avewright/proofwiki-math huggingface dataset:
Theorem: Covariant Hom Functor is Continuous

Let $\mathbf{Set}$ be the category of sets.
Let $\mathbf C$ be a locally small category.
Let $C$ be an object of $\mathbf C$, and let $\map {\operatorname{Hom}_{\mathbf C} } {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the covariant hom functor based at $C$.
Then $\map {\operatorname{Hom}_{\mathbf C} } {C, \cdot}$ is a continuous functor.

Proof:
Let $F := \map {\operatorname{Hom}_{\mathbf C} } {C, \cdot}$.
Let $D: \mathbf J \to \mathbf C$ be a diagram.
Suppose the  cone $\paren { L, \sequence {p_j} }$ is a limit for $D$.
We need to show that $\paren { F L, \sequence {F p_j } }$ is a limit for the diagram $F D: \mathbf J \to \mathbf{Set}$.
As $F$ is a covariant functor, for $\alpha: i \to j$ of $\mathbf J$:
::<nowiki>$\begin{xy}\xymatrix@+0.5em@L+2px{
 &F L
  \ar[dl]_*+{F p_i }
  \ar[dr]^*+{F p_j }
&
\\
 F D_i
  \ar[rr]_*+{F D_\alpha}
&&
 F D_j
}\end{xy}$</nowiki>
That is, $\ds \paren { F L, \sequence {F p_j } }$ is a cone to $F D$.
Now, we need to check that this is indeed a terminal object in $\map {\mathbf{Cone} }{F D}$.
Let $\paren {A, \sequence {q_j} }$ be a cone to $F D$.
That is, for $\alpha: i \to j$ of $\mathbf J$:
::<nowiki>$\begin{xy}\xymatrix@+0.5em@L+2px{
 &A
  \ar[dl]_*+{q_i}
  \ar[dr]^*+{q_j}
&
\\
 F D_i
  \ar[rr]_*+{F D_\alpha}
&&
 F D_j
}\end{xy}$</nowiki>
Note that $A$ is a set and:
:$\ds q_j : A \to \map {\operatorname{Hom}_{\mathbf C} } {C, D_j}$
are mappings.
Let $a \in A$.
Then $\paren {C , \sequence {\map {q_j} a} }$ is a cone to $D$, since:
:$\map {q_j} a : C \to D_j$
and, for $\alpha: i \to j$ of $\mathbf J$:
{{begin-eqn}}
{{eqn | l = D_\alpha \circ \paren {\map {q_i} a}
      | r = \map {\paren {\map {\operatorname{Hom}_{\mathbf C} } {C, D_\alpha} \circ q_i} } a
}}
{{eqn | r = \map {\paren {F D_\alpha \circ q_i} } a
}}
{{eqn | r = \map {q_j} a
}}
{{end-eqn}}
That is, for $\alpha: i \to j$ of $\mathbf J$:
::<nowiki>$\begin{xy}\xymatrix@+0.5em@L+2px{
 &C
  \ar[dl]_*+{\map {q_i} a}
  \ar[dr]^*+{\map {q_j} a}
&
\\
 D_i
  \ar[rr]_*+{D_\alpha}
&&
 D_j
}\end{xy}$</nowiki>
As $\paren { L, \sequence {p_j} }$ is a terminal object in $\map {\mathbf{Cone} } D$, there is a unique morphism:
:$ \ds \pi_a : C \to L$
in $\mathbf C$ such that:
:$\map {q_j} a = p_j \circ \pi_a$
That is:
::<nowiki>$\begin{xy}\xymatrix@+0.5em@L+2px{
 &C
  \ar[ddl]_*+{\map {q_i} a}
  \ar[ddr]^*+{\map {q_j} a}
  \ar[d]^*+{\pi_a}
&
\\
& { L}
  \ar[dl]_*+{p_i}
  \ar[dr]^*+{p_j}
&
\\
 D_i
  \ar[rr]_*+{D_\alpha}
&&
 D_j
}\end{xy}$</nowiki>
This means, however, the mapping $A \to \map {\operatorname{Hom}_{\mathbf C} } {C, L}$ defined by:
:$\pi : a \mapsto \pi_a$
is the unique morphism in $\mathbf{Set}$ such that:
:$q_j = F p_j \circ \pi$
That is:
::<nowiki>$\begin{xy}\xymatrix@+0.5em@L+2px{
&A
  \ar[ddl]_*+{q_i}
  \ar[ddr]^*+{q_j}
  \ar[d]^*+{\pi}
&
\\
 &F L
  \ar[dl]_*+{F p_i }
  \ar[dr]^*+{F p_j }
&
\\
 F D_i
  \ar[rr]_*+{F D_\alpha}
&&
 F D_j
}\end{xy}$</nowiki>
{{qed}}
Theorem: Composite Mapping is Mapping

