nyuuzyou/wikis huggingface dataset:
Euclidean Borel Sigma-Algebra Closed under Scalar Multiplication

== Theorem ==
Let $\map \BB {\R^n}$ be the Borel $\sigma$-algebra on $\R^n$.
Let $B \in \BB$, and let $t \in \R_{>0}$.
Then also $t \cdot B := \set {t \mathbf b: \mathbf b \in B} \in \BB$.
== Proof ==
Define $f: \R^n \to \R^n$ by $\map f {\mathbf x} := \dfrac 1 t \mathbf x$.
Then for all $\mathbf x \in \R^n$, $\map {f^{-1} } {\mathbf x} = \set {t \mathbf x}$, where $f^{-1}$ denotes the preimage of $f$.
Thus $\ds t \cdot B = \bigcup_{\mathbf b \mathop \in B} \map {f^{-1} } {\mathbf b} = \map {f^{-1} } B$, where the last equality holds by definition of preimage.
It follows that the statement of the theorem comes down to showing that $f$ is $\BB \, / \, \BB$-measurable.
By Characterization of Euclidean Borel Sigma-Algebra the half-open $n$-rectangles $\JJ_{ho}^n$ generate $\BB$.
Applying Mapping Measurable iff Measurable on Generator, it suffices to demonstrate:
:$\forall J \in \JJ_{ho}^n: \map {f^{-1} } J \in \BB$
Now for a half-open $n$-rectangle $J = \horectr {\mathbf a} {\mathbf b}$, it follows from the consideration on $f^{-1}$ above, that:
:$\map {f^{-1} } J = \horectr {t \mathbf a} {t \mathbf b}$
which is again in $\JJ_{ho}^n$ and so, in particular, in $\BB$.
Thus $f$ is $\BB \, / \, \BB$-measurable, that is:
:$\forall B \in \BB: t \cdot B = \map {f^{-1} } B \in \BB$
which was to be demonstrated.
== Sources ==
* : $\S 5$: Problem $8$
Category:Measure Theory
Euclidean Domain/Euclidean Algorithm/Examples/5 i and 3 + i in Gaussian Integers

== Examples of Use of Euclidean Algorithm in Euclidean Domain ==
The GCD of $5 i$ and $3 + 1$ in the ring of Gaussian integers is found to be:
:$\gcd \set {5 i, 3 + 1} = 1 + 2 i$
and its associates $-1 - 2 i$, $-2 + i$ and $2 - i$.
== Proof ==
Let $x = 5 i$ and $y = 3 + i$.
We need to find $q$ and $r$ such that:
:$x = y q + r$
with:
:$\map \nu r < \map \nu y$
where $\map \nu x := \cmod x^2$
Thus we calculate:
$q$ is to be set to one of the Gaussian integers nearest to it.
Thus let $q = i$.
Hence:
Then:
Thus a GCD of $5 i$ and $3 + i$ is $1 + 2 i$.
From Euclidean Domain is GCD Domain, its associates are also GCDs of $5 i$ and $3 + i$.
Hence the result.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 28$. Highest Common Factor: Example $56$
Category:Examples of Euclidean Algorithm
Euclidean Domain is GCD Domain

== Theorem ==
Let $\struct {D, +, \times}$ be a Euclidean domain.
Then any two elements $a, b \in D$ have a greatest common divisor $d$ such that:
:$d \divides a \land d \divides b$
:$x \divides a \land x \divides b \implies x \divides d$
and $d$ is written $\gcd \set {a, b}$.
For any $a, b \in D$:
:$\exists s, t \in D: s a + t b = d$
Any two greatest common divisors of any $a, b$ are associates.
== Proof ==
Let $a, b \in D$.
Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$.
Then $U$ is an ideal of $D$.
Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are Principal Ideal.
By Sum of Ideals is Ideal, $U$ is an ideal.
By Euclidean Domain is Principal Ideal Domain, $U$ is a principal ideal, $\ideal d$ say.
As $a, b \in U$ it follows that $d$ is a divisor of $a$ and $b$, that is:
:$d \divides a \land d \divides b$
Since $d$ itself is in $U$, we have:
:$\exists s, t \in D: s a + t b = d$
By Common Divisor in Integral Domain Divides Linear Combination:
:$x \divides a \land x \divides b \implies x \divides d$
So $d$ is a greatest common divisor of $a$ and $b$.
If $d$ and $d'$ are both greatest common divisors of $a$ and $b$, then $d \divides a \land d \divides b$ and so $d \divides d'$.
Similarly $d' \divides d$.
So $d$ and $d'$ are associates.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 28$. Highest Common Factor: Theorem $54$
Category:Euclidean Domains
Category:GCD Domains
Euclidean Domain is Principal Ideal Domain

