avewright/proofwiki-math huggingface dataset:
Theorem: Eigenvectors of Symmetric Matrix are Orthogonal

Let $K$ be a ring. 
Let $A$ be a symmetric matrix over $K$.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $A$.
Let $\mathbf v_1, \mathbf v_2$ be eigenvectors of $A$ corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$ respectively.
Then $\mathbf v_1$ and $\mathbf v_2$ are orthogonal.

Proof:
We have:
:$A \mathbf v_1 = \lambda_1 \mathbf v_1$
and:
:$A \mathbf v_2 = \lambda_2 \mathbf v_2$
We also have:
{{begin-eqn}}
{{eqn | l = \mathbf v_1 \cdot \paren {A \mathbf v_2}
      | r = \mathbf v_1 \cdot \paren {\lambda_2 \mathbf v_2}
}}
{{eqn | r = \lambda_2 \mathbf v_1 \cdot \mathbf v_2
      | c = Dot Product Operator is Bilinear
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \mathbf v_1 \cdot \paren {A \mathbf v_2}
      | r = \mathbf v_1^\intercal \paren {A \mathbf v_2}
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = \paren {A^\intercal \mathbf v_1}^\intercal \mathbf v_2
      | c = Transpose of Matrix Product
}}
{{eqn | r = \paren {A \mathbf v_1}^\intercal \mathbf v_2
      | c = {{Defof|Symmetric Matrix}}
}}
{{eqn | r = \paren {A \mathbf v_1} \cdot \mathbf v_2
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = \paren {\lambda_1 \mathbf v_1} \cdot \mathbf v_2
}}
{{eqn | r = \lambda_1 \mathbf v_1 \cdot \mathbf v_2
      | c = Dot Product Operator is Bilinear
}}
{{end-eqn}}
We therefore have:
:$\lambda_2 \mathbf v_1 \cdot \mathbf v_2 = \lambda_1 \mathbf v_1 \cdot \mathbf v_2$
and therefore:
:$\paren {\lambda_2 - \lambda_1} \innerprod {\mathbf v_1} {\mathbf v_2} = 0$
Since $\lambda_1 \ne \lambda_2$, we have:
:$\innerprod {\mathbf v_1} {\mathbf v_2} = 0$
Hence $\mathbf v_1$ and $\mathbf v_2$ are orthogonal.
{{qed}}
Category:Eigenvectors of Square Matrices
Category:Symmetric Matrices
lpcagecg6k47tj4ucwkli8fwebybpnb
Theorem: Product of Diagonal Matrices is Diagonal

Let $A$ and $B$ be $n \times n$ diagonal matrices. 
Then the matrix product $A B$ is an $n \times n$ diagonal matrix.
Further: 
:$\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$

Proof:
We have: 
:$\ds \paren {A B}_{ij} = \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j}$
Since $A$ and $B$ are diagonal:
:$\paren A_{i k} = 0$ for $i \ne k$,
and:
:$\paren B_{k j} = 0$ for $k \ne j$.
If $i \ne j$, for each $k$ we either have $i \ne k$ or $k \ne j$, so:
:$\paren A_{i k} \paren B_{k j} = 0$ for each $1 \le k \le n$.
Hence if $i \ne j$:
:$\ds \paren {A B}_{i j} = \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j} = 0$
so $A B$ is diagonal.
Let $i = j$.
Then, if $k \ne i$ we have: 
:$\paren A_{i k} \paren B_{k j} = \paren A_{i k} \paren B_{k i} = 0$
so:
:$\ds \sum_{k \mathop = 1}^n \paren A_{i k} \paren B_{k j} = \paren A_{i i} \paren B_{i i}$
We can conclude:
:$\paren {A B}_{i j} = \begin {cases} \paren A_{i i} \paren B_{i i} & i = j \\ 0 & i \ne j \end {cases}$
{{qed}}
Category:Diagonal Matrices
Category:Conventional Matrix Multiplication
0e7x9d3pk7yuar3kubpg041l1g0n1uq
Theorem: Open Real Interval is Homeomorphic to Real Number Line

Let $\R$ be the real number line with the Euclidean topology.
Let $I := \openint a b$ be a non-empty open real interval.
Then $I$ and $\R$ are homeomorphic.

