445: For Covering Map, 2 Lifts of Continuous Map from Connected Topological Space Totally Agree or Totally Disagree

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A description/proof of that for covering map, 2 lifts of continuous map from connected topological space totally agree or totally disagree

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of covering map.
• The reader knows a definition of lift of continuous map with respect to covering map.
• The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
• The reader admits the proposition that any open subset of any locally path-connected topological space is locally path-connected.
• The reader admits the proposition that any connected component on any locally path-connected topological space is open.
• The reader admits the proposition that any open set on any open topological subspace is open on the base space.
• The reader admits the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.

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Target Context

• The reader will have a description and a proof of the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any connected and locally path-connected topological spaces, $T_1,T_2$, any covering map, $\pi :T_1\to T_2$, which means that $\pi$ is continuous and surjective and around any point, $p\in T_2$, there is a neighborhood, $N_p\subseteq T_2$, that is evenly covered by $\pi$, any connected topological space, $T_3$, and any continuous map, $f:T_3\to T_2$, any 2 lifts of $f$, $f^{\prime },f^{″}:T_3\to T_1$, agree at each point of $T_3$ or disagree at each point of $T_3$.

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2: Proof

The subspace, ${\pi }^{-1}(N_p)$, may consist of multiple connected components, each denoted as ${{\pi }^{-1}(N_p)}_{\alpha }$ where $\alpha \in A_p$ where $A_p$ is a possibly uncountable indices set.

We can take an open neighborhood, $U_p\subseteq T_2$, as $N_p$, because if $N_p$ is not open, there is an open neighborhood, $U_p\subseteq N_p$, which is homeomorphic to each ${{\pi }^{-1}(U_p)}_{\alpha }$ by ${\pi }_{p,\alpha }:=\pi {|}_{{{\pi }^{-1}(U_p)}_{\alpha }}:{{\pi }^{-1}(U_p)}_{\alpha }\to U_p$, because ${\pi }_{p,\alpha }$ is obviously bijective, is continuous with the domain and the codomain regarded as the subspaces of ${{\pi }^{-1}(N_p)}_{\alpha }$ and $N_p$ respectively as a restriction of continuous $\pi {|}_{{{\pi }^{-1}(N_p)}_{\alpha }}:{{\pi }^{-1}(N_p)}_{\alpha }\to N_p$, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, and its inverse is continuous with the domain and the codomain regarded likewise as a restriction of continuous ${\pi {|}_{{{\pi }^{-1}(N_p)}_{\alpha }}}^{-1}$, likewise, but by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, those maps are continuous also with the domain and the codomain regarded as subspaces of $T_1$ and $T_2$ respectively.

${{\pi }^{-1}(U_p)}_{\alpha }$ is open on $T_1$, because ${\pi }^{-1}(U_p)$ is open on $T_1$ and is locally path-connected, by the proposition that any open subset of any locally path-connected topological space is locally path-connected, ${{\pi }^{-1}(U_p)}_{\alpha }$ is open on ${\pi }^{-1}(U_p)$, by the proposition that any connected component on any locally path-connected topological space is open, and ${{\pi }^{-1}(U_p)}_{\alpha }$ is open on $T_1$, by the proposition that any open set on any open topological subspace is open on the base space.

$\pi \circ f^{\prime }=\pi \circ f^{″}=f$ by the definition of lift of continuous map with respect to covering map.

For any point, $p\in T_3$, there are an $U_{f(p)}\subseteq T_2$ and $\{{{\pi }^{-1}(U_{f(p)})}_{\alpha }\subseteq T_1\}$. There are 2 possibilities: 1) $f^{\prime }(p)=f^{″}(p)\in {{\pi }^{-1}(U_{f(p)})}_{\alpha }$ for an $\alpha$; 2) $f^{\prime }(p)\ne f^{″}(p)$ where $f^{\prime }(p)\in {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{\prime }}$ and $f^{″}(p)\in {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{″}}$ for some ${\alpha }^{\prime }\ne {\alpha }^{″}$.