== Theorem ==
A Euclidean domain is a principal ideal domain.
== Proof ==
Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$ and whose Euclidean valuation is $\nu$.
We need to show that every ideal of $\struct {D, +, \times}$ is a principal ideal.
Let $U$ be an ideal of $\struct {D, +, \times}$ such that $U \ne \set 0$.
Let $d \in U$ such that $d \ne 0$ and $\map \nu d$ is as small as possible for elements of $U$.
By definition, $\nu$ is defined as $\nu : D \setminus \set 0 \to \N$, so the codomain of $\nu$ is a subset of the natural numbers.
By the Well-Ordering Principle, such an element $d$ exists as an element of the preimage of the least member of the image of $U$.
Let $a \in U$.
Let us write $a = d q + r$ where either $r = 0$ or $\map \nu r < \map \nu d$.
Then $r = a - d q$ and so $r \in U$.
Suppose $r \ne 0$.
That would mean $\map \nu r < \map \nu d$ contradicting $d$ as the element of $U$ with the smallest $\nu$.
So $r = 0$, which means $a = q d$.
That is, every element of $U$ is a multiple of $d$.
So $U$ is the principal ideal generated by $d$.
This deduction holds for all ideals of $D$.
Hence the result.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Theorem $53$
* 
Category:Euclidean Domains
Category:Principal Ideal Domains
Euclidean Metric and Chebyshev Distance on Real Metric Space give rise to Same Topological Space

== Theorem ==
For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $d_2$ be the Euclidean metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Let $T_2 = \struct {\R^n, \tau_2}$ denote the topological space which is induced by $d_2$.
Let $T_\infty = \struct {\R^n, \tau_\infty}$ denote the topological space which is induced by $d_\infty$.
Then $T_2$ and $T_\infty$ are the same.
That is:
:$\tau_2 = \tau_\infty$
== Proof ==
From P-Product Metrics on Real Vector Space are Topologically Equivalent, $\tau_2$ and $\tau_\infty$ are topologically equivalent metrics.
The result follows from Topologically Equivalent Metrics induce Equal Topologies.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $3$
Category:Chebyshev Distance
Category:Euclidean Metric
Category:Topologically Equivalent Metrics
Euclidean Metric induces Product Topology

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $d$ be the Euclidean metric on $M_1 \times M_2$.
The topology induced by $d$ on $M_1 \times M_2$ is the product topology on $M_1 \times M_2$.
== Proof ==
== Sources ==
* 
Category:Euclidean Metric induces Product Topology
Category:Euclidean Metric
Category:Product Topology
Euclidean Metric is Metric

== Theorem ==
Let $M_{1'} = \struct {A_{1'}, d_{1'} }, M_{2'} = \struct {A_{2'}, d_{2'} }, \ldots, M_{n'} = \struct {A_{n'}, d_{n'} }$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
The Euclidean metric on $\AA$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Pythagoras' Theorem
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.7$
Category:Euclidean Metric
Category:Euclidean Metric is Metric
Euclidean Metric is Metric/Proof 2

== Theorem ==
== Proof ==
We have that the Euclidean metric on $\AA$ is defined as:
:$\ds \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2}$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
=== Proof of ===
So holds for $d_2$.
=== Proof of ===
Let:
:$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
:$(3): \quad \map {d_{i'} } {x_i, y_i} = r_i$
:$(4): \quad \map {d_{i'} } {y_i, z_i} = s_i$.
Thus we need to show that:
:$\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^2}^{\frac 1 2} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^2}^{\frac 1 2} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^2}^{\frac 1 2}$
We have:
So holds for $d_2$.
=== Proof of ===
So holds for $d_2$.
=== Proof of ===
So holds for $d_2$.
Category:Euclidean Metric is Metric
Euclidean Metric on Real Number Line is Metric