Proof:
By definition of open real interval, for $I$ to be non-empty it must be the case that $a < b$.
In particular it is noted that $a \ne b$.
Thus $a - b \ne 0$.
Let $I' := \openint {-1} 1$ denote the open real interval from $-1$ to $1$.
From Open Real Intervals are Homeomorphic, $I$ and $I'$ are homeomorphic.
Consider the real function $f: I' \to \R$ defined as:
:$\forall x \in I': \map f x = \dfrac x {1 - \size x}$
Then after some algebra:
:$\forall x \in \R: \map {f^{-1} } x = \dfrac x {1 + \size x}$
Both of these are defined, as $\size x < 1$.
By the Combination Theorem for Continuous Real Functions, both $f$ and $f^{-1}$ are continuous on the open real intervals on which they are defined.
Thus by definition $I'$ and $\R$ are homeomorphic.
From Homeomorphism Relation is Equivalence it follows that $I$ and $\R$ are homeomorphic.
{{qed}}
Theorem: Open Real Interval is Homeomorphic to Real Number Line

Let $\R$ be the real number line with the Euclidean topology.
Let $I := \openint a b$ be a non-empty open real interval.
Then $I$ and $\R$ are homeomorphic.

Proof:
Let $I := \openint {-\dfrac \pi 2} {\dfrac \pi 2}$ denote the open real interval from $-\dfrac \pi 2$ to $\dfrac \pi 2$.
Consider the real function $f: I \to \R$ defined as:
:$\forall x \in I: \map f x = \tan x$
Then we have:
:$\forall x \in \R: \map {f^{-1} } x = \arctan x$
From Homeomorphism Relation is Equivalence it follows that $I$ and $\R$ are homeomorphic.
Then by Open Real Intervals are Homeomorphic, $I$ is homeomorphic to every other open real interval.
{{qed}}
Theorem: Closed Unit Interval is Homeomorphic to Letter L

Let $\R$ be the real number line under the Euclidean metric.
Let $\Bbb I := \closedint 0 1$ be the closed unit interval.
Let $\mathsf L \subseteq \R^2$ denote the letter $L$:
:$\mathsf L := \closedint 0 1 \times \set 0 \cup \set 0 \times \closedint 0 1$
Then $\Bbb I$ and $\mathsf L$ are homeomorphic.

Proof:
:thumbright
Consider the mapping $f: \Bbb I \to \mathsf L$ defined as:
:$\forall x \in \Bbb I: \map f x = \begin {cases} \tuple {0, 1 - 2 x} & : x \in \closedint 0 {\dfrac 1 2} \\ \tuple {2 x - 1, 0} & : x \in \closedint {\dfrac 1 2} 1 \end {cases}$
It is seen that:
:$f \closedint 0 {\dfrac 1 2} = \set 0 \times \closedint 0 1$
and:
:$f \closedint {\dfrac 1 2} 1 = \closedint 0 1 \times \set 0$
such that:
:$\map f {\dfrac 1 2} = \tuple {0, 0}$
By construction, $f$ is a bijection.
Its continuity follows from the Combination Theorem for Continuous Real Functions.
{{qed}}
Theorem: Letter L and Letter T are not Homeomorphic

Let $\R^2$ denote the real number plane under the Euclidean topology.
Let $\mathsf L \subseteq \R^2$ denote the letter $L$:
:$\mathsf L := \closedint 0 1 \times \set 0 \cup \set 0 \times \closedint 0 1$
Let $\mathsf T \subseteq \R^2$ denote the letter $T$:
:$\mathsf T := \closedint {-1} 1 \times \set 0 \cup \set 0 \times \closedint 0 1$
Then $\mathsf L$ and $\mathsf T$ are not homeomorphic.

