avewright/proofwiki-math huggingface dataset:
Theorem: Metacompact Space is not necessarily Paracompact

A metacompact topological space is not necessarily also a paracompact space.

Proof:
Let $T$ be the Dieudonné plank.
From Dieudonné Plank is Metacompact, $T$ is a metacompact space.
From Dieudonné Plank is not Paracompact, $T$ is not a paracompact space.
Hence the result.
{{qed}}
Theorem: Paracompact Countably Compact Space is Compact

Let $T = \struct {S, \tau}$ be a countably compact space which is also paracompact.
Then $T$ is compact.

Proof:
From the definition of paracompact space, a paracompact space is also a metacompact space.
The result follows from Metacompact Countably Compact Space is Compact.
{{qed}}
Theorem: Minimal Hausdorff Space is not necessarily Compact

Let $S$ be a set.
Let $\tau$ be the minimal subset of the power set $\powerset S$ such that $\struct {S, \tau}$ is a Hausdorff space.
Then it is not necessarily the case that $\struct {S, \tau}$ is compact.

Proof:
Let $T = \struct {S, \tau}$ be the canonical minimal Hausdorff non-compact space.
This space has been so named on {{ProofWiki}} in order to allow reference to it without needing to describe it whenever it is mentioned.
By Canonical Minimal Hausdorff Non-Compact Space is Minimal Hausdorff, $\tau$ is the minimal subset of the power set $\powerset S$ such that $T$ is a Hausdorff space.
By Canonical Minimal Hausdorff Non-Compact Space is not Compact, $T$ is not a compact topological space.
Hence the result.
{{qed}}
Theorem: Maximal Compact Topological Space is not necessarily Hausdorff

Let $S$ be a set.
Let $\tau$ be the maximal subset of the power set $\powerset S$ such that $\struct {S, \tau}$ is a compact topological space.
Then it is not necessarily the case that $\struct {S, \tau}$ is a Hausdorff space.

Proof:
Let $T = \struct {S, \tau}$ be the canonical maximal compact non-Hausdorff space.
This space has been so named on {{ProofWiki}} in order to allow reference to it without needing to describe it whenever it is mentioned.
By Canonical Maximal Compact Non-Hausdorff Space is Maximal Compact, $\tau$ is the maximal subset of the power set $\powerset S$ such that $T$ is compact.
By Canonical Maximal Compact Non-Hausdorff Space is not Hausdorff, $T$ is not a  Hausdorff space.
Hence the result.
{{qed}}
Theorem: Infinite Product of Sigma-Compact Spaces is not always Sigma-Compact

Let $I$ be an indexing set with infinite cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be $\sigma$-compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also $\sigma$-compact.

Proof:
Let $T = \struct {\Z_{\ge 0}, \tau}$ be the topological space formed by the discrete topology on the set of positive integers.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \Z_{\ge 0} } \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the countable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $\Z_{\ge 0}$ with the product topology $\tau'$.
From Countable Discrete Space is Sigma-Compact, $T$ is $\sigma$-compact.
From Countable Product of Countable Discrete Spaces is not Sigma-Compact, $T'$ is not $\sigma$-compact.
Hence the result.
{{qed}}
Theorem: Uncountable Product of Sequentially Compact Spaces is not always Sequentially Compact

Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be sequentially compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also sequentially compact.

Proof:
Let $\mathbb I$ denote the closed unit interval.
Let $T = \struct {\mathbb I, \tau}$ be the topological space consisting of $\mathbb I$ under the usual (Euclidean) topology.
Let $T' = \mathbb I^{\mathbb I} = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\mathbb I, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\mathbb I, \tau}$ indexed by $\mathbb I$ with the product topology $\tau'$.
From Closed Real Interval is Sequentially Compact, $T$ is  sequentially compact.
{{explain|The above has not been proven yet, but I think it is: $\mathbb I$ is a metric space, and a compact metric space is sequentially compact.}}
But from Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.
Hence the result.
{{qed}}
Theorem: Product of Countably Compact Spaces is not always Countably Compact

Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be countably compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also countably compact.

Proof:
Let $T$ denote the Novak space.
Let $T \times T$ denote the Cartesian product of the Novak space with itself under the product topology.
From Novak Space is Countably Compact, $T$ is a countably compact space.
But from Cartesian Product of Novak Spaces is not Countably Compact, $T \times T$ is not a countably compact space.
Hence the result.
{{qed}}
Theorem: Infinite Product of Weakly Locally Compact Spaces is not always Weakly Locally Compact

Let $I$ be an indexing set with infinite cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be weakly locally compact-compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also weakly locally compact.