== Theorem ==
The Euclidean metric on the real number line $\R$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : One: Metric Spaces: $1$: Open and Closed Sets
Category:Euclidean Metric
Category:Real Number Line with Euclidean Metric
Category:Euclidean Metric on Real Number Line is Metric
Euclidean Metric on Real Number Line is Metric/Proof 2

== Theorem ==
== Proof ==
Consider the real number line under the Euclidean metric:
:$M = \struct {\R, d}$
where $d$ is the distance function given by:
:$\map d {x, y} = \size {x - y}$
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
Category:Euclidean Metric on Real Number Line is Metric
Euclidean Metric on Real Number Plane is Rotation Invariant

== Theorem ==
Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.
Let $d_2$ denote the Euclidean metric on $\R^2$.
Then $d_2$ is unchanged by application of $r_\alpha$:
:$\forall x, y \in \R^2: \map {d_2} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_2} {x, y}$
== Proof ==
Let $x = \tuple {x_1, x_2}$ and $y = \tuple {y_1, y_2}$ be arbitrary points in $\R^2$.
Note that $\paren {\map {d_2} {x, y} }^2$ can be expressed as:
:$\paren {\map {d_2} {x, y} }^2 = \paren {\mathbf x - \mathbf y}^\intercal \paren {\mathbf x - \mathbf y}$
where:
:$x$ and $y$ are expressed in vector form: $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \end {pmatrix}, y = \begin {pmatrix} y_1 \\ y_2 \end {pmatrix}$
:$\paren {\mathbf x - \mathbf y}^\intercal$ denotes the transpose of $\paren {\mathbf x - \mathbf y}$
Then:
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 21$
Category:Geometric Rotations
Category:Euclidean Metric
Euclidean Metric on Real Number Space is Translation Invariant

== Theorem ==
Let $\tau_{\mathbf t}: \R^n \to \R^n$ denote the translation of the real Euclidean space of $n$ dimensions by the vector $\mathbf t = \tuple {t_1, t_2, \ldots, t_n}$.
Let $d_2$ denote the Euclidean metric on $\R^n$.
Then $d_2$ is unchanged by application of $\tau$:
:$\forall \mathbf x, \mathbf y \in \R^n: \map {d_2} {\map \tau {\mathbf x}, \map \tau {\mathbf y} } = \map {d_2} {\mathbf x, \mathbf y}$
== Proof ==
Let $\mathbf x = \tuple {x_1, x_2, \ldots, x_n}$ and $\mathbf y = \tuple {y_1, y_2, \ldots, y_n}$ be arbitrary points in $\R^n$.
Then:
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 21$
:::where the exercise is limited to the Euclidean plane
Category:Translation Mappings
Category:Euclidean Metric
Euclidean Metric on Real Vector Space is Metric

== Theorem ==
The Euclidean metric on a real vector space $\R^n$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Euclidean Metric on Real Number Line is Metric
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Pythagoras' Theorem
* : One: Metric Spaces: $1$: Open and Closed Sets
Category:Euclidean Metric
Category:Euclidean Metric on Real Vector Space is Metric
Euclidean Metric on Real Vector Space is Metric/Proof 2

== Theorem ==
== Proof ==
Consider the Euclidean space $M = \struct {\R^n, d_2}$ where $d_2$ is the distance function given by:
:$\ds \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{\frac 1 2}$
where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
=== Proof of ===
So holds for $d_2$.
=== ===
It is required to be shown:
:$\map {d_2} {x, y} + \map {d_2} {y, z} \ge \map {d_2} {x, z}$
for all $x, y, z \in \R^n$.
Let:
:$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
:$(3): \quad x_i - y_i = r_i$
:$(4): \quad y_i - z_i = s_i$.
Thus we need to show that:
:$\ds \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2} \ge \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}$
We have:
So holds for $d_2$.
=== ===
So holds for $d_2$.
=== ===
So holds for $d_2$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.1$
Category:Euclidean Metric on Real Vector Space is Metric
Euclidean Plane is Abstract Geometry

