nyuuzyou/wikis huggingface dataset:
Cantor-Bernstein-Schröder Theorem/Lemma/Proof 2

== Theorem ==
== Proof ==
Define a mapping $E: \powerset S \to \powerset S$ as:
:$\map E K = S \setminus \paren {T \setminus f \sqbrk K}$
where $f \sqbrk K$ is the image of $K$ under $f$.
Then:
:$\map E K = \paren {S \setminus T} \cup f \sqbrk K$
By Image of Subset under Relation is Subset of Image and the corollary to Set Union Preserves Subsets, $E$ is increasing.
Thus by Knaster-Tarski Lemma: Power Set, $E$ has a fixed point $X$.
Then by definition of fixed point:
:$\map E X = X$
That is:
:$\paren {S \setminus \paren {T \setminus f \sqbrk X} } = X$
Taking the set difference from $S$:
:$T \setminus f \sqbrk X = S \setminus X$
Let $f'$ be the restriction of $f$ to $X \times f \sqbrk X$.
By Injection to Image is Bijection, $f'$ is a bijection.
By Identity Mapping is Bijection, the identity mapping $I_{S \mathop \setminus X}$ from $S \setminus X$ to $T \setminus f \sqbrk X$ is a bijection.
Thus by Union of Bijections with Disjoint Domains and Codomains is Bijection, $g = f' \cup i$ is the bijection we seek.
Category:Cantor-Bernstein-Schröder Theorem
Cantor-Bernstein-Schröder Theorem/Lemma/Proof 3

== Theorem ==
== Proof == 
Define $C = \{f^k(x) | k \in \N, x \in S \setminus T \}$.
Clearly, $C = C_0 \cup C_1$, where:
$C_0 = S \setminus T$, the difference between $S$ and $T$, 
$C_1 = \{f^k(x) | k \in \N_{> 0}, x \in S \setminus T \}$.
Note, that $S \setminus C_0 = S \setminus (S \setminus T) = S \cap T = T$ (use this theorem and $T \subseteq S$).
Obviously, $im(f|C) = C_1$.
Define a mapping $h: S \to S$ as follows:
$\map h x = \begin {cases} \map f x & : x \in C \\ x & : x \notin C \end {cases}$
Clearly, $h = f|C \cup I_{S \setminus C}$, where $I_{S \setminus C}$ is the identity mapping on the set $S \setminus C$.
Note, that $dom(h) = S$;
$im(h) = im(f|C) \cup im(I_{S \setminus C}) =$
$C_1 \cup (S \setminus C) =$
$C_1 \cup (S \setminus (C_0 \cup C_1)) =$
$S \setminus C_0 =$
$T$.
Also, $h$ is an injection because $h$ is the union of the injections $f|C$ (Restriction of Injection is Injection) and $I_{S \setminus C}$,
and also $dom(f|C) \cap dom(I_{S \setminus C}) = \{\}$
and $im(f|C) \cap im(I_{S \setminus C}) = C_1 \cap (S \setminus C) = \{\}$ because $C_1 \subseteq C$.
Now $h$ is a bijection $S \to T$ because $dom(h) = S, im(h) = T$ and $h$ is an injection.
Category:Cantor-Bernstein-Schröder Theorem
Cantor-Bernstein-Schröder Theorem/Proof 1

== Theorem ==
== Proof ==
From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two bijections:
:$f: S \to T_1$
:$g: T \to S_1$
Let:
:$S_2 = g \sqbrk {f \sqbrk S} = g \sqbrk {T_1} \subseteq S_1$
and:
:$T_2 = f \sqbrk {g \sqbrk T} = f \sqbrk {S_1} \subseteq T_1$
So $S_2 \subseteq S_1$ and $S_2 \sim S$, while $T_2 \subseteq T_1$, and $T_2 \sim T$.
For each natural number $k$, let $S_{k + 2} \subseteq S$ be the image of $S_k$ under the mapping $g \circ f$.
Then:
:$S \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_k \supseteq S_{k + 1} \ldots$
Let $\ds D = \bigcap_{k \mathop = 1}^\infty S_k$.
Now we can represent $S$ as:
where $S \setminus S_1$ denotes set difference.
Similarly, we can represent $S_1$ as:
Now let:
and rewrite $(1)$ and $(2)$ as:
Now:
and so on.
So $N \sim N_1$.
It follows from $(3)$ and $(4)$ that a bijection can be set up between $S$ and $S_1$.
But $S_1 \sim T$.
Therefore $S \sim T$.
== Sources ==
* : $\S 2.6$: The Cantor-Bernstein theorem: Theorem $7$
Category:Cantor-Bernstein-Schröder Theorem
Cantor-Bernstein-Schröder Theorem/Proof 3