Let $S_1, S_2, S_3$ be sets.
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.
Then the composite mapping $g \circ f$ is also a mapping.

Proof:
The composite of $f$ and $g$ is defined as:
:$g \circ f := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f \land \tuple {y, z} \in g}$
It is necessary to show that $g \circ f$ is both left-total and many-to-one.
Theorem: Normal Space is Regular

Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a regular space.

Proof:
Let $T = \struct {S, \tau}$ be a normal space.
From Normal Space is $T_3$, we have that $T$ is a $T_3$ space.
We also have by definition of normal space that $T$ is a $T_1$ space.
From $T_1$ Space is $T_0$ we have that $T$ is a $T_0$ space
So $T$ is both a $T_3$ space and a $T_0$ space.
Hence $T$ is a regular space by definition.
{{qed}}
Theorem: Product Space is T2 iff Factor Spaces are T2/Necessary Condition

Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }$ be an indexed family of non-empty topological spaces for $\alpha$ in some indexing set $I$.
Let $\ds T = \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\SS$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ for $\alpha \in I$ be $T_2$ (Hausdorff) space.
Then $T$ is a $T_2$ (Hausdorff) space.

Proof:
Let each of $\struct {S_\alpha, \tau_\alpha}$ for $\alpha \in I$ be $T_2$ (Hausdorff) space.
Let $x, y \in S : x \ne y$.
Then $x_\alpha \ne y_\alpha$ for some $\alpha \in I$.
Since $\struct {S_\alpha, \tau_\alpha}$ is Hausdorff then:
:$\exists U, V \in \tau_\alpha: x_\alpha \in U, y_\alpha \in V : U \cap V = \O$
Let $\pr_\alpha: S \to S_\alpha$ be the projection of $S$ to $S_\alpha$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\pr_\alpha^\gets} U \cap \map {\pr_\alpha^\gets} V
      | r = \map {\pr_\alpha^\gets} {U \cap V}
      | c = Preimage of Intersection under Mapping
}}
{{eqn | r = \map {\pr_\alpha^\gets} \O
}}
{{eqn | r = \O
}}
{{end-eqn}} 
By definition of the projection $\pr_\alpha$:
:$\map {\pr_\alpha} x = x_\alpha \in U$
By definition of the preimage under $\pr_\alpha$:
:$x \in \map {\pr_\alpha^\gets} U$
Similarly:
:$y \in \map {\pr_\alpha^\gets} V$
By definition of the product topology $\tau$:
:$\map {\pr_\alpha^\gets} U, \map {\pr_\alpha^\gets} V \in \tau$
Since $x, y \in S$ were arbitrary, it follows that $T$ is a $T_2$ (Hausdorff) space.
Theorem: Product Space is T2 iff Factor Spaces are T2/Necessary Condition/Product of 2 Spaces

Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $T = T_\alpha \times T_\beta$ be the product space of $T_\alpha$ and $T_\beta$
Let $T_\alpha$ and $T_\beta$ both be $T_2$ (Hausdorff) space.
Then $T$ is also a $T_2$ (Hausdorff) space.

