description/proof of that for metric space, closed subset is \(G_\delta\)
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of metric space.
- The reader knows a definition of \(G_\delta\) subset of topological space.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that for any metric space, any closed subset is \(G_\delta\).
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
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(C\): \(\in \{\text{ the closed subsets of } T\}\)
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Statements:
\(C \in \{\text{ the } G_\delta \text{ subsets of } T\}\)
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2: Proof
Whole Strategy: Step 1: for each \(j \in \mathbb{N} \setminus \{0\}\), take \(U_j := \cup_{c \in C} B_{c, 1 / j}\), and take \(U := \cap_{j \in \mathbb{N} \setminus \{0\}} U_j\); Step 2: see that \(C \subseteq U\); Step 3: see that \(U \subseteq C\); Step 4: conclude the proposition.
Step 1:
For each \(j \in \mathbb{N} \setminus \{0\}\), let us take \(U_j := \cup_{c \in C} B_{c, 1 / j}\), where \(B_{c, 1 / j}\) is the \(1 / j\)-'open ball' around \(c\).
\(U_j\) is an open subset of \(T\) as a union of open subsets.
Let us take \(U := \cap_{j \in \mathbb{N} \setminus \{0\}} U_j\).
\(U\) is a \(G_\delta\) subset.
Step 2:
Let us see that \(C \subseteq U\).
For each \(c \in C\), for each \(j \in \mathbb{N} \setminus \{0\}\), \(c \in U_j\), which means that \(c \in U\).
So, \(C \subseteq U\).
Step 3:
Let us see that \(U \subseteq C\).
Let \(u \in U\) be any.
Let \(U_u \subseteq T\) be any open neighborhood of \(u\).
There is an \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\) and \(B_{u, \epsilon} \subseteq U_u\).
There is a \(j \in \mathbb{N} \setminus \{0\}\) such that \(1 / j \lt \epsilon\).
As \(u \in U\), \(u \in U_j\), which means that there is a \(c \in C\) such that \(u \in B_{c, 1 / j}\), which means that \(dist (u, c) \lt 1 / j\), which means that \(c \in B_{u, \epsilon}\).
So, \(B_{u, \epsilon} \cap C \neq \emptyset\), so, \(U_u \cap C \neq \emptyset\).
So, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, \(u \in \overline{C}\).
But as \(C\) is closed, \(\overline{C} = C\), so, \(u \in C\).
So, \(U \subseteq C\).
Step 4:
So, \(U = C\), which means that \(C\) is a \(G_\delta\) subset.