Proof:
{{AimForCont}} $f: \mathsf T \to \mathsf L$ is a homeomorphism.
Let $g$ be the restriction of $f$ to $\mathsf T \setminus \set \bszero$, where $\bszero := \tuple {0, 0}$ denotes the origin of $\R^2$.
Then from Restriction of Homeomorphism is Homeomorphism, $g$ is also a homeomorphism.
But $\bszero$ is the junction point of $\mathsf T$, which means that $\mathsf T \setminus \set \bszero$ consists of $3$ disjoint half-open intervals.
However, no matter where $\map f x$ is located in $\mathsf L$, the set $\mathsf L \setminus \set {\map f x}$ consists either of $1$ or $2$ half-open intervals.
{{finish|It remains to be shown that spaces consisting of different numbers of half-open intervals are not homeomorphic.}}
This contradicts the supposition that such a homeomorphism $f$ exists.
Hence, by Proof by Contradiction, $\mathsf L$ and $\mathsf T$ are not homeomorphic.
{{qed}}
Theorem: Real Symmetric Positive Definite Matrix has Positive Eigenvalues

Let $A$ be a symmetric positive definite matrix over $\mathbb R$.
Let $\lambda$ be an eigenvalue of $A$. 
Then $\lambda$ is real with $\lambda > 0$.

Proof:
Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf v$ be a corresponding eigenvector.
From Eigenvalues of Symmetric Matrix are Real, $\lambda$ is real.
From the definition of a positive definite matrix, we have: 
:$\mathbf v^\intercal A \mathbf v > 0$
That is: 
{{begin-eqn}}
{{eqn | l = 0
      | o = <
      | r = \mathbf v^\intercal A \mathbf v
}}
{{eqn | r = \mathbf v^\intercal \paren {\lambda \mathbf v}
      | c = {{Defof|Eigenvector of Real Square Matrix}}
}}
{{eqn | r = \lambda \paren {\mathbf v^\intercal \mathbf v}
}}
{{eqn | r = \lambda \paren {\mathbf v \cdot \mathbf v}
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = \lambda \norm {\mathbf v}^2
      | c = Dot Product of Vector with Itself
}}
{{end-eqn}}
From Euclidean Space is Normed Vector Space, we have: 
:$\norm {\mathbf v}^2 > 0$
so:
:$\lambda > 0$
{{qed}}
Category:Symmetric Matrices
Category:Positive Definite Matrices
nkkc7wel5v7i91gettowe6i5mjo0p4d
Theorem: Characterisation of Real Symmetric Positive Definite Matrix/Necessary Condition

Let $A$ be an $n \times n$ positive definite symmetric matrix over $\R$.
Then:
:there exists a nonsingular matrix $C$ such that $A = C^\intercal C$.

Proof:
Let $A$ be positive definite. 
From Spectral Theorem for Real Symmetric Matrices:
:there exists an orthogonal matrix $P$ and diagonal matrix $D$ such that $A = P^\intercal D P$.
Further, from Characterization of Diagonalizable Matrices:
:the diagonal entries of $D$ are the eigenvalues of $A$. 
From Real Symmetric Positive Definite Matrix has Positive Eigenvalues:
:the diagonal entries of $D$ are positive. 
We can therefore construct a real diagonal matrix $S$ by: 
:$\paren S_{i j} = \begin{cases} \sqrt {\paren D_{i i} } & i = j \\ 0 & i \ne j \end{cases}$
From Product of Diagonal Matrices is Diagonal, we have: 
:$\paren {S^2}_{i j} = \begin{cases} \paren D_{i i} & i = j \\ 0 & i \ne j \end{cases}$
so:
:$S^2 = D$
We also have: 
{{begin-eqn}}
{{eqn | l = \det S
      | r = \prod_{i \mathop = 1}^n \sqrt {\paren D_{i i} }
}}
{{eqn | r = \sqrt {\prod_{i \mathop = 1}^n \paren D_{i i} }
}}
{{eqn | o = >
      | r = 0
      | c = as $\paren D_{i i} > 0$ for each $i$
}}
{{end-eqn}}
We therefore have: 
{{begin-eqn}}
{{eqn | l = \map \det {P^\intercal S P}
      | r = \map \det {P^\intercal} \det S \det P
      | c = Determinant of Matrix Product
}}
{{eqn | r = \paren {\det P}^2 \det S
      | c = Determinant of Transpose
}}
{{eqn | r = \det S
      | c = Determinant of Orthogonal Matrix is Plus or Minus One
}}
{{eqn | o = >
      | r = 0
}}
{{end-eqn}}
So from Matrix is Nonsingular iff Determinant has Multiplicative Inverse:
:$P^\intercal S P$ is nonsingular.
Let $C = P^\intercal S P$. 
Then:
{{begin-eqn}}
{{eqn | l = C^\intercal C
      | r = \paren {P^\intercal S P}^\intercal P^\intercal S P
}}
{{eqn | r = P^\intercal \paren {P^\intercal S}^\intercal P^\intercal S P
      | c = Transpose of Matrix Product
}}
{{eqn | r = P^\intercal S^\intercal P P^\intercal S P
      | c = Transpose of Matrix Product
}}
{{eqn | r = P^\intercal S^\intercal S P
      | c = as $P$ is orthogonal
}}
{{eqn | r = P^\intercal S^2 P
      | c = Diagonal Matrix is Symmetric
}}
{{eqn | r = P^\intercal D P
}}
{{eqn | r = A
}}
{{end-eqn}}
As $C$ is nonsingular, the proof is complete.
Theorem: Characterisation of Real Symmetric Positive Definite Matrix/Sufficient Condition

