A description/proof of that for map from topological space into metric space, image of closed set is closed on image of domain, if for any sequence on closed set for which image of sequence converges on image of domain, convergent point is on image of closed set
Topics
About: topological space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of sequence on topological space.
- The reader knows a definition of convergence of net with directed index set.
- The reader admits the proposition that for any metric space, the closure of any subset equals the set of the convergent points of all the convergent sequences on the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any topological space into any metric space, the image of any closed set is closed on the image of the domain, if for any sequence on the closed set for which the image of the sequence converges on the image of the domain, the convergent point is on the image of the closed set.
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: Description
For any topological space, \(T_1\), any metric space, \(T_2\), and any map, \(f: T_1 \rightarrow T_2\), the image of any closed set, \(f (C)\), is closed on \(f (T_1)\), if for any sequence, \(s: N \rightarrow C\), on \(C\) for which the image of \(s\), \(f (s (n))\), converges on \(f (T_1)\), the convergent point, \(lim_{n \rightarrow \infty} f (s (n))\), is on \(f (C)\), which is \(lim_{n \rightarrow \infty} f (s (n)) \in f (C)\).
2: Proof
The collection of the images of all the sequences on \(C\), \(\{f (s)\}\), represents all the sequences on \(f (C)\), because for any \(p \in f (C)\), \(p = f (p')\) for a \(p' \in C\), and any sequence on \(f (C)\), \(s': N \rightarrow f (C)\), can be realized by \(s' (n) = f (s (n))\). \(f (T_1)\) is a metric space as a subspace of \(T_2\), and we talk in terms of \(f (T_1)\), not of \(T_2\), hereafter.
Let us suppose that \(lim_{n \rightarrow \infty} f (s (n)) \in f (C)\) for any sequence, \(s: N \rightarrow C\), on \(C\) for which the image of \(s\), \(f (s (n))\), converges on \(f (T_1)\).
By the proposition that for any metric space, the closure of any subset equals the set of the convergent points of all the convergent sequences on the subset, the closure of \(f (C)\) on \(f (T_1)\), \(\overline{f (C)}\), is the set of the convergent points of all the convergent sequences on \(f (C)\). \(f (C) \subseteq \overline{f (C)}\), because any \(p \in f (C)\) is the convergent point of the constant sequence, \(f (s (n)) = p\). \(\overline{f (C)} \subseteq f (C)\), because for any \(p \in \overline{f (C)}\), \(p\) is the convergent point of a sequence, and \(p = f (p')\) for a \(p' \in C\) by the supposition of the proposition, so, \(p \in f (C)\). So, \(f (C) = \overline{f (C)}\), closed.
3: Note
We are talking about the closeness of \(f (C)\) on \(f (T_1)\), not on \(T_2\), so, we have talked about the sequences that converge on \(f (T_1)\), not on \(T_2\), and \(\overline{f (C)}\) is the closure of \(f (C)\) on \(f (T_1)\), not on \(T_2\), while \(f (T_1)\) is a metric space as a subspace of \(T_2\).
