description/proof of for homeomorphism from metric space with induced topology, codomain topology is induced by metric induced by homeomorphism
Topics
About: topological space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of topology induced by metric.
Target Context
- The reader will have a description and a proof of the proposition that for any homeomorphism from any metric space with the induced topology, the codomain topology is induced by the metric induced by the homeomorphism.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\) with metric, \(dist_1\), with the induced topology, \(O_1\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\), with topology, \(O_2\)
\(f\): \(: M_1 \to T_2\), \(\in \{\text{ the homeomorphisms }\}\)
\(dist_2\): \(= \text{ the metric on } T_2\) such that \(\forall t_2, t'_2 \in T_2 (dist_2 (t_2, t'_2) = dist_1 (f^{-1} (t_2), f^{-1} (t'_2)))\)
\(O'_2\): \(= \text{ the topology induced by } dist_2\)
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Statements:
\(O_2 = O'_2\)
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2: Proof
Whole Strategy: Step 1: see that \(dist_2\) is a metric; Step 2: see that for each \(U_2 \in O_2\), \(U_2 \in O'_2\); Step 3: see that for each \(U'_2 \in O'_2\), \(U'_2 \in O_2\); Step 4: conclude the proposition.
Step 1:
\(dist_2\) is indeed a metric, because for each \(t_{2, 1}, t_{2, 2}, t_{2, 3} \in T_2\), 1) \(0 \le dist_2 (t_{2, 1}, t_{2, 2})\) and \(dist_2 (t_{2, 1}, t_{2, 2}) = 0 \iff t_{2, 1} = t_{2, 2}\): \(dist_2 (t_{2, 1}, t_{2, 2}) = dist_1 (f^{-1} (t_{2, 1}), f^{-1} (t_{2, 2}))\), but \(0 \le dist_1 (f^{-1} (t_{2, 1}), f^{-1} (t_{2, 2}))\), and \(dist_1 (f^{-1} (t_{2, 1}), f^{-1} (t_{2, 2})) = 0 \iff f^{-1} (t_{2, 1}) = f^{-1} (t_{2, 2}) \iff t_{2, 1} = t_{2, 2}\); 2) \(dist_2 (t_{2, 1}, t_{2, 2}) = dist_2 (t_{2, 2}, t_{2, 1})\): \(dist_2 (t_{2, 1}, t_{2, 2}) = dist_1 (f^{-1} (t_{2, 1}), f^{-1} (t_{2, 2})) = dist_1 (f^{-1} (t_{2, 2}), f^{-1} (t_{2, 1})) = dist_2 (t_{2, 2}, t_{2, 1})\); 3) \(dist_2 (t_{2, 1}, t_{2, 3}) \le dist_2 (t_{2, 1}, t_{2, 2}) + dist_2 (t_{2, 2}, t_{2, 3})\): \(dist_2 (t_{2, 1}, t_{2, 3}) = dist_1 (f^{-1} (t_{2, 1}), f^{-1} (t_{2, 3})) \le dist_1 (f^{-1} (t_{2, 1}), f^{-1} (t_{2, 2})) + dist_1 (f^{-1} (t_{2, 2}), f^{-1} (t_{2, 3})) = dist_2 (t_{2, 1}, t_{2, 2}) + dist_2 (t_{2, 2}, t_{2, 3})\).
Let us see that for each open ball, \(B_{m_1, \epsilon} \subseteq M_1\), \(f (B_{m_1, \epsilon}) = B_{f (m_1), \epsilon} \subseteq T_2\).
Let \(t_2 \in f (B_{m_1, \epsilon})\) be any.
\(t_2 = f (b_1)\) for a \(b_1 \in B_{m_1, \epsilon}\).
\(dist_2 (t_2, f(m_1)) = dist_1 (f^{-1} (t_2), f^{-1} (f(m_1))) = dist_1 (b_1, m_1) \lt \epsilon\).
So, \(t_2 \in B_{f (m_1), \epsilon}\).
So, \(f (B_{m_1, \epsilon}) \subseteq B_{f (m_1), \epsilon}\).
Let \(b_2 \in B_{f (m_1), \epsilon}\) be any.
\(dist_2 (b_2, f (m_1)) \lt \epsilon\).
But \(dist_2 (b_2, f (m_1)) = dist_1 (f^{-1} (b_2), f^{-1} (f (m_1))) = dist_1 (f^{-1} (b_2), m_1)\).
So, \(f^{-1} (b_2) \in B_{m_1, \epsilon}\).
So, \(b_2 \in f (B_{m_1, \epsilon})\).
So, \(B_{f (m_1), \epsilon} \subseteq f (B_{m_1, \epsilon})\).
So, \(f (B_{m_1, \epsilon}) = B_{f (m_1), \epsilon}\).
Also, for each open ball, \(B_{t_2, \epsilon} \subseteq T_2\), \(f^{-1} (B_{t_2, \epsilon}) = B_{f^{-1} (t_2), \epsilon} \subseteq M_1\), because \(B_{t_2, \epsilon} = f (B_{f^{-1} (t_2), \epsilon})\).
Step 2:
Let \(U_2 \in O_2\) be any.
Let \(u_2 \in U_2\) be any.
\(f^{-1} (u_2) \in f^{-1} (U_2)\).
As \(f\) is a homeomorphism, \(f^{-1} (U_2) \subseteq M_1 \in O_1\), and as \(O_1\) is induced by \(dist_1\), there is an open ball, \(B_{f^{-1} (u_2), \epsilon} \subseteq M_1\), such that \(B_{f^{-1} (u_2), \epsilon} \subseteq f^{-1} (U_2)\).
\(f (B_{f^{-1} (u_2), \epsilon}) \subseteq f (f^{-1} (U_2)) = U_2\).
But \(f (B_{f^{-1} (u_2), \epsilon}) = B_{u_2, \epsilon}\), so, \(B_{u_2, \epsilon} \subseteq U_2\).
That means that \(U_2 \in O'_2\).
Step 3:
Let \(U'_2 \in O'_2\) be any.
\(f^{-1} (U'_2) \in O_1\), because for each \(m_1 \in f^{-1} (U'_2)\), \(f (m_1) \in U'_2\), so, there is an open ball, \(B_{f (m_1), \epsilon} \subseteq T_2\), such that \(B_{f (m_1), \epsilon} \subseteq U'_2\), and \(f^{-1} (B_{f (m_1), \epsilon}) \subseteq f^{-1} (U'_2)\), but \(f^{-1} (B_{f (m_1), \epsilon}) = B_{m_1, \epsilon}\), so, \(f^{-1} (U'_2)\) is open in the topology induced by \(dist_1\), which is nothing but \(O_1\).
\(U'_2 = f (f^{-1} (U'_2)) \in O_2\), because \(f\) is a homeomorphism.
Step 4:
So, \(O_2 = O'_2\).
