A description/proof of that set of subsets around each point with conditions generates unique topology with each set being neighborhood basis
Topics
About: topological space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any set, any set of some subsets around each point with certain conditions generates the unique topology with each set being a neighborhood basis.
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 set, \(S\), and any set, \(S_p\), of some subsets around each point, \(p \in S\), which means that each element of \(S_p\) contains \(p\), that satisfies the conditions: [ 1) \(S_p\) is not empty; 2) for any \(S_1, S_2 \in S_p\), there is an \(S_3 \in S_p\) such that \(S_3 \subseteq S_1 \cap S_2\); 3) for any \(S_1 \in S_p\), there is a subset around \(p\), \(p \in S_2 \subseteq S_1\), such that for each \(p' \in S_2\), there is an \(S_3 \in S_{p'}\) such that \(S_3 \subseteq S_2\)], \(\{S_p\}\) generates the unique topology on \(S\) such that \(S_p\) is a neighborhood basis at \(p\).
2: Proof
Let us define a topology such that any subset \(S_1 \subseteq S\) is open if and only if at each point, \(p \in S_1\), there is an \(S_2 \in S_p\) such that \(S_2 \subseteq S_1\). It is really a topology, because \(S\) is open as \(S_p\) is not empty and any element of \(S_p\) is contained in \(S\); the empty set is vacuously open; any union of open sets, \(\cup_\alpha U_\alpha\), is open, because at each point, \(p \in \cup_\alpha U_\alpha\), \(p \in U_\alpha\) for an \(\alpha\), there is an \(S_1 \in S_p\) such that \(S_1 \subseteq U_\alpha\) as \(U_\alpha\) is open, and \(S_1 \subseteq \cup_\alpha U_\alpha\); any finite intersection of open sets, \(\cap_i U_i\), is open, because at each point, \(p \in \cap_i U_i\), \(p \in U_i\) for each \(i\), there is an \(S_i \in S_p\) such that \(S_i \subseteq U_i\) for each \(i\), \(\cap_i S_i \subseteq \cap_i U_i\), but there is an \(S_{1'} \in S_p\) such that \(S_{1'} \subseteq \cap_i S_i\) by the condition 2), and \(S_{1'} \subseteq \cap_i S_i \subseteq \cap_i U_i\).
\(S_p\) is a neighborhood basis at \(p\), because each element, \(S_1 \in S_p\), is a neighborhood, because by the condition 3), \(S_1\) contains an \(p \in S_2\) that is open; for each neighborhood, \(N_p\), of \(p\), there is an open neighborhood, \(U_p \subseteq N_p\), of \(p\), and there is an \(S_1 \in S_p\) such that \(S_1 \subseteq U_p \subseteq N_p\), by the definition of the topology.
The topology is unique by the proposition that for any topological space, any set of neighborhood bases at all points determines the topology.
3: Note
The condition 3) is necessary in order to guarantee that \(S_p\) is a neighborhood basis at \(p\), although the topology (which may not be the only possible topology) can be defined without it.
As an example that \(S_p\) cannot be any neighborhood basis for any topology without the condition 3), let us think of \(S = \mathbb{R}\), \(S_p = \{(p - 1, p + 1)\}\). \(\{S_p\}\) actually satisfies the conditions 1) and 2) where 2) is only \((p - 1, p + 1) \subseteq (p - 1, p + 1) \cap (p - 1, p + 1) = (p - 1, p + 1)\) as there is only 1 element in \(S_p\). Then, in order for \(S_p\) to be a neighborhood basis at \(p\), \((p - 1, p + 1)\) has to be a neighborhood of \(p\), which has to contain an open neighborhood of \(p\), which has to contain an element of \(S_p\), which can be only \((p - 1, p + 1)\), so, \((p - 1, p + 1)\) has to be the open neighborhood. Then, \((p - 1, p + 1) \cap (p + 1 - 1, p + 1 + 1) = (p, p + 1)\) has to be an open neighborhood of \(p + 1/2\), which has to contain an element of \(S_{p + 1/2}\), which can be only \((p + 1/2 - 1, p + 1/2 + 1) = (p - 1/2, p + 3/2)\), which is not really contained in \((p, p + 1)\). So, whatever topology we choose, \(S_p\) cannot be any neighborhood basis.