== Theorem ==
Let $S$ and $T$ be sets, such that:
:$\exists f: S \to T$ such that $f$ is an injection
:$\exists g: T \to S$ such that $g$ is an injection.
Then there exists a bijection from $S$ to $T$.
== Proof ==
Let $S, T$ be sets.
Let $\powerset S, \powerset T$ denote their power sets.
Let $f: S \to T$ and $g: T \to S$ be injections that we know to exist between $S$ and $T$.
Consider the relative complements of elements of $\powerset S$ and $\powerset T$ as mappings:
:$\complement_S: \powerset S \to \powerset S: \forall X \in \powerset S: \relcomp S X = S \setminus X$
:$\complement_T: \powerset T \to \powerset T: \forall Y \in \powerset T: \relcomp T Y = T \setminus Y$
which follow directly from the definition of relative complement.
Let $\alpha$ and $\beta$ denote the direct image mappings of $f$ and $g$, respectively.
Consider the mapping $z: \powerset S \to \powerset S$ defined by the composition:
:$z = \complement_S \circ \beta \circ \complement_T \circ \alpha$
=== $z$ is an increasing mapping ===
Let $A, B \in \powerset S$ with $A \subseteq B$.
We have:
By the Knaster-Tarski Lemma: Power Set, there is a $\mathbb G \in \powerset S$ such that:
:$\map z {\mathbb G} = \mathbb G$
From Relative Complement of Relative Complement we have that $\complement_S \circ \complement_S$ is the identity mapping on $\powerset S$.
Thus we obtain:
At this stage, a diagram can be helpful:
:
Let $h: S \to T$ be the mapping defined as:
:$\forall x \in S: \map h x = \begin {cases}
\map f x & : x \in \mathbb G \\
\map {g^{-1} } x & : x \in \relcomp S {\mathbb G}
\end {cases}$
From the above, we have that:
:$\relcomp S {\mathbb G} \subseteq g \sqbrk T$
Therefore, as $g$ is an injection, it follows that the preimage $\map {g^{-1} } x$ is a singleton.
So $h$ is a bijection by dint of the injective nature of both $f$ and $g^{-1}$.
== Sources ==
* : $\S 8$: Theorem $8.2$
Category:Cantor-Bernstein-Schröder Theorem
Cantor-Bernstein-Schröder Theorem/Proof 5

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ and $g: T \to S$ be injections.
Then there exists a bijection $\phi: S \to T$.
== Proof ==
By Injection to Image is Bijection:
:$g: T \to g \sqbrk T$ is a bijection.
Thus $T$ is equivalent to $g \sqbrk T$.
By Composite of Injections is Injection $g \circ f: S \to g \sqbrk T \subset S$ is also an injection (to a subset of the domain of $g \circ f$).
Then by Cantor-Bernstein-Schröder Theorem: Lemma:
:There exists a bijection $h: S \to g \sqbrk T$.
Thus $S$ is equivalent to $g \sqbrk T$.
We already know that $T$ is equivalent to $g \sqbrk T$.
Thus by Set Equivalence behaves like Equivalence Relation, $S$ is equivalent to $T$.
By the definition of set equivalence:
:There is a bijection $\phi: S \to T$.
Category:Cantor-Bernstein-Schröder Theorem
Cantor-Bernstein-Schröder Theorem/Proof 6

