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Let $X$ be a set:

Let us write $int(F)$ the interior of $F$

I want to show that $int(U \cap V) \subset int(U) \cap int(V)$

My reasonning goes like so:

  • $U \cap V \subset U \implies int(U \cap V) \subset int(U)$

  • $U \cap V \subset V \implies int(U \cap V) \subset int(V)$

Therefore:

  • $int(U \cap V) \cap int(U \cap V) \subset int(U) \cap int(V) \implies int(U \cap V) \subset int(U) \cap int(V) $

Does this constitute a rigorous proof

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    $\begingroup$ Yes, provided that you’ve already shown that $A\subseteq B$ implies that $\operatorname{int}A\subseteq\operatorname{int}B$. And you can go directly from the first two bullet points to the conclusion: if $A\subseteq B$ and $A\subseteq C$, then $A\subseteq B\cap C$ without any need for an intermediate $A\cap A\subseteq B\cap C$. $\endgroup$ – Brian M. Scott Sep 23 '16 at 18:07

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