== Theorem ==
The Euclidean plane $\struct {\R^2, L_E}$ is an abstract geometry.
== Proof ==
We will show that the axioms for an abstract geometry hold.
=== Axiom 1 ===
Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be two distinct points in $\struct {\R^2, L_E}$.
If $x_1 = x_2 = a$ then $P, Q \in L_a$.
If $x_1 \ne x_2$ then let:
:$m = \dfrac {y_2 - y_1} {x_2 - x_1}$
:$b = y_2 - m x_2$
Then $P, Q \in L_{m,b}$, since:
:$b + m x_1 = y_2 - m \paren {x_2 - x_1} = y_2 - \paren {y_2 - y_1} = y_1$
:$b + m x_2 = y_2 - m \paren {x_2 - x_2} = y_2$
So any two points in $\R^2$ lie on a line in $L_E$.
=== Axiom 2 ===
Let $a \in \R$ be arbitrary.
Then:
:$\tuple {a, 0}, \tuple {a, 1} \in L_a$
Let $m, b \in \R$ be arbiitrary.
Then:
:$\tuple {0, b}, \tuple {1, m + b} \in L_{m, b}$
So every line in $L_E$ has at least two distinct points.
Hence $\struct {\R^2, L_E}$ is an abstract geometry.
== Sources ==
* : $\S 2.1$
Category:Abstract Geometries
Category:Euclidean Geometry
Euclidean Plus Metric is Metric

== Theorem ==
Let $\R$ be the set of real numbers.
Let $d: \R \times \R \to \R$ be the Euclidean plus metric:
:$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j}} - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$
Then $d$ is indeed a metric.
== Proof ==
Recall that $\set {r_j}_{j \mathop \in \N}$ is an enumeration of the rational numbers $\Q$.
Also, we note that:
meaning that this is a convergent series and so the definition is meaningful.
Next the axioms for a metric are checked in turn.
=== Proof of ===
=== Proof of ===
Let $i \in \N$ be fixed.
Define:
:$\map {f_i} x := \ds \max_{j \mathop \le i} \frac 1 {\size {x - r_j} }$
Then:
Now since $\inf \set {1, \size {\map {f_i} x - \map {f_i} z} } \le 1$, it follows that:
Suppose now that $\size {\map {f_i} x - \map {f_i} z} \le 1$.
Then:
On the other hand, if:
:$\size {\map {f_i} x - \map {f_i} z} > 1$
then also:
:$\size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z} > 1$
and:
Combining both cases with the estimates above, we conclude:
:$\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} } \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} }$
Finally, now, we have:
=== Proof of ===
We have that:
Hence:
=== Proof of ===
Suppose that $x \ne y$.
Then:
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $5$
Category:Euclidean Plus Metric
Euclidean Space is Banach Space

== Theorem ==
Let $m$ be a positive integer.
Then the Euclidean space $\R^m$, along with the Euclidean norm, forms a Banach space over $\R$.
== Proof 1 ==
== Proof 2 ==
Category:Euclidean Space
Category:Banach Spaces
Category:Euclidean Space is Banach Space
Euclidean Space is Banach Space/Proof 1

== Theorem ==
== Proof ==
The Euclidean space $\R^m$ is a vector space over $\R$.
That the norm axioms are satisfied is proven in Euclidean Space is Normed Vector Space.
Then we have Euclidean Space is Complete Metric Space.
The result follows by the definition of a Banach space.
Category:Euclidean Space is Banach Space
Euclidean Space is Banach Space/Proof 2

== Theorem ==
== Proof ==
By definition, Euclidean norm is the same as $p$-norm with $p = 2$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m} } }_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$.
Let $k \in \N_{> 0} : k \le m$.
Then:
Hence, $\sequence {x_n^{\paren k} }_{n \mathop \in \N}$ is a Cauchy sequence in $\R$.
Then:
:$\ds \lim_{n \mathop \to \infty} x_n^{\paren k} = L^{\paren k}$
Let $\mathbf L = \tuple {L^{\paren 1}, \ldots, L^{\paren m}} \in \R^m$
We have that for all $n > N$:
Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges to $\mathbf L$ in $\struct {\R^m, \norm {\, \cdot \,}_2}$.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Euclidean Space is Banach Space
Euclidean Space is Complete Metric Space