Let $A$ be an $n \times n$ symmetric matrix over $\mathbb R$ such that: 
:there exists a nonsingular matrix $C$ such that $A = C^\intercal C$.
Then $A$ is positive definite.

Proof:
Let $A$ be a symmetric matrix such that: 
:there exists an nonsingular matrix $C$ such that $A = C^\intercal C$.
Let $\mathbf v$ be a non-zero vector. 
Then:
{{begin-eqn}}
{{eqn | l = \mathbf v^\intercal A \mathbf v
      | r = \mathbf v^\intercal C^\intercal C \mathbf v
}}
{{eqn | r = \paren {C \mathbf v}^\intercal C \mathbf v
      | c = Transpose of Matrix Product
}}
{{eqn | r = \paren {C \mathbf v} \cdot \paren {C \mathbf v}
      | c = {{Defof|Dot Product}}
}}
{{eqn | r = \norm {C \mathbf v}^2
      | c = Dot Product of Vector with Itself
}}
{{eqn | o = >
      | r = 0
      | c = Euclidean Space is Normed Vector Space: {{NormAxiomVector|1}}
}}
{{end-eqn}}
So $A$ is positive definite.
Theorem: Characterisation of Real Symmetric Positive Definite Matrix

Let $A$ be an $n \times n$ symmetric matrix over $\mathbb R$.
Then $A$ is positive definite {{iff}}:
:there exists a nonsingular matrix $C$ such that $A = C^\intercal C$.

Proof:
=== Necessary Condition ===
{{:Characterisation of Real Symmetric Positive Definite Matrix/Necessary Condition}}{{qed|lemma}}
Theorem: Coffee Mug and Doughnut are Homeomorphic

A {{WP|Doughnut|doughnut}} and a {{WP|Coffee mug|coffee mug}} are homeomorphic.

Proof:
They are both solid objects with one hole.
A classical {{WP|Doughnut|doughnut}} is in the shape of a torus.
That is, it is a solid figure with one hole.
The classic coffee-cup shape is a receptacle with a closed loop of china forming a handle.
The hole through which you put your fingers is homeomorphic to the hole in the doughnut.
{{qed}}
Theorem: Trefoil Knot is Homeomorphic to Circle

The trefoil knot is homeomorphic to the circle.

Proof:
Despite the fact that you cannot actually rearrange a trefoil knot actually into a circle in the usual $\R^3$ space, you can set up a mappping from one to the other.
{{finish}}
Theorem: Homeomorphism between Topological Spaces may not be Unique

Let $T_1$ and $T_2$ be topological spaces.
Let $f$ be a homeomorphism from $T_1$ to $T_2$.
Then $f$ may not necessarily be unique.

Proof:
Let $\R$ be the real number line with the Euclidean topology.
Let $I_1 := \openint a b$ and $I_2 := \openint c d$ be non-empty open real intervals.
From Open Real Intervals are Homeomorphic, $I_1$ and $I_2$ are homeomorphic.
The example given of a homeomorphism was the real function $f: I_1 \to I_2$ defined as:
:$\forall x \in I_1: \map f x = c + \dfrac {\paren {d - c} \paren {x - a} } {b - a}$
However, note that the following real functions for all $n \in \Z_{>0}$:
{{begin-eqn}}
{{eqn | lo= f_n: I_1 \to I_2
      | l = x
      | o = \mapsto
      | r = c + \dfrac {\paren {d - c} \paren {x - a}^n} {\paren {b - a}^n}
      | c = 
}}
{{eqn | lo= g_n: I_1 \to I_2
      | l = x
      | o = \mapsto
      | r = d + \dfrac {\paren {d - c} \paren {x - a}^n} {\paren {b - a}^n}
      | c = 
}}
{{end-eqn}}
are all homeomorphisms from $I_1$ to $I_2$.
{{qed}}
Theorem: Cardinality of Set is Topological Property

Let $T = \struct {S, \tau}$ be a topological space.
The cardinality of $S$ is a topological property of $T$.