Proof:
Let $T_\alpha$ and $T_\beta$ be $T_2$ (Hausdorff) space.
Let $\tuple {a, b}$ and $\tuple {c, d}$ be two distinct points of the product space $T$.
Let $a = c$.
Then as $\tuple {a, b} \ne \tuple {c, d}$ it follows that $b \ne d$.
As $T_\beta$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\beta$ such that $b \in U, d \in V$.
Then $T_\alpha \times U$ and $T_\alpha \times V$ are two open disjoint sets in the product space $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$.
Let $a \ne c$.
As $T_\alpha$ is Hausdorff, there exists two disjoint open sets $U, V \subseteq T_\alpha$ such that $a \in U, c \in V$.
Then $U \times T_\beta$ and $V \times T_\beta$ are two open disjoint sets in the product space $T$ containing $\tuple {a, b}$ and $\tuple {c, d}$.
{{qed}}
Theorem: Domain of Continuous Injection to Hausdorff Space is Hausdorff

Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $f: S_\alpha \to S_\beta$ be a continuous mapping which is an injection.
If $T_\beta$ is a $T_2$ (Hausdorff) space, then $T_\alpha$ is also a $T_2$ (Hausdorff) space.

Proof:
Let $x, y \in S_\alpha$ be distinct points.
We want to find disjoint open sets $U, V \in \tau_\alpha$ containing $x$ and $y$ respectively.
Since $f$ is injective the points $\map f x, \map f y \in S_\beta$ are distinct.
By assumption $T_\beta$ is Hausdorff.
Therefore we can choose disjoint open sets $U', V'$ of $T_\beta$ such that $\map f x \in U'$ and $\map f y \in V'$.
Since $f$ is continuous, the sets $U = f^{-1} \sqbrk {U'}$ and $V = f^{-1} \sqbrk {V'}$ are open.
Moreover we have $x \in U$ and $y \in V$.
Finally we show that the sets $U$ and $V$ are disjoint.
{{AimForCont}} $z \in U \cap V$.
Then by the definition of $U$, $V$ we have $\map f z \in U'$ and $\map f z \in V'$.
This is a contradiction, since $U' \cap V' = \O$.
{{Qed}}
Theorem: Equivalence of Definitions of Homeomorphisms between Topological Spaces

Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $f: T_\alpha \to T_\beta$ be a bijection.
{{TFAE|def = Homeomorphism (Topological Spaces)|view = homeomorphism}}

Proof:
=== Definition 1 iff Definition 2 ===
Let $f$ and $f^{-1}$ both be continuous.
As $f$ is continuous, then by definition:
:$V \in \tau_\beta \implies f^{-1} \sqbrk V \in \tau_\alpha$
and as $f^{-1}$ is continuous, then by definition:
:$U \in \tau_\alpha \implies \paren {f^{-1} }^{-1} \sqbrk U = f \sqbrk U \in \tau_\beta$
That is:
:$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$
Conversely, suppose that:
:$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$
By definition:
:$U \in \tau_\alpha \implies f \sqbrk U \in \tau_\beta$ {{iff}} $f^{-1}$ is continuous.
Also by definition:
:$f \sqbrk U \in \tau_\beta \implies U = f^{-1} \sqbrk {f \sqbrk U} \in \tau_\alpha$ {{iff}} $f$ is continuous.
That is, both $f$ and $f^{-1}$ are continuous.
{{qed|lemma}}
Theorem: Quotient Space of Hausdorff Space is not necessarily Hausdorff

Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $T_\RR := \struct {S / \RR, \tau_\RR}$ be the quotient space of $S$ by $\RR$.
Then $T_\RR$ is not necessarily also a $T_2$ (Hausdorff) space.