Proof:
Let $T = \struct {\Z_{\ge 0}, \tau}$ be the topological space formed by the discrete topology on the set of positive integers.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \Z_{\ge 0} } \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the countable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $\Z_{\ge 0}$ with the product topology $\tau'$.
From Discrete Space is Strongly Locally Compact, $T$ is strongly locally compact.
From Strongly Locally Compact Space is Weakly Locally Compact, $T$ is weakly locally compact.
From Countable Product of Countable Discrete Spaces is not  Weakly Locally Compact, $T'$ is not weakly locally compact.
Hence the result.
{{qed}}
Theorem: Product of Lindelöf Spaces is not always Lindelöf

Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a Lindelöf space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also Lindelöf space.

Proof:
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space.
From Sorgenfrey's Half-Open Square Topology is Not Lindelöf, $T'$ is not a Lindelöf space.
Hence the result.
{{qed}}
Theorem: Uncountable Product of First-Countable Spaces is not always First-Countable

Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a first-countable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a first-countable space.

Proof:
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Discrete Space is First-Countable, $T$ is a first-countable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not First-Countable, $T$ is not a first-countable space.
Hence the result.
{{qed}}
Theorem: Uncountable Product of Second-Countable Spaces is not always Second-Countable

Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a second-countable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a second-countable space.

Proof:
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Second-Countable, $T$ is a second-countable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
{{qed}}
Theorem: Uncountable Product of Separable Spaces is not always Separable

Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a separable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a separable space.

Proof:
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $\ds T' = \struct {\prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Separable, $T$ is a separable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Separable, $T$ is not a separable space.
Hence the result.
{{qed}}
Theorem: Product of Paracompact Spaces is not always Paracompact

Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a paracompact space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also paracompact space.

Proof:
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line is Paracompact, $T$ is a paracompact space.
From Sorgenfrey's Half-Open Square Topology is Not Paracompact, $T'$ is not a paracompact space.
Hence the result.
{{qed}}
Theorem: Product of Metacompact Spaces is not always Metacompact

Let $I$ be an indexing set.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a metacompact space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also metacompact space.

Proof:
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line is Metacompact, $T$ is a metacompact space.
From Sorgenfrey's Half-Open Square Topology is Not Metacompact, $T'$ is not a metacompact space.
Hence the result.
{{qed}}
Theorem: Paracompactness is not always Preserved under Open Continuous Mapping

Let $T_A = \struct {X_A, \tau_A}$ be a topological space which is paracompact.
Let $T_B = \struct {X_B, \tau_B}$ be another topological space.
Let $\phi: T_A \to T_B$ be a mapping which is both continuous and open.
Then it is not necessarily the case that $T_B$ is also paracompact.

Proof:
We have Open Continuous Image of Paracompact Space is not always Countably Metacompact.
We also have:
:Paracompact Space is Metacompact
:Metacompact Space is Countably Metacompact
Hence the result.
{{qed}}
Theorem: Union of Closure with Closure of Complement is Whole Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $S \setminus H$ denote the complement of $H$ relative to $S$.
Then:
:$H^- \cup \paren {S \setminus H}^- = S$

Proof:
We have that:
:$H^- \cup \paren {S \setminus H}^- \subseteq S$
by definition of $S$.
From Union with Relative Complement:
:$H \cup \paren {S \setminus H} = S$
From Set is Subset of its Topological Closure:
{{begin-eqn}}
{{eqn | l = H
      | o = \subseteq
      | r = H^-
}}
{{eqn | l = S \setminus H
      | o = \subseteq
      | r = \paren {S \setminus H}^-
}}
{{end-eqn}}
From Set Union Preserves Subsets:
:$H \cup \paren {S \setminus H} \subseteq H^- \cup \paren {S \setminus H}^-$
which means:
:$S \subseteq H^- \cup \paren {S \setminus H}^-$
The result follows by definition of equality of sets.
{{qed}}
Category:Set Closures
Category:Set Union
Category:Relative Complement
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Theorem: Component of Point is not always Intersection of its Clopen Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $\map {\operatorname {Comp}_x} T$ denote the component of $x$ in $T$.
Let $K_x = \ds \bigcap_{x \mathop \in K} K$ clopen in $T$.
Then it is not always the case that $\map {\operatorname {Comp}_x} T = K_x$

