401: On T_1 Topological Space, Point Is \omega-Accumulation Point of Subset iff It Is Accumulation Point of Subset

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A description/proof of that on $T_1$ topological space, point is $\omega$-accumulation point of subset iff it is accumulation point of subset

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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 $T_1$ topological space.
• The reader knows a definition of $\omega$-accumulation point of subset.
• The reader knows a definition of accumulation point of subset.

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Target Context

• The reader will have a description and a proof of the proposition that on any $T_1$ topological space, any point is an $\omega$-accumulation point of any subset if and only if the point is an accumulation point of the 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 $T_1$ topological space, $T$, and any subset, $S\subseteq T$, any point, $p\in T$, is an $\omega$-accumulation point of $S$ if and only if $p$ is an accumulation point of $S$.

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2: Proof

Let us suppose that $p$ is an accumulation point of $S$. For any neighborhood, $U_p$, of $p$, there is a point, $p_1\in U_p\cap S$ such that $p_1\ne p$. There is a neighborhood, $U_{p,1}$, of $p$ such that $p_1\notin U_{p,1}$, because $T$ is a $T_1$ space. There is a point, $p_2\in U_p\cap U_{p,1}\cap S$ such that $p_2\ne p$. $p_1\ne p_2$, because $p_1\notin U_{p,1}$. There is a neighborhood, $U_{p,2}$, of $p$ such that $p_2\notin U_{p,2}$. There is a point, $p_3\in U_p\cap U_{p,1}\cap U_{p,2}\cap S$ such that $p_3\ne p$. $p_1,p_2\ne p_3$, because $p_1,p_2\notin U_{p,1}\cap U_{p,2}$, and so on. After all, there is a point, $p_i\in U_p\cap S$, that is distinct from $p,p_1,p_2,...,p_{i-1}$, for any natural number $i$. So, $U_p\cap S$ has at least countably infinite number of points excluding $p$.

Let us suppose that $p$ is an $\omega$-accumulation point of $S$. Obviously, $p$ is an accumulation point of $S$.

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References

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