Proof:
Consider the real number line with the Euclidean topology $\struct {\R, \tau}$.
By Real Number Line satisfies all Separation Axioms, $\struct {\R, \tau}$ is a $T_2$ (Hausdorff) space.
By Quotient Space of Real Line may not be T0, there is a relation $\RR$ on $\R$ such that the quotient space $\struct {\R / \RR, \tau_\RR}$ is not a $T_0$ space.
Thus the theorem holds by Sequence of Implications of Separation Axioms.
{{qed}}
Theorem: Existence and Uniqueness of Direct Limit of Sequence of Groups

Let $\sequence {G_n}_{n \mathop \in \N}$ be a sequence of groups.
Let $\sequence {g_n}_{n \mathop \in \N}: g_n: G_n \to G_{n + 1}$ be a sequence of group homomorphisms.
Then their direct limit $G_\infty$ exists and is unique up to unique isomorphism.

Proof:
{{refactor|Transclude the lemmata|level = basic}}
Theorem: Topological Product of Compact Spaces/Finite Product

Let $T_1, T_2, \ldots, T_n$ be topological spaces.
Let $\ds \prod_{i \mathop = 1}^n T_i$ be the product space of $T_1, T_2, \ldots, T_n$.
Then $\ds \prod_{i \mathop = 1}^n T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_n$ are compact.

Proof:
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \prod_{i \mathop = 1}^n T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_n$ are compact
Theorem: Uniform Continuity on Metric Space does not imply Compactness

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a uniformly continuous mapping on $A_1$.
Then it is not necessarily the case that $M_1$ is a compact metric space.

Proof:
Let $M_1 = \struct {A_1, d_1}$ be any metric space which is not compact.
Let $I_{M_1}: M_1 \to M_1$ be the identity mapping.
From Identity Mapping is Uniformly Continuous, $I_{M_1}$ is uniformly continuous on $M_1$.
Hence the result.
{{qed}}
Theorem: Uniformly Continuous Function is Continuous/Metric Space

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_1, d_1}\right)$ be metric spaces.
Let the mapping $f: M_1 \to M_2$ be uniformly continuous on $M_1$.
Then $f$ is continuous on $M_1$.

Proof:
Let $f$ be uniformly continuous on $M_1$.
Let $x \in M_1$.
Let $\epsilon > 0$.
As $f$ is uniformly continuous, $\exists \delta > 0$ such that:
:$\forall y \in M_1: d_1 \left({x, y}\right) < \delta: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$
Thus by definition $f$ is continuous at $x$.
{{qed}}
Category:Metric Spaces
Category:Continuous Mappings on Metric Spaces
Category:Uniformly Continuous Mappings
qdpnhtcjgrywzny4gfvtlykqmhh0bj9
Theorem: Uniformly Continuous Function is Continuous/Real Function

Let $I$ be an interval of $\R$.
Let $f: I \to \R$ be a uniformly continuous real function on $I$.
Then $f$ is continuous on $I$.

Proof:
From Real Number Line is Metric Space, $\R$ under the Euclidean metric is a metric space.
The result follows by Uniformly Continuous Function is Continuous: Metric Space.
{{qed}}
Theorem: Uniformly Continuous Function is Continuous/Real Function

Let $I$ be an interval of $\R$.
Let $f: I \to \R$ be a uniformly continuous real function on $I$.
Then $f$ is continuous on $I$.

Proof:
Let $x \in I$.
Let $\epsilon \in \R_{>0}$.
As $f$ is uniformly continuous: 
:$\exists \delta \in \R_{>0}: \paren {x, y \in I, \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon}$
Then, for all $y \in I$ such that $\size {x - y} < \delta$:
:$\size {\map f x - \map f y} < \epsilon$
Thus by definition $f$ is continuous at $x$.
As $x$ was arbitrary, $f$ is continuous on all of $I$.
{{qed}}
Theorem: Young's Inequality for Increasing Functions/Equality

Let $a_0$ and $b_0$ be strictly positive real numbers.
Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a strictly increasing bijection.
Let $a$ and $b$ be real numbers such that $0 \le a \le a_0$ and $0 \le b \le b_0$.
Then $b = \map f a$ {{iff}}:
:$\ds a b = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
where $\ds \int$ denotes the Darboux integral.