Let us suppose the possibility 1). As ${{\pi }^{-1}(U_{f(p)})}_{\alpha }$ is open on $T_1$ and $f^{\prime }$ is continuous, there is an open neighborhood, $U_p^{\prime }\subseteq T_3$, such that $f^{\prime }(U_p^{\prime })\subseteq {{\pi }^{-1}(U_{f(p)})}_{\alpha }$. Likewise, there is an open neighborhood, $U_p^{″}\subseteq T_3$, such that $f^{″}(U_p^{″})\subseteq {{\pi }^{-1}(U_{f(p)})}_{\alpha }$. Let us define $U_p:=U_p^{\prime }\cap U_p^{″}$. For any $p^{\prime }\in U_p$, $f^{\prime }(p^{\prime }),f^{″}(p^{\prime })\in {{\pi }^{-1}(U_{f(p)})}_{\alpha }$, ${\pi }_{p,\alpha }\circ f^{\prime }(p^{\prime })={\pi }_{p,\alpha }\circ f^{″}(p^{\prime })$, but as ${\pi }_{p,\alpha }$ is bijective, $f^{\prime }(p^{\prime })=f^{″}(p^{\prime })$. So, for any point, $p\in T_3$, if $f^{\prime }(p)=f^{″}(p)$, $f^{\prime }$ and $f^{″}$ agree on a neighborhood of $p$.

Let us suppose the possibility 2). As ${{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{\prime }}$ is open on $T_1$ and $f^{\prime }$ is continuous, there is an open neighborhood, $U_p^{\prime }\subseteq T_3$, such that $f^{\prime }(U_p^{\prime })\subseteq {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{\prime }}$. Likewise, there is an open neighborhood, $U_p^{″}\subseteq T_3$, such that $f^{″}(U_p^{″})\subseteq {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{″}}$. Let us define $U_p:=U_p^{\prime }\cap U_p^{″}$. For any $p^{\prime }\in U_p$, $f^{\prime }(p^{\prime })\in {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{\prime }}$ and $f^{″}(p^{\prime })\in {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{″}}$. As ${{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{\prime }}\cap {{\pi }^{-1}(U_{f(p)})}_{{\alpha }^{″}}=\mathrm{\varnothing }$, $f^{\prime }(p^{\prime })\ne f^{″}(p^{\prime })$. So, for any point, $p\in T_3$, if $f^{\prime }(p)\ne f^{″}(p)$, $f^{\prime }$ and $f^{″}$ disagree on a neighborhood of $p$.

So, $f^{\prime }$ and $f^{″}$ agree at each point of $T_3$ or disagree at each point of $T_3$, by the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

444: Banach Space

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A definition of Banach space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition
• 2: Note

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Starting Context

• The reader knows a definition of normed vectors space.
• The reader knows a definition of metric induced by norm.
• The reader knows a definition of complete metric space.

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Target Context

• The reader will have a definition of Banach space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

Any normed vectors space that is a complete metric space with the metric induced by the norm

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2: Note

Being complete is about metric space, so, the vectors space has to be given the metric.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

443: Separable Topological Space

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A definition of separable topological space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of dense subset of topological space.

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Target Context

• The reader will have a definition of separable topological space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

Any topological space that has a countable dense subset

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

442: Nowhere Dense Subset of Topological Space

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A definition of nowhere dense subset of topological space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of closure of subset.
• The reader knows a definition of interior of subset of topological space.

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Target Context

• The reader will have a definition of nowhere dense subset of topological space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

Any subset of any topological space whose (the subset's) closure interior is the empty set

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

441: Dense Subset of Topological Space

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A definition of dense subset of topological space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of closure of subset.

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Target Context

• The reader will have a definition of dense subset of topological space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

Any subset of any topological space whose (the subset's) closure is the topological space

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

440: Rotation in n-Dimensional Euclidean Vectors Space Is Same 2-Dimensional Rotations Along (n - 2)-Dimensional Subspace Axis

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A description/proof of that rotation in $n$-dimensional Euclidean vectors space is same $2$-dimensional rotations along $(n-2)$-dimensional subspace axis

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Topics

About: Euclidean vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of Euclidean vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that any rotation in the $n$-dimensional Euclidean vectors space is any same $2$-dimensional rotations along any $(n-2)$-dimensional subspace axis.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any Euclidean vectors space, ${\mathbb{R}}^n$, any rotation in ${\mathbb{R}}^n$ is any same $2$-dimensional rotations along any $(n-2)$-dimensional subspace axis.

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2: Note

This is not any rigorous proof but an intuitive understanding what rotation in any higher-dimensional Euclidean vectors space is.