Proof:
Note that from Clopen Set contains Components of All its Points:
:$\map {\operatorname {Comp}_x} T \subseteq K_x$
It remains to be demonstrated that it is not always the case that $K_x \subseteq \map {\operatorname {Comp}_x} T$.
Let $T$ be the nested rectangle space in the Euclidean plane.
Let $L_1$ and $L_2$ be the boundary lines of $T$.
Let $x \in L_1$.
From Boundary Line in Nested Rectangle Space is Component $L_1$ is a component of $T$.
That is:
:$L_1 = \map {\operatorname {Comp}_x} T$ 
From Union of Boundary Lines in Nested Rectangle Space is Quasicomponent, $L_1 \cup L_2$ is a quasicomponent of $T$.
By Quasicomponent is Intersection of Clopen Sets:
:$L_1 \cup L_2 = \ds \bigcap_{x \mathop \in K} K$ clopen in $T$
Thus, while $\map {\operatorname {Comp}_x} T \subseteq L_1 \cup L_2$, it is not the case that $L_1 \cup L_2 \subseteq \map {\operatorname {Comp}_x} T$.
Hence the result.
{{qed}}
Theorem: Complement of Clopen Set is Clopen

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a clopen set of $T$.
Let $\relcomp S H$ denote the complement of $H$ relative to $S$.
Then $\relcomp S H$ is also a clopen set of $T$.

Proof:
By definition of clopen, $H$ is open in $T$.
By definition of closed set, $\relcomp S H$ is closed in $T$.
By definition of clopen, $H$ is closed in $T$. 
By definition of closed set, $\relcomp S H$ is open in $T$.
Thus $\relcomp S H$ is both open in $T$ and closed in $T$.
Hence the result, by definition of clopen set.
{{qed}}
Category:Relative Complement
Category:Clopen Sets
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Theorem: Clopen Set and Complement form Separation

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a clopen set of $T$.
Let $\relcomp S H$ be the complement of $H$ relative to $S$.
Then $H$ and $\relcomp S H$ form a separation of $T$.

Proof:
By Set with Relative Complement forms Partition, $H$ and $\relcomp S H$ form a partition of $S$.
By Complement of Clopen Set is Clopen, $\relcomp S H$ is also a clopen set of $T$.
By definition of clopen set, both $H$ and $\relcomp S H$ are open in $T$.
Thus $H$ and $\relcomp S H$ are a pair of open sets in $T$ forming a partition of $S$.
Hence the result, by definition of separation.
{{qed}}
Category:Clopen Sets
Category:Separations
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Theorem: Path Component is not necessarily Injective Path Component

Let $T = \struct {S, \tau}$ be a topological space.
Let $P$ be a path component of $T$.
Then it is not necessarily the case that $P$ is also an injective path component of $T$.

Proof:
Let $T = \struct {S, \tau_p}$ be a finite particular point space.
From Particular Point Space is Path-Connected, $T$ is path-connected.
Therefore $S$ is a path component in $T$.
But from Particular Point Space is not Injectively Path-Connected, $T$ is not injectively path-connected.
Therefore $S$ is not an injective path component in $T$.
Hence the result.
{{qed}}
Theorem: Component is not necessarily Path Component

Let $T = \struct {S, \tau}$ be a topological space.
Let $C$ be a component of $T$.
Then it is not necessarily the case that $C$ is also an path component of $T$.

Proof:
Let $C$ be the closed topologist's sine curve embedded in the real Euclidean plane.
From Closed Topologist's Sine Curve is Connected, $C$ is connected in $T$
Therefore $C$ is a component in the subspace of $T$ induced by $C$.
From Closed Topologist's Sine Curve is not Path-Connected, $C$ is not path-connected.
Therefore $C$ is not a path component in the subspace of $T$ induced by $C$.
Hence the result.
{{qed}}
Theorem: Quasicomponent is not necessarily Component

Let $T = \struct {S, \tau}$ be a topological space.
Let $Q$ be a quasicomponent of $T$.
Then it is not necessarily the case that $C$ is also a component of $T$.

Proof:
From Component of Point is not always Intersection of its Clopen Sets, the set intersection of the clopen sets containing a point $x$ may not always be contained in the component of $x$.
The result follows from Quasicomponent is Intersection of Clopen Sets.
{{qed}}
Theorem: Simple Infinite Continued Fraction is Uniquely Determined by Limit

Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be simple infinite continued fractions in $\R$.
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same limit.
Then they are equal.