Proof:
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T_1$ and $T_2$ be homeomorphic.
Then by definition there exists a homeomorphism $f: T_1 \to T_2$.
Hence by definition $f$ is a bijection.
Hence by definition $S$ and $T$ are equivalent.
That is, they have the same cardinality.
Thus cardinality is preserved by homeomorphism.
Hence the result by definition of topological property.
{{qed}}
Theorem: Boundedness is not Topological Property

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be homeomorphic.
Then it is not necessarily the case that:
:$M_1$ is bounded {{iff}} $M_2$ is bounded.
That is, boundedness is not a topological property.

Proof:
;Proof by Counterexample
Let the metric space $M_1 = \struct {S_1, d}$ such that:
:$S_1 = \openint 0 1$ is the open unit interval
:$d$ is the usual (Euclidean) metric on $S_1$.
Let the metric space $M_2 = \struct {\R, d}$ such that:
:$\R$ is the set of real numbers
:$d$ is again the usual (Euclidean) metric on $\R$.
Then $M_1$ is bounded by, for example, $1$.
However, $M_2$ is not bounded.
But from Open Real Interval is Homeomorphic to Real Number Line, we have that $M_1$ and $M_2$ are homeomorphic.
Hence the result.
{{qed}}
Theorem: Minimum of Exponential Random Variables has Exponential Distribution

Let $\beta_1, \beta_2, \ldots, \beta_n$ be positive real numbers.
Let $X_1, X_2, \ldots, X_n$ be independent random variables.
For each $i$, let $X_i \sim \Exponential {\beta_i}$, where $\Exponential {\beta_i}$ is the exponential distribution with parameter $\beta_i$. 
Let: 
:$\ds M = \map {\min_{1 \mathop \le i \mathop \le n} } {X_i} $
Then: 
:$\ds M \sim \Exponential {\paren {\sum_{i \mathop = 1}^n \frac 1 {\beta_i} }^{-1} }$

Proof:
We aim to show that: 
:$\ds \map \Pr {M \le m} = 1 - \map \exp {-m \sum_{i \mathop = 1}^n \frac 1 {\beta_i} }$
for each $m > 0$.
Note that: 
:$\ds M > m$
{{iff}}:
:$\ds X_i > m$
for each $i$. 
We therefore have: 
{{begin-eqn}}
{{eqn	| l = \map \Pr {M > m}
	| r = \map \Pr {\bigcap_{i \mathop = 1}^n \set {X_i > m} }
}}
{{eqn	| r = \prod_{i \mathop = 1}^n \map \Pr {X_i > m}
	| c = {{Defof|Independent Random Variables}}
}}
{{eqn	| r = \prod_{i \mathop = 1}^n \map \exp {-\frac m {\beta_i} }
	| c = {{Defof|Exponential Distribution}}
}}
{{eqn	| r = \map \exp {-m \sum_{i \mathop = 1}^n \frac 1 {\beta_i} }
	| c = Exponential of Sum
}}
{{end-eqn}}
so:
:$\ds \map \Pr {M \le m} = 1 - \map \exp {-m \sum_{i \mathop = 1}^n \frac 1 {\beta_i} }$
{{qed}}
Category:Exponential Distribution
bur1wrpajccixjyafijp9m46hcad8kj
Theorem: Multiple of Exponential Random Variable has Exponential Distribution

Let $\beta, k$ be real numbers with $\beta > 0$.
Let $X$ be a random variable.
Let $X \sim \Exponential \beta$, where $\Exponential \beta$ is the exponential distribution with parameter $\beta$. 
Then:
:$k X \sim \Exponential {k \beta}$