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3: Proof

Any rotation is a kind of map, $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$.

Let us always think of rotations centered at the origin, which means that $f(0)=0$. Any rotation centered at another point, $p_0\in {\mathbb{R}}^n$, can be regarded to be $f^{\prime -1}\circ f\circ f^{\prime }$, where $f^{\prime }:{\mathbb{R}}^n\to {\mathbb{R}}^n$, $p\mapsto p-p_0$, the translation.

Any rotation in ${\mathbb{R}}^3$ is a rotation around an axis, which is a line (that passes the origin, which we will not mention here after, because our lines are always so, because we are thinking only about rotations centered at the origin).

As we tend to imagine a rotation as in ${\mathbb{R}}^3$, we tend to imagine a line as the axis.

But what is the axis for any rotation in ${\mathbb{R}}^2$? The line perpendicular to ${\mathbb{R}}^2$? But the line is not in ${\mathbb{R}}^2$, which seems a problem: why do we need to introduce the ambient ${\mathbb{R}}^3$ when we are thinking of the ${\mathbb{R}}^2$ space? At least, we did not need to introduce the ambient ${\mathbb{R}}^4$ in order to think of rotations in ${\mathbb{R}}^3$, which seems not symmetric.

What is 'axis' indeed? The axis of any rotation is the set of points that are fixed by the rotation.

In fact, the axis for any rotation in ${\mathbb{R}}^2$ is not the line in ${\mathbb{R}}^3$, but the origin, the point.

What is the axis for any rotation in ${\mathbb{R}}^4$? $(x^1,x^2,x^3,x^4)\mapsto (cos\theta x^1-sin\theta x^2,sin\theta x^1+cos\theta x^2,x^3,x^4)$ is a rotation, which fixes the $x^3,x^4$ coordinates, which means that the fixed set is $\{(0,0)\}\times {\mathbb{R}}^2$, which is the axis. So, the axis for any rotation in ${\mathbb{R}}^4$ is a $2$-dimensional subspace of ${\mathbb{R}}^4$.

Generally, the axis for any rotation in ${\mathbb{R}}^n$ is an $(n-2)$-dimensional subspace of ${\mathbb{R}}^n$.

Let us think of a rotation in ${\mathbb{R}}^3$. The rotation has the axis as a $(3-2)$-dimensional subspace, a line. At each point on the axis, there is the $(3-(3-2)=2)$-dimensional plane perpendicular to the axis. Do the rotation for such each plane centered at the intersection with the axis by any same angle, which is nothing but the rotation in ${\mathbb{R}}^3$. The reason why the axis of any rotation in ${\mathbb{R}}^3$ is a line is that $3-2=1$, not that axis in general is a line, which is wrong.

Let us think of a rotation in ${\mathbb{R}}^n$. The rotation has the axis as an $(n-2)$-dimensional subspace. At each point on the axis, there is the $(n-(n-2)=2)$-dimensional plane perpendicular to the axis. Do the rotation for such each plane centered at the intersection with the axis by any same angle, which is nothing but the rotation in ${\mathbb{R}}^n$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

439: Continuous Image of Path-Connected Subspace of Domain Is Path-Connected on Codomain

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A description/proof of that continuous image of path-connected subspace of domain is path-connected on codomain

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of path-connected topological space.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that any continuous map image of any path-connected subspace of the domain is path-connected on the codomain.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any topological spaces, $T_1,T_2$, any continuous map, $f:T_1\to T_2$, and any path-connected subspace, $T_3\subseteq T_1$, the image, $f(T_3)$, is path-connected on $T_2$.

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2: Proof

For any points, $p_1,p_2\in f(T_3)$, is there a path, $\lambda :[0,r]\to f(T_3)$ such that $p_1=\lambda (0)$ and $p_2=\lambda (r)$? We can take any point from $f^{-1}(p_i)\cap T_3$ as $p_i^{\prime }$. As $T_3$ is path-connected, there is a path, ${\lambda }^{\prime }:[0,1]\to T_3$, such that ${\lambda }^{\prime }(0)=p_1^{\prime }$ and ${\lambda }^{\prime }(1)=p_2^{\prime }$. Let us define $\lambda :=f{|}_{T_3}\circ {\lambda }^{\prime }:[0,1]\to T_3\to f(T_3)$, which is continuous as a composition of continuous maps ($f{|}_{T_3}$ is continuous by the proposition that any restriction of any continuous map on the domain and the codomain is continuous), and $\lambda (0)=p_1$ and $\lambda (1)=p_2$.