Proof:
Follows immediately from Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself.
{{qed}}
Theorem: Simple Infinite Continued Fraction is Uniquely Determined by Limit

Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be simple infinite continued fractions in $\R$.
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same limit.
Then they are equal.

Proof:
Recall that by Simple Infinite Continued Fraction Converges, they do indeed have a limit.
The result will be achieved by the Second Principle of Mathematical Induction.
Suppose $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ have the same value.
First we note that if $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ then:
:$a_0 = b_0$
since both are equal to the integer part of the common value.
{{explain|a result proving the above}}
This is our basis for the induction.
Now suppose that for some $k \ge 1$, we have:
:$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$
Then all need to do is show that $a_{k + 1} = b_{k + 1}$.
Now:
:$\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {a_0, a_1, \ldots, a_k, \sqbrk {a_{k + 1}, a_{k + 2}, \ldots} }$
and similarly:
:$\sqbrk {b_0, b_1, b_2, \ldots} = \sqbrk {b_0, b_1, \ldots, b_k, \sqbrk {b_{k + 1}, b_{k + 2}, \ldots} }$
{{explain|this needs to be proved}}
As these have the same value and have the same first $k$ partial denominators, it follows that:
:$\sqbrk {a_{k + 1}, a_{k + 2}, \ldots, } = \sqbrk {b_{k + 1}, b_{k + 2}, \ldots}$
But now $a_{k + 1} = b_{k + 1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.
Hence the result.
{{qed}}
Theorem: Continued Fraction Expansion of Irrational Number Converges to Number Itself

Let $x$ be an irrational number.
Then the continued fraction expansion of $x$ converges to $x$.

Proof:
Let $\sequence {a_0, a_1, \ldots}$ be its continued fraction expansion.
Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be its numerators and denominators.
Then $C_n = p_n / q_n$ is the $n$th convergent.
By Accuracy of Convergents of Continued Fraction Expansion of Irrational Number, for $n \ge 2$:
:$\size {x - \dfrac {p_n} {q_n} } < \dfrac 1 {q_n q_{n + 1} }$
By Lower Bounds for Denominators of Simple Continued Fraction:
:$q_n q_{n + 1} \ge n$ for $n \ge 5$
So from Basic Null Sequences and the Squeeze Theorem:
:$\dfrac 1 {q_n q_{n + 1} } \to 0$
as $n \to \infty$.
Thus $C_n = p_n / q_n$ converges to $x$.
That is, $(a_0, a_1, \ldots)$ converges to $x$.
{{qed}}
Theorem: Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself

Let $\sequence {a_n}_{n \mathop \ge 0}$ be a simple infinite continued fractions in $\R$.
Then $\sequence {a_n}_{n \mathop \ge 0}$ converges to an irrational number, whose continued fraction expansion is $\sequence {a_n}_{n \mathop \ge 0}$.

Proof:
By Simple Infinite Continued Fraction Converges to Irrational Number, the value of $\sequence {a_n}_{n \mathop \ge 0}$ exists and is irrational.
Let $\sequence {b_n}_{n \mathop \ge 0}$ be its continued fraction expansion.
By Continued Fraction Expansion of Irrational Number Converges to Number Itself, $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ have the same value.
The result will be achieved by the Second Principle of Mathematical Induction.
First we note that if $\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {b_0, b_1, b_2, \ldots}$ then $a_0 = b_0$ since both are equal to the integer part of the common value.
{{explain|a result proving the above}}
This is our basis for the induction.
Now suppose that for some $k \ge 1$, we have:
:$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$.
Then all need to do is show that $a_{k + 1} = b_{k + 1}$.
Now:
:$\sqbrk {a_0, a_1, a_2, \ldots} = \sqbrk {a_0, a_1, \ldots, a_k, \sqbrk {a_{k + 1}, a_{k + 2}, \ldots} }$
and similarly:
:$\sqbrk {b_0, b_1, b_2, \ldots} = \sqbrk {b_0, b_1, \ldots, b_k, \sqbrk {b_{k + 1}, b_{k + 2}, \ldots} }$
{{explain|this needs to be proved}}
As these have the same value and have the same first $k$ partial denominators, it follows that:
:$\sqbrk {a_{k + 1}, a_{k + 2}, \ldots} = \sqbrk {b_{k + 1}, b_{k + 2}, \ldots}$
But now $a_{k + 1} = b_{k + 1}$ as each is equal to the integer part of the value of this simple infinite continued fraction.
Hence the result.
{{qed}}