Proof:
Let:
:$Y \sim k X$
We aim to show that: 
:$\ds \map \Pr {Y \le y} = 1 - \map \exp {-\frac y {k \beta} }$
for each $y > 0$. 
We have: 
{{begin-eqn}}
{{eqn	| l = \map \Pr {Y \le y}
	| r = \map \Pr {k X \le y}
}}
{{eqn	| r = \map \Pr {X \le \frac y k}
}}
{{eqn	| r = 1 - \map \exp {-\frac y {k \beta} }
	| c = {{Defof|Exponential Distribution}}
}}
{{end-eqn}}
{{qed}}
Category:Exponential Distribution
csymtekv3nu91vf0tt8web3gj0fhgji
Theorem: Exponential of Negative of Exponential Random Variable has Beta Distribution

Let $\beta$ be a positive real number.
Let $X \sim \Exponential \beta$ where $\Exponential \beta$ is the exponential distribution with parameter $\beta$. 
Then:
:$e^{-X} \sim \BetaDist {\dfrac 1 \beta} 1$

Proof:
Note that if:
:$Y \sim \BetaDist {\dfrac 1 \beta} 1$
then the probability density function of $Y$, $f_Y$ is given by:
{{begin-eqn}}
{{eqn	| l = \map {f_Y} y
	| r = \frac {y^{\frac 1 \beta - 1} \paren {1 - y}^{1 - 1} } {\map \Beta {\frac 1 \beta, 1} }
	| c = {{Defof|Beta Distribution}}
}}
{{eqn	| r = \frac {y^{\frac 1 \beta - 1} } {\frac {\map \Gamma {\frac 1 \beta} \map \Gamma 1} {\map \Gamma {\frac 1 \beta + 1} } }
	| c = {{Defof|Beta Function}}
}}
{{eqn	| r = \frac 1 {\frac {\map \Gamma {\frac 1 \beta} } {\frac 1 \beta \map \Gamma {\frac 1 \beta} } } y^{\frac 1 \beta}
	| c = Gamma Difference Equation
}}
{{eqn	| r = \frac 1 \beta y^{\frac 1 \beta}
}}
{{end-eqn}}
for each $y > 0$. 
Let:
:$Z = e^{-X}$
It suffices to show that $Z$ has the same probability density function as $Y$.
We have: 
{{begin-eqn}}
{{eqn	| l = \map \Pr {Z \le z}
	| r = \map \Pr {e^{-X} \le z}
}}
{{eqn	| r = \map \Pr {-X \le \ln z}
}}
{{eqn	| r = \map \Pr {X \ge -\ln z}
}}
{{eqn	| r = \map \exp {-\frac {-\ln z} {\beta} }
	| c = {{Defof|Exponential Distribution}}
}}
{{eqn	| r = z^{\frac 1 \beta}
}}
{{end-eqn}}
By Derivative of Power, the probability density function of $Z$, $f_Z$ is therefore given by: 
:$\map {f_Z} z = \dfrac 1 \beta z^{\frac 1 \beta - 1}$
for each $z > 0$.
So:
:$f_Y = f_Z$
{{qed}}
Category:Exponential Distribution
Category:Beta Distribution
3nt9qo4e6up9kjq329lcqoqey5pqeho
Theorem: Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact/Necessary Condition

Let $X$ be a finite dimensional normed vector space.
Let $\Bbb S = \map {\Bbb S_1} 0$ be the unit sphere centred at $0$ in $X$.
Then $\Bbb S$ is compact.

