Inclusion with open, closed sets and border

Question

I have tree exercises in which I'm asked to determine whether or not the following propositions are true and, to prove them if they are true.

In the case they are are false, I have to find a counterexample.

1) $\ A \text{ open}$ , $\ B \text{ open}$,$\ A \subset B $ $\to $ $\partial A \subset B $

2) $\ A \text{ open}$ , $\ B \text{ closed}$, $A \subset B $ $\to $ $\partial A \subset B $

Here I don't know how to proceed. I know that:

$\ \partial A $ is not included in $\ A $ and $\ \partial A \subset A^{\complement}$, where $A^{\complement}$ is the complementary subset of $A$.

How can I go on?

Answer 1

Counter-examples to (1). Let $A=B=(0,1)$ in the space $\Bbb R.$ Then $\partial A =\overline A \cap \overline {A^c}=\{0,1\}$ which is non-empty and disjoint from $B.$... If you want $A \ne B$ let $A=(0,1)$ and $B=(0,2)$ in the space $\Bbb R.$ Then $0\in (\partial A) \setminus B.$ ... The "point" is that $A$ and $B$ can share a common boundary point $p$ when$A\subset B$, and that if $B$ is open then $ \partial B=\overline B \setminus B,$ so $p\not \in B$.

For (2) if $A\subset B=\overline B$ then $$B=\overline B\supset \overline A\supset (\overline A\cap \overline {A^c})=\partial A.$$ Regardless of whether or not $A$ is open.

Answer 2

1) As a counterexample you can take $A=(0,1)$ and $B=(0,2)$ both as subsets of $\mathbb R$

2) This is true. If $A\subset B$ and $B$ is closed then $\partial A\subseteq\mathsf{cl}(A)\subseteq B$.

Further you say that: "I know that $\partial A$ is not included in $A$..." But there are cases where $\partial A\subseteq A$. This if $A$ is closed, and the fact that $A$ is open does not imply that $A$ is not closed. In e.g. the discrete topology every set is open and closed.