== Theorem ==
Let $A$ and $B$ be sets.
Let $f: A \to B$ and $g: B \to A$ be injections.
Then there is a bijection $h: A \to B$; so that $A$ and $B$ are equivalent.
Furthermore:
:For all $x \in A$ and $y \in B$, if $y = \map h x$ then either $y = \map f x$ or $x = \map g y$.
== Proof ==
Let $\powerset A$ be the power set of $A$.
Define a mapping $E: \powerset A \to \powerset A$ thus:
:$\map E S = A \setminus g \sqbrk {B \setminus f \sqbrk S}$
=== $E$ is increasing ===
Let $S, T \in \powerset A$ such that $S \subseteq T$.
Then:
That is, $\map E S \subseteq \map E T$.
By the Knaster-Tarski Lemma, $E$ has a fixed point $X$.
By the definition of fixed point:
:$\map E X = X$
Thus by the definition of $E$:
:$A \setminus g \sqbrk {B \setminus f \sqbrk X} = X$
Therefore:
:$(1): \quad A \setminus \paren {A \setminus g \sqbrk {B \setminus f \sqbrk X} } = A \setminus X$
Since $g$ is a mapping into $A$:
:$g \sqbrk {B \setminus f \sqbrk X} \subseteq A$
Thus by Relative Complement of Relative Complement:
:$A \setminus \paren {A \setminus g \sqbrk {B \setminus f \sqbrk X} } = g \sqbrk {B \setminus f \sqbrk X}$
Thus by $(1)$:
:$g \sqbrk {B \setminus f \sqbrk X} = A \setminus X$
Let $f' = f \restriction_{X \times f \sqbrk X}$ be the restriction of $f$ to $X \times f \sqbrk X$.
Similarly, let $g' = g \restriction_{\paren {B \setminus f \sqbrk X} \times \paren {A \setminus X} } = g \restriction_{\paren {B \setminus f \sqbrk X} \times g \sqbrk {B \setminus f \sqbrk X} }$.
By Injection to Image is Bijection, $f'$ and $g'$ are both bijections.
Define a relation $h: A \to B$ by $h = f' \cup {g'}^{-1}$.
We will show that $h$ is a bijection from $A$ onto $B$.
The domain of $f'$ is $X$, which is disjoint from the codomain, $A \setminus X$, of $g'$.
The domain of $g'$ is $B \setminus f \sqbrk X$, which is disjoint from the codomain, $f \sqbrk X$, of $f'$.
Let $h = f' \cup {g'}^{-1}$.
By the corollary to Union of Bijections with Disjoint Domains and Codomains is Bijection:
: $h$ is a bijection from $X \cup \paren {A \setminus X}$ onto $f \sqbrk X \cup \paren {B \setminus f \sqbrk X}$.
By Union with Relative Complement, $h$ is a bijection from $A$ onto $B$.
Since $f' \subseteq f$ and $g' \subseteq g$, each element of $h$ is an element of $f$ or of $g^{-1}$.
That is, if $y = \map h x$ then either $y = \map f x$ or $x = \map g y$.
== Sources ==
* : Theorem $9.2.1$: Proof $2$
Category:Cantor-Bernstein-Schröder Theorem
Cantor Pairing Function is Primitive Recursive

== Theorem ==
The Cantor pairing function is primitive recursive.
== Proof ==
The Cantor pairing function $\pi : \N^2 \to \N$ is defined as:
:$\map \pi {m, n} = \frac 1 2 \paren {m + n} \paren {m + n + 1} + m$
As Cantor Pairing Function is Well-Defined, the function could also be defined as:
:$\map \pi {m, n} = \map {\operatorname{quot}} {\paren {m + n} \paren {m + n + 1}, 2} + m$
which is primitive recursive, as it is obtained by substitution from:
* Addition is Primitive Recursive
* Quotient is Primitive Recursive
* Constant Function is Primitive Recursive
Category:Cantor Pairing Function
Cantor Pairing Function is Well-Defined

== Theorem ==
The Cantor pairing function is well-defined.
== Proof ==
By definition, the Cantor pairing function $\pi : \N^2 \to \N$ is:
:$\ds \map \pi {m, n} = \frac 1 2 \paren {m + n} \paren {m + n + 1} + m$
It suffices to show that, for every $m, n \in \N$:
:$\paren {m + n} \paren {m + n + 1}$
is divisible by $2$.
Suppose that $m + n$ is even.
Then, by definition, $2 \divides \paren {m + n}$.
Thus, by Divisor Divides Multiple:
:$2 \divides \paren {m + n} \paren {m + n + 1}$
Suppose that $m + n$ is not even.
Then, by definition, $m + n$ is odd.
But then, by definition, there exists some $k \in \Z$ such that:
:$m + n = 2k + 1$
Then:
By definition, $2 \paren {k + 1}$ is even.
But then, by definition, $2 \divides 2 \paren {k + 1}$.
Therefore, by Divisor Divides Multiple:
:$2 \divides \paren {2k + 1} \paren {2 \paren {k + 1} } = \paren {m + n} \paren {m + n + 1}$
By Proof by Cases, $\paren {m + n} \paren {m + n + 1}$ is divisible by $2$.
Category:Cantor Pairing Function
Cantor Set has Zero Lebesgue Measure