== Theorem ==
Let $m$ be a positive integer.
Then the Euclidean space $\R^m$, along with the Euclidean metric, forms a complete metric space.
== Proof ==
Let $\sequence {\tuple {x_{n, 1}, x_{n, 2}, \ldots, x_{n, m} } }_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$.
Let $\epsilon > 0$.
Then $\dfrac \epsilon m > 0$.
We have that Real Number Line is Complete Metric Space.
Therefore:
:$\exists y_1, y_2, \ldots, y_m \in \R$ and $N_1, N_2, \ldots, N_m \in \N$ (depending on $\epsilon$)
such that:
:$\forall k \in \N: 1 \le k \le m: \forall n_k > N_k: \sequence {x_{n, k} – y_k} < \dfrac \epsilon m$
From Euclidean Space is Normed Vector Space:
:$\ds \norm {\tuple {x_{n, 1}, x_{n, 2}, \ldots, x_{n, m} } - \tuple {y_1, y_2, \ldots, y_m} } \le \sum_{k \mathop = 1}^m \size {x_{n, k} – y_k} < \epsilon$
Hence the Euclidean space is a complete metric space.
== Sources ==
* : $3.11 \text c$
Category:Euclidean Space
Category:Complete Metric Spaces
Euclidean Space is Normed Vector Space

== Theorem ==
The Euclidean space $\R^n$ is a normed vector space.
== Proof ==
Let $\norm {\, \cdot \,}$ denote the Euclidean norm on $\R^n$.
We prove that $\norm {\, \cdot \,}$ is indeed a norm on $\R^n$ by proving it fulfils the norm axioms.
=== ===
As $\mathbf 0 = \tuple {0, \ldots, 0}$, it follows that:
:$\norm {\mathbf 0} = \sqrt {0^2 + \ldots + 0^2} = 0$
Suppose instead that:
:$\norm {\mathbf v} = 0$ 
for some $\mathbf v = \tuple {v_1, \ldots, v_n}$ with $v_1, \ldots, v_n \in \R$.
As $\ds \sqrt {\sum_{i \mathop = 1}^n v_n^2 } = 0$, it follows from squaring both sides that:
:$\ds \sum_{i \mathop = 1}^n v_n^2 = 0$
From Even Power is Non-Negative, it follows that:
:$\forall i \in \N_{>0}: v_i = 0$
Hence:
:$\mathbf v = \mathbf 0$
Thus holds.
=== ===
Let $\mathbf v = \tuple {v_1, \ldots, v_n} \in \R^n$ with $v_1, \ldots, v_n \in \R$.
Let $r \in \R$.
Then:
Thus holds.
=== ===
Follows from Triangle Inequality for Vectors in Euclidean Space.
Thus holds.
== Sources ==
* : 1. Functions on Euclidean Space: Norm and Inner Product
Category:Examples of Norms
Category:Euclidean Space
Euclidean Space is Path-Connected

== Theorem ==
Let $\R^n$ be the $n$-dimensional Euclidean space for $n \in \N$ a natural number.
Then $\R^n$ is path-connected.
== Proof ==
Let $\mathbf x, \mathbf y \in \R^n$ be arbitrary points of $\R^n$.
Define $l: \closedint 0 1 \to \R^n$ by:
:$\map l t = \paren {1 - t} \mathbf x + t \mathbf y$
Then $\map l 0 = 1 \mathbf x + 0 \mathbf y = \mathbf x$, whereas $\map l 1 = 0 \mathbf x + 1 \mathbf y = \mathbf y$.
Finally, it remains to show that $l$ is continuous. 
Fix $\epsilon > 0$ and suppose that $t, t' \in \closedint 0 1$ are such that $\size {t - t'} < \dfrac {\epsilon} {1 + \norm {\mathbf x} + \norm {\mathbf y} }$.
Then:
We can now estimate the norm of this last expression:
Since $\epsilon$ was arbitrary, we conclude that $l$ is continuous.
Therefore, it forms a path from $\mathbf x$ to $\mathbf y$.
Since $\mathbf x$ and $\mathbf y$ were arbitrary, it follows that $\R^n$ is path-connected.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
* : $6.3$: Path-connectedness: Examples $6.3.4$
Category:Path-Connected Spaces
Category:Metric Spaces
Euclidean Space is Subspace of Extended Real Number Space

