A description/proof of that closure of difference of subsets is not necessarily difference of closures of subsets, but is contained in closure of minuend
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of closure of subset.
- The reader admits the proposition that the closure of the intersection of any finite number of subsets is contained in the intersection of the closures of the subsets.
Target Context
- The reader will have a description and a proof of the proposition that the closure of the difference of any 2 subsets is not necessarily the difference of the closures of the subsets, but is contained in the closure of the minuend.
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, \(T\), and any subsets, \(S_1, S_2 \subseteq T\), the closure of the difference of the subsets, \(\overline{S_1 \setminus S_2}\), is not necessary the difference of the closures of the subsets, \(\overline{S_1} \setminus \overline{S_2}\), but is contained in the closure of the minuend, \(\overline{S_1}\), which means \(\overline{S_1 \setminus S_2} \subseteq \overline{S_1}\).
2: Proof
Suppose that \(T\) is the \(\mathbb{R}^2\) Euclidean topological space, \(S_1\) is the open ball, \(B_{0-2}\), and \(S_2\) is the open ball, \(B_{0-1}\), where \(B_{p-r}\) denotes the open ball centered at \(p\) with the \(r\) diameter. Then, \(\overline{S_1 \setminus S_2}\) contains the border of \(B_{0-1}\), but \(\overline{S_1} \setminus \overline{S_2}\) does not. So, as there is a counterexample, \(\overline{S_1 \setminus S_2}\) is not necessarily \(\overline{S_1} \setminus \overline{S_2}\).
\(S_1 \setminus S_2 = S_1 \cap (T \setminus S_2)\). By the proposition that the closure of the intersection of any finite number of subsets is contained in the intersection of the closures of the subsets, \(\overline{S_1 \setminus S_2} \subseteq \overline{S_1} \cap \overline{(T \setminus S_2)} \subseteq \overline{S_1}\).
