eqn}}
{{eqn | l = x
      | o = \le
      | r = b
      | c = 
}}
{{eqn | r = a + b - a
      | c = 
}}
{{eqn | o = <
      | r = a + \paren {b  + 1 - a}
      | c = 
}}
{{eqn | r = a + \epsilon
      | c = 
}}
{{end-eqn}}
That is:
:$a - \epsilon < x < a + \epsilon$
and so:
:$x \in \map {B_\epsilon} a$
The result follows by definition of bounded space.
{{qed}}
Theorem: Set of Integers is not Bounded

Let $\R$ be the real number line considered as an Euclidean space.
The set $\Z$ of integers is not bounded in $\R$.

Proof:
Let $a \in \R$.
Let $K \in \R_{>0}$.
Consider the open $K$-ball $\map {B_K} a$.
By the Axiom of Archimedes there exists $n \in \N$ such that $n > a + K$.
As $\N \subseteq \Z$:
:$\exists n \in \Z: a + K < n$
and so:
:$n \notin \map {B_K} a$
As this applies whatever $a$ and $K$ are, it follows that there is no $\map {B_K} a$ which contains all the integers.
Hence the result, by definition of bounded space.
{{qed}}
Theorem: Open Real Interval is not Closed Set/Corollary

Let:
:$I_a = \openint \gets a$
:$I_b = \openint b \to$
be unbounded open real intervals.
Then neither $I_a$ nor $I_b$ are closed sets of $\R$.

Proof:
Consider the relative complement of $I_a$ in $\R$:
:$J = \relcomp \R I = \R \setminus I = \hointr a \to$
Let $\epsilon \in \R_{>0}$.
Consider the open $\epsilon$-ball $\map {B_\epsilon} a$.
Whatever the value of $\epsilon$ is, $a - \epsilon$ is not in $\map {B_\epsilon} a$.
So, by definition, $J$ is not an open set of $\R$.
By Relative Complement of Relative Complement, $\relcomp \R J = I$.
By definition, it follows that $I_a$ is not a closed set of $\R$.
And {{mutatis}}, the same proof shows that neither is $I_b$ a closed set of $\R$.
{{qed}}
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
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