Proof:
=== Necessary Condition ===
{{ProofWanted}}
Theorem: Category of Finite Sets is Cartesian Closed

Let $\mathbf{Finset}$ be the category of finite sets.
Then $\mathbf{Finset}$ is Cartesian closed.

Proof:
{{MissingLinks}}
{{Recall|Cartesian Closed Category}}
{{:Definition:Cartesian Closed Category}}
Recall Cartesian Product is Set Product.
By Cardinality of Cartesian Product of Finite Sets, the finite product of sets has finite cardinality.
Hence $\mathbf{C}$ has finite products.
By Cardinality of Set of All Mappings, the exponentials have finite cardinality.
Hence $\mathbf{C}$ has exponentials.
Hence the result.
{{Finish|I think we still need to show that the usual set product is the product in the category theory sense. Same for exponential. We need Category of Sets is Cartesian Closed.}}
Theorem: Components of Separation are Clopen

Let $T = \struct {S, \tau}$ be a topological space.
Let $A \mid B$ be a separation of $T$.
Then both $A$ and $B$ are clopen in $T$.

Proof:
From Set with Relative Complement forms Partition:
:$A = \relcomp S B$
and:
:$B = \relcomp S A$
where $\complement_S$ denotes the complement relative to $S$.
As $A$ and $B$ are both open, it follows by definition that $A$ and $B$ are also both closed.
That is, by definition, they are clopen.
{{qed}}
Theorem: Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets

{{TFAE|def = Connected Topological Space}}
Let $T = \struct {S, \tau}$ be a topological space.

Proof:
From Biconditional Equivalent to Biconditional of Negations it follows that the statement can be expressed as:
:$T$ admits a separation
{{iff}}:
:there exist two closed sets of $T$ which form a (set) partition of $S$.
By definition, a separation of $T$ is a (set) partition of $S$ by $A, B$ which are open in $T$.
From Complements of Components of Two-Component Partition form Partition:
:$A \mid B$ is a (set) partition of $S$ {{iff}} $\relcomp S A \mid \relcomp S B$ is a (set) partition of $S$.
Hence the result by definition of closed set.
Theorem: Empty Set Satisfies Topology Axioms

Let $T = \struct {\O, \set \O}$ where $\O$ denotes the empty set.
Then $T$ satisfies the open set axioms for a topological space.

Proof:
We proceed to verify the open set axioms for $\set \O$ to be a topology on $\O$.
Let $\tau = \set \O$.
Theorem: Complements of Components of Two-Component Partition form Partition

Let $S$ be a set with at least two elements.
Let $A, B \subseteq S$.
Let $\complement_S$ denote the complement relative to $S$.
$A \mid B$ is a partition of $S$ {{iff}} $\relcomp S A \mid \relcomp S B$ is a partition of $S$.

Proof:
=== Necessary Condition ===
Let $A \mid B$ be a partition of $S$.
That is, by definition:
{{begin-eqn}}
{{eqn | n = 1
      | l = A \cap B
      | r = \O
}}
{{eqn | n = 2
      | l = A \cup B
      | r = S
}}
{{eqn | n = 3
      | l = A, B
      | o = \ne
      | r = \O
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | n = 1
      | l = A \cap B 
      | r = \O
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp S {A \cap B}
      | r = S
      | c = Relative Complement of Empty Set
}}
{{eqn | ll= \leadsto
      | l = \relcomp S A \cup \relcomp S B
      | r = S
      | c = De Morgan's Laws: Complement of Intersection
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
      | l = A \cup B 
      | r = S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \relcomp S {A \cup B}
      | r = \O
      | c = Relative Complement with Self is Empty Set
}}
{{eqn | ll= \leadsto
      | l = \relcomp S A \cap \relcomp S B
      | r = \O
      | c = De Morgan's Laws: Complement of Union
}}
{{end-eqn}}
{{WLOG}}, now suppose that $\relcomp S A = \O$.
{{begin-eqn}}
{{eqn | l = \relcomp S A
      | r = \O
      | c = 
}}
{{eqn | ll= \leadsto
      | l = A
      | r = S
      | c = Relative Complement of Empty Set
}}
{{eqn | ll= \leadsto
      | l = B
      | r = \O
      | c = as $A \cap B = \O$
}}
{{end-eqn}}
Thus by Rule of Transposition:
:$B \ne \O \implies \relcomp S A \ne \O$
And {{mutatis}}:
:$A \ne \O \implies \relcomp S B \ne \O$
And {{hypothesis}} $(3)$:
:$A, B \ne \O$
and so:
:$\relcomp S A, \relcomp S B \ne \O$
Thus all three conditions are satisfied for $\relcomp S A \mid \relcomp S B$ to be a partition of $S$.
{{qed|lemma}}
Theorem: Sigma-Algebra Closed under Finite Intersection

