description/proof of that metric space with induced topology is normal
Topics
About: metric space
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of normal topological space.
- The reader knows a definition of open ball around point on metric space.
- The reader admits the proposition that for any metric space and any subset, the set of the points whose distances to the subset are \(0\) is the closure of the subset.
Target Context
- The reader will have a description and a proof of the proposition that any metric space with the topology induced by the metric is normal.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\), with the topology induced by the metric
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Statements:
\(M \in \{\text{ the normal topological spaces }\}\)
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2: Proof
Whole Strategy: Step 1: take any closed subsets, \(C_1, C_2 \subseteq M\), such that \(C_1 \cap C_2 = \emptyset\); Step 2: deal with the case that \(C_1 = \emptyset\) or \(C_1 = \emptyset\), and suppose otherwise thereafter; Step 3: see that \(C_j\) is the set of the points of \(M\) whose distances to \(C_j\) are \(0\); Step 4: take \(U_1 := \cup_{c_1 \in C_1} B_{c_1, dist (c_1, C_2) / 3}\) and \(U_2 := \cup_{c_2 \in C_2} B_{c_2, dist (c_2, C_1) / 3}\), and see that \(U_1 \cap U_2 = \emptyset\).
Step 1:
Let \(C_1, C_2 \subseteq M\) be any closed subsets such that \(C_1 \cap C_2 = \emptyset\).
Step 2:
When \(C_1 = \emptyset\), \(U_1 := \emptyset\) and \(U_2 := M\) satisfy that \(C_1 \subseteq U_1\), \(C_2 \subseteq U_2\), and \(U_1 \cap U_2 = \emptyset\).
When \(C_2 = \emptyset\), \(U_1 := M\) and \(U_2 := \emptyset\) satisfy that \(C_1 \subseteq U_1\), \(C_2 \subseteq U_2\), and \(U_1 \cap U_2 = \emptyset\).
Let us suppose otherwise, hereafter.
Step 3:
\(C_j\) is the set of the points of \(M\) whose distances to \(C_j\) are \(0\), by the proposition that for any metric space and any subset, the set of the points whose distances to the subset are \(0\) is the closure of the subset: \(\overline{C_j} = C_j\).
That means that for each \(m \in M\), \(dist (m, C_j) = 0\) if and only if \(m \in C_j\).
Step 4:
So, for each \(c_1 \in C_1\), \(0 \lt dist (c_1, C_2)\), because \(c_1 \notin C_2\).
For each \(c_2 \in C_2\), \(0 \lt dist (c_2, C_1)\), likewise.
Let us take \(U_1 := \cup_{c_1 \in C_1} B_{c_1, dist (c_1, C_2) / 3}\) and \(U_2 := \cup_{c_2 \in C_2} B_{c_2, dist (c_2, C_1) / 3}\).
\(U_1 \subseteq M\) and \(U_2 \subseteq M\) are some open subsets.
\(C_1 \subseteq U_1\) and \(C_2 \subseteq U_2\).
Let us see that \(U_1 \cap U_2 = \emptyset\).
Let us suppose that there was an \(m \in U_1 \cap U_2\).
\(m \in B_{c_1, dist (c_1, C_2) / 3}\) for a \(c_1 \in C_1\) and \(m \in B_{c_2, dist (c_2, C_1) / 3}\) for a \(c_2 \in C_2\).
\(dist (c_1, c_2) \le dist (c_1, m) + dist (m, c_2) \lt dist (c_1, C_2) / 3 + dist (c_2, C_1) / 3\).
But \(dist (c_1, C_2) \le dist (c_1, c_2)\) and \(dist (c_2, C_1) \le dist (c_1, c_2)\).
So, \(dist (c_1, C_2) \lt dist (c_1, C_2) / 3 + dist (c_2, C_1) / 3\) and \(dist (c_2, C_1) \lt dist (c_1, C_2) / 3 + dist (c_2, C_1) / 3\).
From the former, \(2 dist (c_1, C_2) \lt dist (c_2, C_1)\), from the latter, \(dist (c_2, C_1) \lt 1 / 2 dist (c_1, C_2)\).
So, \(2 dist (c_1, C_2) \lt 1 / 2 dist (c_1, C_2)\), a contradiction.
So, there is no such \(m\), so, \(U_1 \cap U_2 = \emptyset\).
So, \(M\) is normal.
