1049: Convergence of Map from Topological Space Minus Point into Topological Space w.r.t. Point

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definition of convergence of map from topological space minus point into topological space w.r.t. 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: Structured Description
• 2: Note

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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 convergence of map from topological space minus point into topological space with respect to 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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$t_1$: $\in T_1$
$f$: $:T_1\setminus \{t_1\}\to T_2$
$\ast t_2$: $\in T_2$
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Conditions:
$\mathrm{\forall }{N}_{t_2}\subseteq T_2\in \{\text{ the neighborhoods of }{t}_2\}(\mathrm{\exists }{N}_{t_1}\subseteq T_1\in \{\text{ the neighborhoods of }{t}_1\}(f(N_{t_1}\setminus \{t_1\})\subseteq N_{t_2}))$
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There may be some multiple convergences, but when there is the unique convergence, the convergence can be denoted as $li{m}_{t^{\prime }\to t_1}f(t^{\prime })$.

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2: Note

The intention of taking $T_1\setminus \{t_1\}$ instead of $T_1$ as the domain of $f$ is that $f$ is not required to be defined at $t_1$, not that "$f$ must not be defined at $t_1$".

In fact, when there is an $f^{\prime }:T_1\to T_2$, $f:T_1\setminus \{t_1\}\to T_2$ can be canonically defined as the domain restriction of $f^{\prime }$, and the convergence of $f^{\prime }$ with respect to $t_1$ can be regarded to be the convergence of $f$ with respect to $t_1$.

The reason why we need to think of $T_1\setminus \{t_1\}$ is that there are some important cases in which $f$ is not defined at $t_1$: for example, let $T_1=(-\delta ,\delta )\subseteq \mathbb{R}$, $T_2\in \{\text{ the finite-dimensional }\mathbb{R}\text{ vectors spaces }\}$ with the canonical topology, $t_1=0$, and for a $g:T_1\to T_2$, $f:(-\delta ,\delta )\setminus \{0\}\to T_2,t\mapsto (g(t)-g(0))/(t-0)$, where $li{m}_{t\to 0}f(t)$ is the derivative.

When $T_2$ is Hausdorff and $\{t_1\}\subseteq T_1$ is not open, there can be only 1 convergence: let us suppose that there were 2 convergences, $t_2,t_2^{\prime }\in T_2$, such that $t_2\ne t_2^{\prime }$; there would be some open neighborhoods of $t_2$ and $t_2^{\prime }$, $U_{t_2}$ and $U_{t_2^{\prime }}$, such that $U_{t_2}\cap U_{t_2^{\prime }}=\mathrm{\varnothing }$; there would be some neighborhoods of $t_1$, $N_{t_1}$ and $N_{t_1}^{\prime }$, such that $f(N_{t_1}\setminus \{t_1\})\subseteq U_{t_2}$ and $f(N_{t_1}^{\prime }\setminus \{t_1\})\subseteq U_{t_2^{\prime }}$; but $f((N_{t_1}\cap N_{t_1}^{\prime })\setminus \{t_1\})\subseteq U_{t_2}\cap U_{t_2^{\prime }}=\mathrm{\varnothing }$, which would mean that $N_{t_1}\cap N_{t_1}^{\prime }=\{t_1\}$, which would mean that $\{t_1\}$ was open, a contradiction against the supposition.

In fact, when $\{t_1\}\subseteq T_1$ is open, each point, $t_2\in T_2$, is a convergence with respect to $t_1$, because for any $N_{t_2}$, $N_{t_1}:=\{t_1\}$ can be chosen.

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References

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