Let $A_1, \ldots, A_n \in \Sigma$.
Then $\ds \bigcap_{k \mathop = 1}^n A_k \in \Sigma$.

Proof:
Define for $k \in \N, k > n: A_k = X$.
By axiom $(1)$ of a $\sigma$-algebra, it follows that $\forall k \in \N, k > n: A_k \in \Sigma$.
From Sigma-Algebra Closed under Countable Intersection, it follows that $\ds \bigcap_{k \mathop \in \N} A_k = \bigcap_{k \mathop = 1}^n A_k \in \Sigma$.
{{qed}}
Category:Sigma-Algebras
g8566nwukcudqxakk0c0q1uln2gfv2l
Theorem: Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary

Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ have no two disjoint non-empty closed sets whose union is $S$.
Then the only subsets of $S$ whose boundary is empty are $S$ and $\O$.

Proof:
Let $H \subseteq S$ be a non-empty subset whose boundary $\partial H$ is empty.
Thus:
{{begin-eqn}}
{{eqn | l = \partial H
      | r = \O
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = H^- \cap \paren {S \setminus H}^-
      | r = \O
      | c = Boundary is Intersection of Closure with Closure of Complement
}}
{{end-eqn}}
From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed sets of $T$.
From Union of Closure with Closure of Complement is Whole Space:
:$H^- \cup \paren {S \setminus H}^- = S$
Thus $H^-$ and $\paren {S \setminus H}^-$ are two disjoint closed sets of $T$ whose union is $S$.
Hence, {{hypothesis}}, one of them must be empty.
Suppose $H$ is not empty.
It must therefore follow that:
: $S \setminus H = \O$
Therefore $H = S$.
Thus the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
Theorem: Equivalence of Definitions of Connected Topological Space/No Subsets with Empty Boundary implies No Clopen Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ be such that the only subsets of $S$ whose boundary is empty are $S$ and $\O$.
Then the only clopen sets of $T$ are $S$ and $\O$.

Proof:
Let $H \subseteq S$ be a clopen set of $T$.
From Set is Clopen iff Boundary is Empty, $H$ has an empty boundary.
We have {{hypothesis}} that $H = S$ or $H = \O$.
That is, the only clopen sets of $T$ are $S$ and $\O$.
Theorem: Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let the only clopen sets of $T$ be $S$ and $\O$.
Then there are no two non-empty separated sets of $T$ whose union is $S$.