== Theorem ==
Let $\CC$ be the Cantor set.
Let $\lambda$ denote the Lebesgue measure on the Borel $\sigma$-algebra $\map \BB \R$ on $\R$.
Then $\CC$ is $\map \BB \R$-measurable, and $\map \lambda \CC = 0$.
That is, $\CC$ is a $\lambda$-null set.
== Proof ==
Consider the definition of $\CC$ as a limit of a decreasing sequence.
In the notation as introduced there, we see that each $S_n$ is a collection of disjoint closed intervals.
From Closed Set Measurable in Borel Sigma-Algebra, these are measurable sets.
Furthermore, each $S_n$ is finite.
Hence by Sigma-Algebra Closed under Union, it follows that $C_n := \ds \bigcup S_n$ is measurable as well.
Then, as we have:
:$\CC = \ds \bigcap_{n \mathop \in \N} C_n$
it follows from Sigma-Algebra Closed under Countable Intersection that $\CC$ is measurable.
The $C_n$ also form a decreasing sequence of sets with limit $\CC$.
Thus, from Characterization of Measures: $(3')$, it follows that:
:$\map \lambda \CC = \ds \lim_{n \mathop \to \infty} \map \lambda {C_n}$
It is not too hard to show that, for all $n \in \N$:
:$\map \lambda {C_n} = \paren {\dfrac 2 3}^n$
Now we have by Sequence of Powers of Number less than One that:
:$\ds \lim_{n \mathop \to \infty} \paren {\frac 2 3}^n = 0$
and the result follows.
== Sources ==
* : $\S 7$: Problem $10 \ \text{(iv)}$
Category:Cantor Set
Category:Lebesgue Measure
Cantor Set is Closed in Real Number Space

== Theorem ==
Let $\CC$ be the Cantor set.
Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$.
Then $\CC$ is a closed subset of $\struct {\R, \tau_d}$.
== Proof ==
By definition, the Cantor set is the complement of a union of open sets relative to the closed interval $\closedint 0 1$.
By the definition of a topology, that union is itself open in $\R$.
The closed interval $\closedint 0 1$ is itself the complement of a union of open sets $\openint \gets 0 \cup \openint 1 \to$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $2$
Category:Cantor Set
Cantor Set is Uncountable/Proof 1

== Theorem ==
== Proof ==
From the definition as a ternary representation, $\CC$ consists of all the elements of $\closedint 0 1$ which can be written without using a $1$.
So let $x \in \CC$. Then in base $3$ notation, we have (as $0 \le x \le 1$):
:$\ds x = \sum_{i \mathop = 1}^\infty r_j 3^{-j}$
From the definition of the Cantor set, we have $\forall j: r_j \in \set {0, 2}$.
Furthermore, from Representation of Ternary Expansions, the $r_j$ are unique.
Now define the following function:
:$\ds f: \CC \to \closedint 0 1, \quad \map f {\sum_{i \mathop = 1}^\infty r_j 3^{-j} } = \sum_{i \mathop = 1}^\infty \frac {r_j} 2 2^{-j}$
Observe that $\forall j: \dfrac {r_j} 2 \in \set {0, 1}$.
That the expression is in fact an element of $\closedint 0 1$ now follows from binary notation.
Furthermore by Existence of Base-N Representation, any element $y$ of $\closedint 0 1$ may be written in the following form (where $\forall j: b_j \in \set {0, 1}$):
:$\ds y = \sum_{i \mathop = 1}^\infty b_j 2^{-j}$
Obviously, $y = \map f x$, where $x \in \CC$ is defined as follows:
:$\ds x = \sum_{i \mathop = 1}^\infty 2 b_j 3^{-j}$
It follows that $f$ is surjective.
From Closed Interval in Reals is Uncountable, the closed interval $\closedint 0 1$ is uncountable.
The result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $5$
Category:Cantor Set is Uncountable
Cantor Set is Uncountable/Proof 2

== Theorem ==
== Proof ==
It follows from Representation of Ternary Expansions that every string of the form $0.nnnnn \ldots$ where $n \in \set {0, 2}$ is an element of $\CC$.
We also have that every string of the form $0.nnnnn \ldots$ where $n \in \set {0, 1}$ is an element of $\closedint 0 1 \subset \R$ expressed in binary notation.
Let $f: \CC \to \closedint 0 1$ be the function defined by:
:$\forall x \in \CC: \map f x = \text{ the number obtained by replacing every } 2 \text { in } x \text { with a } 1$
where $x$ is expressed in base $3$ notation.
It is clear from the above that $f$ is a surjection.
== Sources ==
* : $\S 7$: Problem $10 \ \text{(vii)}$
Category:Cantor Set is Uncountable
Cantor Space as Countably Infinite Product

