In my textbook there is this exercise: "Let [imath]A,B \subseteq \mathbb{R}^n[/imath]. Prove or disprove that if [imath]A[/imath] is open, [imath]B[/imath] is closed and [imath]B \subset A[/imath], then [imath]\partial B \subset A[/imath]." In the textbook there is this previously proved theorem that if a set is closed then it contains its boundary, hence in this exercise it is [imath]\partial B \subseteq B[/imath] because by hypothesis [imath]B[/imath] is closed. But by hypothesis [imath]B \subset A[/imath], hence it is [imath]\partial B \subseteq B \subset A[/imath] and this implies [imath]\partial B \subset A[/imath]. So I would say that this is true. However, I did not use that [imath]A[/imath] is open; but I can't find mistakes in my proof. So, is the hypothesis "[imath]A[/imath] open" superfluous, or am I doing a mistake?
Question about a (maybe) superfluous hypothesis in a sets inclusion proof
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pka
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I see nothing a-miss about your proof. If [imath]x\in \partial (B)[/imath] then any open that contains [imath]x[/imath] must contain a point of [imath]B[/imath] and a point not in [imath]B[/imath].
Because [imath]B[/imath] is a closed set, it contains its boundary: i.e. [imath]\partial (B)\subseteq B\subset A[/imath].
Then clearly [imath]x\in A.[/imath]