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3: Note

If the domain is a path-connected topological space, the domain is a path-connected subspace of itself, so, the image of any path-connected topological space is path-connected.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

438: Cauchy Sequence on Metric Space

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of Cauchy sequence on metric space

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition
• 2: Note

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Starting Context

• The reader knows a definition of metric space.

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Target Context

• The reader will have a definition of Cauchy sequence on metric space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

For any metric space, $M$, any sequence, $s:\mathbb{N}\to M$, where $\mathbb{N}$ is the positive natural numbers set, such that for any $ϵ\in \mathbb{R}$ such that $0<ϵ$, there is an $N\in \mathbb{N}$ such that for any $n_1,n_2\in \mathbb{N}$ such that $N<n_1,n_2$, $dist(s(n_1),s(n_2))<ϵ$

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2: Note

Sometimes, we may let $\mathbb{N}$ be the natural numbers set (with $0$).

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

437: Convergence of Sequence on Metric Space

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definition of convergence of sequence on metric space

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of metric space.

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Target Context

• The reader will have a definition of convergence of sequence on metric space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the metric spaces }\}$
$s$: $:\mathbb{N}\to M$, $\in \{\text{ the sequences of points on }M\}$
$\ast m$: $\in M$
//

Statements:
$\mathrm{\forall }ϵ\in \mathbb{R}\text{ such that }0<ϵ(\mathrm{\exists }N\in \mathbb{N}(\mathrm{\forall }n\in \mathbb{N}\text{ such that }N<n(dist(m,s(n))<ϵ)))$
//

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2: Note

Sometimes, we may use $\mathbb{N}\setminus \{0\}$ instead of $\mathbb{N}$.

When $M$ is made the topological space with the canonical induced topology, $\mathbb{N}$ is a directed index set, $s$ is a net with the directed index set, and any convergence of $s$ as the sequence is a convergence of $s$ as the net with the directed index set, because for any open neighborhood, $U_m\subseteq M$, of $m$, there is an $ϵ$ open ball, $B_{m,ϵ}\subseteq U_m$, around $m$, and as $s(n)\in B_{m,ϵ}$, $s(n)\in U_m$.

The convergence, $m$, is inevitably unique: let us suppose that there was another convergence, $m^{\prime }\in M$, which implies that $0<dist(m,m^{\prime })$; $dist(m,m^{\prime })\le dist(m,s(j))+dist(s(j),m^{\prime })$, so, $dist(m,m^{\prime })-dist(m,s(j))\le dist(s(j),m^{\prime })$; for each $N<n$, $dist(m,s(j))<dist(m,m^{\prime })/2$; so, $dist(m,m^{\prime })/2=dist(m,m^{\prime })-dist(m,m^{\prime })/2\le dist(s(j),m^{\prime })$, which means that $s$ would not converge to $m^{\prime }$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

436: Product Set

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of product set

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition 1
• 2: Definition 2
• 3: Note

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Starting Context

• The reader knows a definition of function.
• The reader knows a definition of tuple.

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Target Context

• The reader will have a definition of product set.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition 1

For any possibly uncountable indices set, $A$, and any sets, $\{S_{\alpha }|\alpha \in A\}$, indexed with $A$, the set of all the functions, $\{f:A\to {\cup }_{\alpha \in A}{S}_{\alpha }|f(\alpha )\in S_{\alpha }\text{ for each }\alpha \in A\}$, denoted as ${\times }_{\alpha \in A}{S}_{\alpha }$

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2: Definition 2

For any finite number of sets, $S_1,S_2,...,S_n$, the set of all the $n$-tuples, $\{⟨p_1,p_2,...,p_n⟩|p_j\in S_j\text{ for each }j=1\sim n\}$, denoted as $S_1\times S_2\times ...\times S_n$

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3: Note

Definition 1 is indeed a set: the set of all the functions from $A$ into ${\cup }_{\alpha \in A}{S}_{\alpha }$ is a set (see Proof 8 of the proposition that some expressions can be parts of legitimate formulas for the ZFC set theory), and the formula for the subset axiom is legitimate.