== Theorem ==
Let $\struct {\overline \R, \tau}$ be the extended real number space.
Then $\tau {\restriction_\R}$, the subspace topology on $\R$, is the Euclidean topology.
That is, Euclidean $1$-space is a subspace of the extended real number space.
== Proof ==
Category:Extended Real Number Space
Category:Euclidean Space
Euclidean Space without Origin is Path-Connected

== Theorem ==
Let $n \in \Z: n \ge 2$.
Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $\R^n \setminus \set {\mathbf 0}$ be $\R^n$ with the origin removed.
Then $\R^n \setminus \set {\mathbf 0}$ is path-connected.
== Proof ==
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
Category:Path-Connected Spaces
Category:Metric Spaces
Euclidean Topology is Product Topology

== Theorem ==
Let $T_1 = \struct {\R, \tau_1}$ be the topological space such that $\tau_1$ is the Euclidean topology on $\R$.
Let $T_n = \struct {\R^n, \tau_n}$ be the topological space such that $\tau_n$ is the product topology on the cartesian product $\ds \R_n = \prod_{i \mathop = 1}^n \R$.
Then the Euclidean topology on $\R^n$ and the product topology on $\R^n$ are the same.
== Proof 1 ==
== Proof 2 ==
Category:Euclidean Space
Category:Product Topology
Category:Euclidean Topology is Product Topology
Euclidean Topology is Product Topology/Proof 1

== Theorem ==
== Proof ==
Denote the Euclidean topology on $\R^n$ as $\tau$, and denote the product topology on $\R^n$ as $\tau'$.
Let $U \in \tau$, and let $x = \tuple{x_1, \ldots, x_n} \in U$.
Then there exists $\epsilon \in \R_{>0}$ such that the open ball $\map {B_\epsilon} x \subseteq U$.
We show that:
:$\ds B' = \prod_{i \mathop = 1}^n \openint {x_i - \dfrac \epsilon n} {x_i + \dfrac \epsilon n} \subseteq \map {B_\epsilon} x$
For if $y = \tuple {y_1, \ldots, y_n} \in B'$, then:
From Natural Basis of Product Topology of Finite Product, $B' \in \tau'$.
As $B' \subseteq \map {B_\epsilon} x \subseteq U$, Neighborhood Condition for Coarser Topology shows that $\tau' \subseteq \tau$.
Let $U' \in \tau'$.
Let $x = \tuple {x_1, \ldots, x_n} \in U'$.
From Natural Basis of Product Topology of Finite Product, sets of the type $\ds \prod_{i \mathop = 1}^n U'_i$ with $U'_i \in \tau_1$ form an analytic basis for $\tau'$.
From Equivalence of Definitions of Analytic Basis, it follows that we can select $U'_1, \ldots, U'_n \in \tau_1$ such that $\ds x \in \prod_{i \mathop = 1}^n U'_i \subseteq U'$.
By definition of open set, it follows that for all $i \in \set {1, \ldots, n}$, we can find $\epsilon_i \in \R_{>0}$ such that $\openint {x_i - \epsilon_i} {x_i + \epsilon_i} \subseteq U'_i$.
Put $\epsilon = \min \set {\epsilon_i : i = 1, \ldots, n}$.
We show that the open ball $\map {B_\epsilon} x \subseteq U'$.
For if $y = \tuple {y_1, \ldots, y_n} \in \map {B_\epsilon} x$, then $y_i \in \openint {x_i - \epsilon_i} {x_i + \epsilon_i }$, as:
:$\size {x_i - y_i} < \epsilon \le \epsilon_i$
It follows that:
:$\ds \map {B_\epsilon} x \subseteq \prod_{i \mathop = 1}^n \openint {x_i - \epsilon_i} {x_i + \epsilon_i} \subseteq \prod_{i \mathop = 1}^n U'_i \subseteq U'$
Then Neighborhood Condition for Coarser Topology shows that $\tau \subseteq \tau'$.
It follows that $\tau = \tau'$.
== Sources ==
* : $\S 20$
Category:Euclidean Topology is Product Topology