== Theorem ==
Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$.
Let $\ds A = \prod_{n \mathop = 1}^\infty A_n$.
Let $\struct {A, \tau}$ be the product space where $\tau$ is the product topology on $A$.
Then $A$ is homeomorphic to the Cantor space.
== Proof ==
In $\CC$, basis elements are sets of the form $\set {y: \size {x - y} < \epsilon}$ for $x \in \CC$ and some $\epsilon \in \R_{>0}$.
In $\ds \prod_{n \mathop = 1}^\infty A_n$, sets of the form $\set {\sequence {a_i} \in \prod A_n: a_i \text { is fixed for } 1 \le i \le n}$ forms a basis for the product topology.
Consider the function $f$ taking each point from $\tuple {a_1, a_2, \ldots, a_n, \ldots}$ in $\prod A_n$ to the point $0 \cdotp a_1 a_2 \ldots a_n \ldots_3$ in $\CC$.
As $f$ takes basis elements to basis elements, both $f$ and $f^{-1}$ are seen to be continuous.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $8$
Category:Cantor Space
Category:Product Spaces
Cantor Space is Compact

== Theorem ==
Let $\CC$ be the Cantor set.
Let $\struct {\R, \tau_d}$ be the real number space $\R$ under the Euclidean topology $\tau_d$.
Then $\CC$ is a compact subset of $\struct {\R, \tau_d}$.
== Proof ==
We have Cantor Set is Closed in Real Number Space.
Taking, for example, $0 \in \CC$ and $1 \in \R$ it is clear that:
:$\forall x \in \CC: \map d {0, x} \le 1$
and so $\CC$ is bounded.
Hence the result from the Heine-Borel Theorem.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $2$
* : $\S 7$: Problem $10 \ \text{(ii)}$
Category:Cantor Space
Category:Compact Spaces
Cantor Space is Complete Metric Space

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is a complete metric space.
== Proof ==
We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space.
We also have Cantor Space is Compact.
The result follows from Compact Metric Space is Complete.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $2$
Category:Cantor Space
Category:Complete Metric Spaces
Cantor Space is Dense-in-itself

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is dense-in-itself.
== Proof ==
Let $U \in \tau_d$ be open in $T$.
Let $p \in U$.
Then $\exists x \in U: \exists \epsilon \in \R: \map d {x, p} < \epsilon$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $3$
Category:Cantor Space
Category:Denseness
Cantor Space is Meager in Closed Unit Interval

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is meager in $\closedint 0 1$.
== Proof ==
From Cantor Space is Nowhere Dense, $T$ is nowhere dense in $\closedint 0 1$.
So, trivially, $\CC$ is the union of nowhere dense subsets of $\closedint 0 1$.
Hence the result from definition of meager.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $4$
Category:Cantor Space
Category:Meager Spaces
Cantor Space is Non-Meager in Itself

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is non-meager in itself.
== Proof ==
We have that the Cantor Space is Complete Metric Space.
By Baire Category Theorem, a complete metric space is also a Baire space.
The result then follows by Baire Space is Non-Meager.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $4$
Category:Cantor Space
Category:Non-Meager Spaces
Cantor Space is Nowhere Dense

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is nowhere dense in $\closedint 0 1$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 7$: Problem $10 \ \text{(v)}$
Category:Cantor Space
Category:Nowhere Denseness
Category:Cantor Space is Nowhere Dense
Cantor Space is Nowhere Dense/Proof 1

== Theorem ==
== Proof ==
From Cantor Set is Closed in Real Number Space, $\CC$ is closed.
So from Closed Set equals its Closure:
:$\CC^- = \CC$
where $\CC^-$ denotes the closure of $\CC$.
Let $0 \le a < b \le 1$.
Then $I = \openint a b$ is an open interval of $\closedint 0 1$.
Let $\epsilon = b - a$.
Clearly $\epsilon > 0$.
Let $n \in \N$ such that $3^{-n} < \epsilon$.
So there exists an open interval of $\closedint 0 1$ which has been deleted from $\closedint 0 1$ during the process of creating $\CC$.
Thus no open interval of $\closedint 0 1$ is disjoint from all the open intervals deleted from $\closedint 0 1$.
So an open interval of $\closedint 0 1$ can not be a subset of $\CC = \CC^-$.
Hence the result, by definition of nowhere dense.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $4$
Category:Cantor Space is Nowhere Dense
Cantor Space is Nowhere Dense/Proof 2

