375: For 1st Countable Topological Space, Some Facts About Points Sequences and Subset

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A description/proof of that for 1st countable topological space, some facts about points sequences and 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 1st countable topological space.
• The reader knows a definition of limit of points sequence on topological space.
• The reader knows a definition of closure of subset.
• The reader knows a definition of interior of subset.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any 1st countable topological space, 1) any point is in the closure of any subset if and only if the point is the limit of a points sequence on the subset; 2) any point is in the interior of any subset if and only if every points sequence that converges to the point is eventually in the subset; 3) any subset is closed if and only if the subset contains the limit of every converging points sequence on the subset; 4) any subset is open if and only if every points sequence that converges to a point on the subset is eventually in 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 1st countable topological space, $T$, 1) for any point, $p\in T$, and any subset, $S\subseteq T$, $p\in \bar{S}$ if and only if $p$ is the limit of a points sequence on $S$; 2) for any point, $p\in T$, and any subset, $S\subseteq T$, $p\in IntS$ if and only if every points sequence that converges to $p$ is eventually in $S$; 3) for any subset, $S\subseteq T$, $S$ is closed if and only if $S$ contains the limit of every converging points sequence on $S$; 4) for any subset, $S\subseteq T$, $S$ is open if and only if every points sequence that converges to any point on $S$ is eventually in $S$.

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2: Proof

Let us denote the neighborhood basis around $p$, $B_p$, and the counted neighborhoods in $B_p$, $B_1,B_2,...$.

Let us define an infinitely countable decreasing sequence of neighborhoods around $p$, $B_1^{\prime },B_2^{\prime },...$, such that $B_{i+1}^{\prime }\subseteq B_i^{\prime }$, as follows. Let us define $B_1^{\prime }:=B_1$, a neighborhood around $p$. When $B_i^{\prime }$, a neighborhood around $p$, is already determined, there is the smallest, $j$, such that $B_j\subseteq B_1\cap B_2\cap ...\cap B_i\cap B_i^{\prime }$, and let us define $B_{i+1}^{\prime }:=B_j$, which is a neighborhood around $p$. $B_i^{\prime }\subseteq B_i$ and $B_{i+1}^{\prime }\subseteq B_i^{\prime }$, so, $B_j^{\prime }\subseteq B_i$ for any $i\le j$.

Let us prove 1). Let us suppose that $p\in \bar{S}$. If $p\in S$, $p$ is the limit of the constant $p$ sequence, which is on $S$. Let us suppose that $p$ is an accumulation point of $S$. As $B_i^{\prime }\cap S\ne \mathrm{\varnothing }$, let us take a point, $p_i\in B_i^{\prime }\cap S$. $p_1,p_2,...$ is a points sequence on $S$. For any neighborhood, $U_p$, there is a $B_i\subseteq U_p$. Then, $p_j\in B_j^{\prime }\subseteq U_p$ for $i\le j$. So, $p$ is the limit of the points sequence.

Let us suppose that $p$ is the limit of a points sequence on $S$. If $p\in S$, $p\in \bar{S}$. If $p\notin S$, for any neighborhood, $U_p$, $p_i\in U_p$, but $p_i\in S$, so, $p_i\in U_p\cap S$. So, $p$ is an accumulation point of $S$. $p\in \bar{S}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

Let us prove 2). Let us suppose that $p\in IntS$. Let us suppose that $p_1,p_2,...$ is any points sequence that converges to $p$. There is a neighborhood, $U_p\subseteq IntS\subseteq S$, and the points sequence is eventually in $U_p$, and so, in $S$.

Let us suppose that every points sequence that converges to $p$ is eventually in $S$. Let us suppose that $p\notin IntS$. Then, for any neighborhood, $U_p$, $\mathrm{\neg }{U}_p\subseteq S$. Especially, $\mathrm{\neg }{B}_i^{\prime }\subseteq S$, and let us take a point, $p_i\in B_i^{\prime }$ such that $p_i\notin S$. $p_1,p_2,...$ converges to $p$, because for any neighborhood, $U_p$, there is a $B_i\subseteq U_p$, but $p_j\in B_j^{\prime }\subseteq B_i\subseteq U_p$ for any $i\le j$, which (the points sequence) is not eventually in $S$, a contradiction.

Let us prove 3). Let us suppose that $S$ is closed. For the limit, $p$, of any converging points sequence on $S$, $p\in \bar{S}$, by 1). But as $\bar{S}=S$, $p\in S$.

Let us suppose that $S$ contains the limit of every converging points sequence on $S$. For any point, $p\in \bar{S}$, $p$ is the limit of such a sequence, by 1). So, $p\in S$ by the supposition. So, $\bar{S}\subseteq S\subseteq \bar{S}$, so, $\bar{S}=S$.

Let us prove 4). Let us suppose that $S$ is open. For any points sequence that converges to any point, $p\in S$, $p\in IntS=S$, so, the sequence is eventually in $S$, by 2).

Let us suppose that every points sequence that converges to any point on $S$ is eventually in $S$. For any point, $p\in S$, $p\in IntS$, by 2). So, $S\subseteq IntS\subseteq S$, so, $IntS=S$.

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References

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