Prove that for A $\subseteq$ B int(A) $\subseteq$ int(B) and cl(A) $\subseteq$ cl(B). I have to prove that for A$\subseteq$B, int(A) $\subseteq$ int(B)and cl(A) $\subseteq$ cl(B). I hope someone here can help me out, and I apologize for any obvious mistakes. So here is my approach.
By definition, the interior of B int(B) is the largest open set contained in B. It is the union of all open sets in B: $int(A)=_{W\subseteq A:\ W\;is\;closed}W$. Thus if int(B) $\subseteq$ B and A $\subseteq$ B, then int(A) $\subseteq$ int(B).
By definition, cl(B), is the smallest closed set containing B. It is the intersection of all closed sets containing B. Thus it must hold that cl(A) $\subseteq$ cl(B), as A $\subseteq$ cl(A) and B $\subseteq$ cl(B), with A $\subseteq$ B.
 A: Hints:
(a) If $U$ is an open set such that $U \subset A$ and $A\subseteq B$ then $U\subseteq B$. By definition, $\mbox{int}A$ is the union of all such open sets. Now, it is an open set, which is contained in $B$, right? What can you conclude?
(b) Analogously, if $F$ is a closed set containing $B$, then it must contain $A$. Thus, $\mbox{cl}B$ is the intersection of all such closed sets. This a closed set and it contains $B$. What can you conclude?
A: Hint for the 1st. If $x\in{\rm int}A$ then exists an open $W$ with $a\in W\subseteq A$, but $A\subseteq B$ so $x\in W\subseteq B$, therefore $x\in{\rm int}B$. This settles ${\rm int}A\subseteq{\rm int}B$.
A: $\operatorname{int}(A)$ is the maximal open set subset of $A$, so if $A \subseteq B$ it's in particular some open subset of $B$ too and hence a subset of the maximal open subset of $N$ which is by definition $\operatorname{int}(B)$. So $\operatorname{int}(A) \subseteq \operatorname{int}(B)$ follows.
Dually, $\overline{B}$ is the minimal closed superset of $B$, and hence also some closed superset of $A$ when $A \subseteq B$. As $\overline{A}$ is the minimal superset of $A$, again $\overline{A} \subseteq \overline{B}$ follows by minimality.
A: i recommend a pointwise approach. In other words, let $a \in int(A)$. Then by definition of being in the interior, we have that $\exists U$ open, such that,
$a \in U \subseteq A$. Since $A \subseteq B$ it follows that $a \in U \subseteq B$, i.e. $a \in int(B)$.
A: Note that for the second question: $cl(B)$ is a closed set and $A \subset B \subset cl(B)$. So by definition and from what we are given we know that $A \subset cl(A) \subseteq B \subset cl(B)$.
So, $cl(A) \subseteq cl(B)$.
For the first question, it might be easier to approach in the following way to show that given $A \subseteq B$, $int(A) \subseteq int(B)$:
Let $x \in int(A)$. Then by definition we know that for some $\epsilon > 0$ there exists an open ball $B_{\epsilon}(x) \subseteq A$. Since we are given that $A \subseteq B$ and now know that $ B_{\epsilon}(x) \subseteq A$ $\implies$ $B_{\epsilon}(x) \subseteq B$ so $x \in int(B)$. So $int(A) \subseteq int(B)$.