Proof:
Suppose $A$ and $B$ are separated subsets of $T$ such that $A \cup B = S$.
By definition of separated sets:
:$A \cap B^- = \O$
Then:
{{begin-eqn}}
{{eqn | l = S
      | r = A \cup B
      | c = 
}}
{{eqn | o = \subseteq
      | r = A \cup B^-
      | c = Set is Subset of its Topological Closure
}}
{{eqn | o = \subseteq
      | r = S
      | c = by definition of $S$
}}
{{end-eqn}}
Hence $A = S \setminus B^-$.
From Topological Closure is Closed, $B^-$ is closed in $T$.
Thus $A$ is open in $T$.
Also by definition of separated sets:
:$A^- \cap B = \O$
Hence, by the same reasoning, $B$ must also be open.
But:
:$A \cap B \subseteq A \cap B^- = \O$
and $A \cup B = S$, by assumption.
So:
:$A = S \setminus B$ and $B = S \setminus A$
and we conclude that both $A$ and $B$ are clopen.
Therefore, {{hypothesis}}, one of them must be $S$ and the other must be $\O$.
That is, there are no two non-empty separated sets of $T$ whose union is $S$.
Theorem: Equivalence of Definitions of Connected Topological Space/No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ be such that there are no two non-empty separated sets whose union is $S$.
Then there exists no continuous surjection from $T$ onto a discrete two-point space.

Proof:
Let $T = \struct {S, \tau}$ be a topological space such that there are no two non-empty separated sets whose union is $S$.
Let $D = \struct {\set {0, 1}, \tau}$ be the discrete two-point space on $\set {0, 1}$.
{{AimForCont}} $f: T \to \set {0, 1}$ is a continuous surjection.
By definition of continuous mapping:
:$\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are open sets of $T$.
From the definition of a mapping:
:$\map {f^{-1} } 0 \cup \map {f^{-1} } 1 = S$
and
:$\map {f^{-1} } 0 \cap \map {f^{-1} } 1 = \O$
Then:
:$\map {f^{-1} } 0 = S \setminus \map {f^{-1} } 1$
and:
:$\map {f^{-1} } 1 = T \setminus \map {f^{-1} } 0$
are clopen.
From Closed Set equals its Closure they are their respective closures.
It follows from the definition that $\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are separated subsets of $T$ whose union is $S$.
Hence, {{hypothesis}}, one of them must be empty, and the other one must be $S$.
Therefore $f$ is constant, and so is not a surjection.
This contradicts the original hypothesis.
That is, there exists no continuous surjection from $T$ onto a discrete two-point space.
Theorem: Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation

Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ be such that there exists no continuous surjection from $T$ onto a discrete two-point space.
Then there exist no open sets $A, B \in \tau$ such that $A, B \ne \O$, $A \cup B = S$ and $A \cap B = \O$.

Proof:
Let $T = \struct {S, \tau}$ be a topological space such that there exists no continuous surjection from $T$ onto a discrete two-point space.
Let $D = \struct {\set {0, 1}, \tau}$ be the discrete two-point space on $\left\{{0, 1}\right\}$.
Let $A$ and $B$ be disjoint open sets of $T$ such that $A \cup B = S$.
The aim is to show that one of them is empty.
Let us define the mapping $f: S \to \set {0, 1}$ by:
:<nowiki>$\map f x = \begin{cases}
0 & : x \in A \\
1 & : x \in B
\end{cases}$</nowiki>
There are only four open sets in $\set {0, 1}$, namely:
:$\O$
:$\set 0$
:$\set 1$
:$\set {0, 1}$
We have that:
:$f^{-1} \sqbrk \O = \O$
:$f^{-1} \sqbrk {\set 0} = A$
:$f^{-1} \sqbrk {\set 1} = B$
:$f^{-1} \sqbrk {\set {0, 1} } = S$
where $f^{-1} \sqbrk X$ denotes the preimage of the set $X$.
But {{hypothesis}} all of $\O, A, B, S$ are open sets of $T$.
So by definition $f$ is continuous.
Also {{hypothesis}}, $f$ cannot be surjective.
It follows that $f$ must be constant.
So either $A$ or $B$ must be empty, and the other one must be $S$.
Hence the result.
Theorem: Existence of Vector Space Bases implies Axiom of Choice

The supposition that every vector space has a basis, along with the Zermelo-Fraenkel axioms, implies that the {{axiom-link|Choice}} holds.

Proof:
[http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf Blass, 1984]
{{proof wanted}}