Definition 2 is indeed a set, by the proposition that the product of any finite number of sets is a set.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

435: Subset Minus Subset Is Complement of 2nd Subset Minus Complement of 1st Subset

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A description/proof of that subset minus subset is complement of 2nd subset minus complement of 1st subset

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of set.

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Target Context

• The reader will have a description and a proof of the proposition that for any set, any subset (the 1st subset) minus any subset (the 2nd subset) is the complement of the 2nd subset minus the complement of the 1st subset.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any set, $S$, and any subsets, $S_1,S_2\subseteq S$, $S_1\setminus S_2=(S\setminus S_2)\setminus (S\setminus S_1)$.

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2: Proof

For any element, $p\in S_1\setminus S_2$, $p\in S_1$ and $p\notin S_2$, $p\in S\setminus S_2$ and $p\notin S\setminus S_1$, so, $p\in (S\setminus S_2)\setminus (S\setminus S_1)$.

For any element, $p\in (S\setminus S_2)\setminus (S\setminus S_1)$, $p\in S\setminus S_2$ and $p\notin S\setminus S_1$, $p\in S_1$ and $p\notin S_2$, so, $p\in S_1\setminus S_2$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

434: Union of Indexed Subsets Minus Union of Subsets Indexed with Same Indices Set Is Contained in Union of Subset Minus Subset for Each Index

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A description/proof of that union of indexed subsets minus union of subsets indexed with same indices set is contained in union of subset minus subset for each index

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of set.

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Target Context

• The reader will have a description and a proof of the proposition that for any set and any possibly uncountable indices set, the union of any subsets (the 1st set of subsets) indexed with the indices set minus the union of any subsets (the 2nd set of subsets) indexed with the indices set is contained in the union of the subset from the 1st set minus the subset from the 2nd set for each index.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any set, $S$, any possibly uncountable indices set, $A$, and any sets of subsets, $\{S_{\alpha }|\alpha \in A\}$ and $\{S_{\alpha }^{\prime }|\alpha \in A\}$, ${\cup }_{\alpha \in A}{S}_{\alpha }\setminus {\cup }_{\alpha \in A}{S}_{\alpha }^{\prime }\subseteq {\cup }_{\alpha \in A}(S_{\alpha }\setminus S_{\alpha }^{\prime })$.

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2: Proof

For any element, $p\in {\cup }_{\alpha \in A}{S}_{\alpha }\setminus {\cup }_{\alpha \in A}{S}_{\alpha }^{\prime }$, $p\in S_{\alpha }$ for an $\alpha$ and $p\notin S_{\alpha }^{\prime }$ for each $\alpha$. $p\in S_{\alpha }\setminus S_{\alpha }^{\prime }$ for an $\alpha$, $p\in {\cup }_{\alpha \in A}(S_{\alpha }\setminus S_{\alpha }^{\prime })$.

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3: Note

Equality does not hold in general, because for any $p\in {\cup }_{\alpha \in A}(S_{\alpha }\setminus S_{\alpha }^{\prime })$, $p\in S_{\alpha }\setminus S_{\alpha }^{\prime }$ for an $\alpha$, $p\in S_{\alpha }$ and $p\notin S_{\alpha }^{\prime }$ for the $\alpha$, but if $p\in S_{\beta }^{\prime }$ for a $\beta \ne \alpha$, $p\notin {\cup }_{\alpha \in A}{S}_{\alpha }\setminus {\cup }_{\alpha \in A}{S}_{\alpha }^{\prime }$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

433: Bijection

<The previous article in this series | The table of contents of this series | The next article in this series>

A definition of bijection

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of injection.
• The reader knows a definition of surjection.

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Target Context

• The reader will have a definition of bijection.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

For any sets, $S_1$ and $S_2$, any map, $f:S_1\to S_2$, that is an injection and a surjection

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

432: Interior of Subset of Topological Space

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A definition of interior of subset of topological space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of topological space.

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Target Context

• The reader will have a definition of interior of subset of topological space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

For any topological space, $T$, and any subset, $S\subseteq T$, the largest open subset, $intS\subseteq T$, contained in $S$, which is $intS\subseteq S$

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

431: Complete Metric Space

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A definition of complete metric space

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Topics

About: metric space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition

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Starting Context

• The reader knows a definition of metric space.
• The reader knows a definition of Cauchy sequence on metric space.
• The reader knows a definition of convergence of sequence on metric space.

