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> [!proposition] Proposition. ([[closure and intersection]])
> Let $A,B,$ and $A_{\alpha}$ denote subsets of a [[topological space]] $X$.
>
**$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$.**
>
Let $c \in \overline{A \cap B}$. Then $c$ is in the intersection of all [[closed set]]s containing $A \cap B$, hence in $\overline{A}$ and in $\overline{B}$.
>
*The reverse inclusion is not true!* Consider the [[open set]]s $(0,1)$ and $(1,2)$ in $\mathbb{R}$ with the [[standard topology on the real line|standard topology]].
>
More generally, we can say that:
>
**$\overline{\bigcap_{\alpha}^{} A_{\alpha} } \subset \bigcap_{\alpha}^{} \overline{A_{\alpha}}$.**
>
Let $c \in \overline{\bigcap_{\alpha}^{} A_{\alpha} }$. Then $c$ belongs to the intersection of all [[closed set]]s containing $\bigcap_{\alpha}^{}A_{\alpha}$; in particular it belongs to each $\overline{A_{\alpha}}$.
>
**There is no relationship between $\overline{A - B}$ and $\overline{A}-\overline{B}$.** To see this, consider $\mathbb{R}$ with the [[standard topology on the real line|standard topology]]. Letting $\mathbb{I}$ denote the irrational real numbers, we have that $\overline{\mathbb{R}-\mathbb{I}}=\overline{\mathbb{Q}}=\mathbb{R}$, which is not equal to $\overline{\mathbb{R}}-\overline{\mathbb{I}}= \emptyset$. So $\overline{A-B} \not \subset \overline{A} - \overline{B}$. On the other hand, letting $A=(0,1)$ and $B=(1,2)$ we see that $\overline{A}-\overline{B}=[0,1) \neq [0,1]= \overline{A-B}$. So $\overline{A}-\overline{B} \neq \overline{A-B}$.
^a21f89
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> ```
> [!frontlink]
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> ```