== Theorem ==
== Proof ==
Let $S_n$ and $C_n$ be as in the definition of the Cantor set as a limit of a decreasing sequence.
Then the length of every interval in $S_n$ is seen to be $\dfrac 1 {3^n} = 3^{-n}$.
Let $0 \le a < b \le 1$.
Then $\openint a b \subseteq \closedint 0 1$ is an open interval.
Let $n \in \N$ such that $3^{-n} < b - a$, so that the length of every interval in $S_n$ is $3^{-n} < b - a$.
Therefore, as the intervals in $S_n$ do not overlap, no interval of length $b - a$ is contained in $C_n = \ds \bigcup S_n$. 
Consequently, as $\CC \subseteq C_n$, no interval of length $b - a$ is contained in $\CC$.
Since the interval $\openint a b$ was of arbitrary length, there do not exist any open intervals in $\CC$.
Hence the result, by definition of nowhere dense.
Category:Cantor Space is Nowhere Dense
Cantor Space is Perfect

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $\CC$ is a perfect set of the real number space $\R$ under the usual (Euclidean) topology $\tau_d$.
== Proof ==
From Cantor Space is Dense-in-itself, $\CC$ contains no isolated points.
We also have that the Cantor Set is Closed in Real Number Space.
The result follows from the definition of perfect set.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $3$
Category:Cantor Space
Category:Perfect Sets
Cantor Space is Second-Countable

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is a second-countable space.
== Proof ==
We have that the Cantor space is a topological subspace of the real number space with the usual (Euclidean) topology $\struct {\R, \tau_d}$.
We also have that the Real Number Line is Second-Countable.
The result follows from Second-Countability is Hereditary.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $2$
Category:Cantor Space
Category:Second-Countable Spaces
Cantor Space is Totally Separated

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is totally separated.
== Proof ==
Let $a, b \in \CC$ such that $a < b$.
Then $b - a = \epsilon$.
Consider $n \in \N$ such that $3^{-n} < \epsilon$.
So $\exists r \in \R: a < r < b, r \notin \CC$.
Let $A = \CC \cap \hointr 0 r$ and $B = \CC \cap \hointl r 1$.
Thus $A \mid B$ is a separation of $\CC$ such that $a \in A, b \in B$.
Such a separation can be found for any two distinct $a, b \in \CC$.
Hence the result by definition of totally separated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $6$
Category:Cantor Space
Category:Totally Separated Spaces
Cantor Space is not Extremally Disconnected

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is not extremally disconnected.
== Proof ==
From Cantor Space satisfies all Separation Axioms we have that $T$ is a $T_2$ (Hausdorff) space.
Consider the real number $\dfrac 1 4 = 0.020202 \ldots_3$.
We have that:
:$C_1 := \CC \cap \hointr 0 {\dfrac 1 4}$
:$C_2 := \CC \cap \hointl {\dfrac 1 4} 1$
are disjoint sets both of which are open sets of $T$.
However, as $\dfrac 1 4 \in \CC$ we have that:
:$\dfrac 1 4 \in C_1^-$
and
:$\dfrac 1 4 \in C_2^-$
where $C_1^-$ and $C_2^-$ are the closures of $C_1$ and $C_2$ in $T$.
Thus $C_1^- \cap C_2^- \ne \O$.
The result follows from Extremally Disconnected by Disjoint Open Sets.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $7$
Category:Cantor Space
Category:Extremally Disconnected Spaces
Cantor Space is not Locally Connected

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is not locally connected.
== Proof ==
Let $\BB$ be a basis of $T$.
Let $A \in \BB$.
By definition of $\BB$, $A$ is an open set of $T$.
But the Cantor Space is Totally Separated.
Therefore $A$ is not a connected set.
Hence the result from definition of a locally connected space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $9$
Category:Cantor Space
Category:Locally Connected Spaces
Cantor Space is not Scattered

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ is not scattered.
== Proof ==
By definition, $T$ is scattered it contains no non-empty subset which is dense-in-itself.
We have that Cantor Space is Dense-in-itself.
Hence the result by definition of a scattered space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $3$
Category:Cantor Space
Category:Scattered Spaces
Cantor Space satisfies all Separation Axioms

== Theorem ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Then $T$ satisfies all the separation axioms.
== Proof ==
We have that the Cantor space is a metric subspace of the real number space $\R$, and hence a metric space.
The result follows from Metric Space fulfils all Separation Axioms.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $2$
Category:Cantor Space
Category:Separation Axioms