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Target Context

• The reader will have a definition of complete metric space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

Any metric space, $M$, such that any Cauchy sequence on $M$ converges to a point on $M$

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

430: 2 Continuous Maps from Connected Topological Space into Hausdorff Topological Space Such That, for Any Point, if They Agree at Point, They Agree on Neighborhood, Totally Agree or Totally Disagree

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A description/proof of that 2 continuous maps from connected topological space into Hausdorff topological space such that, for any point, if they agree at point, they agree on neighborhood, totally agree or totally disagree

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of Hausdorff topological space.
• The reader knows a definition of continuous map.
• The reader knows a definition of neighborhood of point.
• The reader admits the proposition that any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the point.
• The reader admits the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.

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Target Context

• The reader will have a description and a proof of the proposition that any 2 continuous maps from any connected topological space into any Hausdorff topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood, totally agree or totally disagree on the whole domain.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

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1: Description

For any connected topological space, $T_1$, any Hausdorff topological space, $T_2$, and any continuous maps, $f_1,f_2:T_1\to T_2$, such that if $f_1(p)=f_2(p)$ at any point, there is a neighborhood, $N_p$, of $p$ such that $f_1(p^{\prime })=f_2(p^{\prime })$ for each $p^{\prime }\in N_p$, $f_1(p)=f_2(p)$ for each $p\in T_1$ or $f_1(p)\ne f_2(p)$ for each $p\in T_1$.

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2: Proof

$f_1$ and $f_2$ disagree on a neighborhood if they disagree at any point, by the proposition that any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the point. So, $f_1$ and $f_2$ totally agree or totally disagree on whole $T_1$, by the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.

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3: Note

Any Euclidean topological space is Hausdorff, so, any 2 continuous maps from any connected topological space into any Euclidean topological space that (the maps) agree on a neighborhood if they agree at any point totally agree or totally disagree on the whole domain.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

429: 2 Continuous Maps into Hausdorff Topological Space That Disagree at Point Disagree on Neighborhood of Point

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A description/proof of that 2 continuous maps into Hausdorff topological space that disagree at point disagree on neighborhood of point

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of Hausdorff topological space.
• The reader knows a definition of continuous map.
• The reader knows a definition of neighborhood of point.

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Target Context

• The reader will have a description and a proof of the proposition that any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the point.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

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1: Description

For any topological space, $T_1$, any Hausdorff topological space, $T_2$, and any continuous maps, $f_1,f_2:T_1\to T_2$, if $f_1(p)\ne f_2(p)$ at any point, $p\in T_1$, there is a neighborhood, $N_p$, of $p$ such that $f_1(p^{\prime })\ne f_2(p^{\prime })$ for each point, $p^{\prime }\in N_p$.

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2: Proof

Let us suppose that $f_1(p)\ne f_2(p)$. There are some disjoint open neighborhoods, $U_{f_1(p)},U_{f_2(p)}\in T_2$, of $f_1(p),f_2(p)$, respectively, such that $U_{f_1(p)}\cap U_{f_2(p)}=\mathrm{\varnothing }$, because $T_2$ is Hausdorff. As $f_i$ is continuous, there is a neighborhood, $N_{p,i}$, of $p$ such that $f_i(N_{p,i})\subseteq U_{f_i(p)}$. $N_p:=N_{p,1}\cap N_{p,2}$ is a neighborhood, and $f_i(N_p)\subseteq U_{f_i(p)}$. $f_1(N_p)\cap f_2(N_p)=\mathrm{\varnothing }$, which means that $f_1(p^{\prime })\ne f_2(p^{\prime })$ for each $p^{\prime }\in N_p$.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>

428: Euclidean-Normed Euclidean Vectors Space

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A definition of Euclidean-normed Euclidean vectors space

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Definition
• 2: Note

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Starting Context

• The reader knows a definition of Euclidean vectors space.
• The reader knows a definition of Euclidean-norm on Euclidean vectors space.

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Target Context

• The reader will have a definition of Euclidean-normed Euclidean vectors space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Definition

Any Euclidean vectors space, ${\mathbb{R}}^n$, together with its Euclidean-norm

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2: Note

Although a Euclidean vectors space tends to be implicitly supposed to have the Euclidean norm, it is not necessarily so.

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References

<The previous article in this series | The table of contents of this series | The next article in this series>