Proof:
Let $X$ be a finite dimensional vector space $\R^d$.
We have that all norms on finite-dimensional vector space are equivalent.
Choose Euclidean norm $\norm {\, \cdot \,}_2$.
Let $\struct {\R^d, \norm {\, \cdot \,}_2}$ be the normed finite-dimensional real vector space with Euclidean norm.
Let $\map {\Bbb S^{d - 1}_1} 0$ be a unit sphere in $\struct {\R^d, \norm {\, \cdot \,}_2}$.
By definition, $\Bbb S^{d - 1}$ is bounded.
Let $\mathbf L := \tuple {L_1, \dots, L_d} \in \R^d$.
Let $\mathbf x_n := \tuple {x_n^{\paren 1}, \dots, x_n^{\paren d}}$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a sequence in $\Bbb S^{d - 1}$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converge to $\mathbf L$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {\mathbf x_n - \mathbf L}_2 < \epsilon$
Furthermore, for all $k \in \N : k \le d$ we have that:
{{begin-eqn}}
{{eqn | l = \size {x_n^{\paren k} - L_k}
      | r = \sqrt {\size {x_n^{\paren k} - L_k}^2 }
}}
{{eqn | o = \le
      | r = \sqrt{\sum_{j \mathop = 1}^d \size {x_n^{\paren j} - L_k}^2}
}}
{{eqn | r = \norm {\mathbf x_n - \mathbf L}_2
      | c = {{defof|Euclidean Norm}}
}}
{{end-eqn}}
Thus:
:$\forall k \in \N : k \le d : \forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {x_n^{\paren k} - L_k} < \epsilon$
In other words, $\ds \lim_{n \mathop \to \infty} x_n^{\paren k} = L_k$.
We have that $\mathbf x_n \in \Bbb S^{d - 1}$.
Therefore:
:$\ds \norm {\mathbf x_n}_2^2 = \sum_{j \mathop = 1}^d \paren {x_n^{\paren j}}^2 = 1$
$\map f x = x^2$ is a continuous function in its whole range.
By Limit of Composite Function, taking the limit $n \to \infty$ results in:
:$\ds \norm {\mathbf L}_2^2 = \sum_{j \mathop = 1}^d \paren {L_j}^2 = 1$
Therefore, $\mathbf L \in \Bbb S^{d - 1}$.
Hence, $\Bbb S^{d - 1}$ is closed.
By Heine-Borel theorem, $\Bbb S^{d - 1}$ is compact.
Theorem: Limit Point of Set may or may not be Element of Set

Let $S$ be a set.
Let $H \subseteq S$ be a subset of $S$.
Let $T = \struct {H, \tau}$ be a topological space on the underlying set $H$.
Let $a \in S$ be a limit point of $T$.
Then $a$ may or may not be an element of $H$.
Whether it is or not depends upon the nature of both $a$ and $T$.

Proof:
Consider:
:the open real interval $\openint a b$
:the closed real interval $\closedint a b$.
Both of these are subsets of the set of real numbers $\R$.
From Limit Point Examples: End Points of Real Interval, $a$ is a limit point of both $\openint a b$ and $\closedint a b$.
But $a \in \closedint a b$ while $a \notin \openint a b$.
Hence the result.
{{qed}}
Theorem: Element of Topological Space may or may not be Limit Point

Let $S$ be a set.
Let $H \subseteq S$ be a subset of $S$.
Let $T = \struct {H, \tau}$ be a topological space on the underlying set $H$.
Let $a \in H$.
Then $a$ may or may not be a limit point of $T$.
Whether it is or not depends upon the nature of both $a$ and $T$.

Proof:
Let $\R$ be the set of real numbers.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
:$H = \set 0 \cup \openint 1 2$
From Limit Point Examples: Union of Singleton with Open Real Interval, $0$ is not a limit point of $H$, although $0 \in H$.
From Limit Point Examples: End Points of Real Interval, $a$ is a limit point of $\closedint a b$ where $a \in \closedint a b$.
Hence the result.
{{qed}}
Theorem: Countable Set may have Uncountable Limit Points

Let $S$ be an uncountable set.
Let $H \subseteq S$ be a countable subset of $S$.
Let $T = \struct {H, \tau}$ be a topological space on the underlying set $H$.
Then despite the fact that $H$ is countable, the set of limit points of $T$ may be uncountable.

Proof:
Let $\R$ be the set of real numbers.
Let $\Q$ be the set of rational numbers.
Let $x \in \R$.
Then from Real Number is Limit Point of Rational Numbers in Real Numbers, $x$ is a limit point of $T$.
As $x$ is arbitrary, it follows that every element of $\R$ is a limit point of $T$.
From Rational Numbers are Countably Infinite, $\Q$ is a countable set.
From Real Numbers are Uncountable, $\R$ is an uncountable set.
Hence the result.
{{qed}}
Theorem: Condition for Point being in Closure/Metric Space

Let $M = \struct {S, d}$ be a metric space.
Let $H \subseteq S$.
Let $\map \cl H$ denote the closure of $H$ in $M$.
Let $x \in S$.
Then $x \in \map \cl H$ {{iff}}:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.

Proof:
By definition of closure of $H$ in $M$:
:$\map \cl H = H^i \cup H'$
where:
:$H^i$ denotes the set of isolated points of $H$
:$H'$ denotes the set of limit points of $H$.