361: Closure of Difference of Subsets Is Not Necessarily Difference of Closures of Subsets, But Is Contained in Closure of Minuend

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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

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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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 $\overline{S_1\setminus S_2}\subseteq \overline{S_1}$.

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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, $\overline{S_1\setminus S_2}\subseteq \overline{S_1}\cap \overline{(T\setminus S_2)}\subseteq \overline{S_1}$.

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References

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