588: Normed Vectors Spaces Map Continuous at Point

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definition of normed vectors spaces map continuous at point

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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: Structured Description
• 2: Natural Language Description
• 3: Note

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Starting Context

• The reader knows a definition of normed vectors space.
• The reader knows a definition of map.
• The reader knows a definition of limit of normed vectors spaces map at point.

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Target Context

• The reader will have a definition of normed vectors spaces map continuous at 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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F_1$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$F_2$: $\in \{\mathbb{R},\mathbb{C}\}$, with the canonical field structure
$V_1$: $\in \{\text{ the normed vectors spaces over }{F}_1\}$
$V_2$: $\in \{\text{ the normed vectors spaces over }{F}_2\}$
$p$: $\in V_1$
$\ast f$: $:V_1\to V_2$
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Conditions:
$\mathrm{\exists }li{m}_{p}f\in V_2\wedge li{m}_{p}f=f(p)$.
//

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2: Natural Language Description

For $\mathbb{R}$ or $\mathbb{C}$ with the canonical field structure, $F_1$, $\mathbb{R}$ or $\mathbb{C}$ with the canonical field structure, $F_2$, any normed vectors space over $F_1$, $V_1$, any normed vectors space over $F_2$, $V_2$, and any point, $p\in V_1$, any map, $f:V_1\to V_2$, such that the limit at $p$, $li{m}_{p}f$, exists and $li{m}_{p}f=f(p)$

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3: Note

This definition is equivalent with this definition:for any sequence, $p_1,p_2,...$, where $p_j\in V_1$, that converges to $p$, the sequence, $f(p_1),f(p_2),...$, converges to $f(p)$.

Let us confirm that fact.

Let us suppose that $f$ is continuous at $p$ by the definition of this article. For any $ϵ$, there is a $\delta$ such that for each $p^{\prime }$ such that $\Vert p^{\prime }-p\Vert <\delta$, $\Vert f(p^{\prime })-f(p)\Vert <ϵ$. For any sequence that converges to $p$, $p_1,p_2,...$, there is a $k$ such that for each $k<j$, $\Vert p_j-p\Vert <\delta$. Then, for each $k<j$, $\Vert f(p_j)-f(p)\Vert <ϵ$, which means that $f(p_1),f(p_2),...$ converges to $f(p)$.

For the other direction, let us think of the contraposition. Let us suppose that $f$ is not continuous at $p$. That means that for an $ϵ$, there is no $\delta$ such that for each $p^{\prime }\in V_1$ such that $\Vert p^{\prime }-p\Vert <\delta$, $\Vert f(p^{\prime })-f(p)\Vert <ϵ$, which implies that for each $\delta$, there is a $p^{\prime }$ such that $\Vert p^{\prime }-p\Vert <\delta$ but $ϵ\le \Vert f(p^{\prime })-f(p)\Vert$. Let us take ${\delta }_1=1/1,{\delta }_2=1/2,...$ and $p_1^{\prime },p_2^{\prime },...$ as $\Vert p_j^{\prime }-p\Vert <{\delta }_j$ and $ϵ\le \Vert f(p_j^{\prime })-f(p)\Vert$. Then, $p_1^{\prime },p_2^{\prime },...$ converges to $p$, but $f(p_1^{\prime }),f(p_2^{\prime }),...$ does not converge to $f(p)$, which is the negation of 'for any sequence, $p_1,p_2,...$, where $p_j\in V_1$, that converges to $p$, the sequence, $f(p_1),f(p_2),...$, converges to $f(p)$'. So, the contraposition is true, and so, the other direction is true.

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References

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