Interior, Closure, and Derived Sets of A and B

The following question is from Fred H. Croom's book "Principles of Topology"

Let $A, B$ be subsets of a metric space with $A \subset B$. Prove that:

1. int $A$ $\subset$ int $B$;
2. $\bar{A} \subset \bar{B}$;
3. $A^{'}\subset B^{'}$.

The following are my attempts at proving the above statements.

Part 1. By definition, int$A$ is the set of all interior points of $A$. Let $x$ be such an interior point. Thus $A$ is a neighborhood of $x$ provided there is an open set $O$ which contains $x$ and is contained in $A$. Since $x \in O \subset A$, then $x \in O \subset B.$ Therefore int$A$ $\subset$ int$B$.

Part 2. Let $x\in \bar{A}$. Then every open set containing $x$ contains a point of $A$ by definition of the closure of a set $A$. Since $A \subset B$, then every open set containing $x$ contains a point of $B$. Thus $x\in\bar{B}$. Therefore $\bar{A} \subset \bar{B}$

Part 3. Let $x\in A^{'}$ Thus $x$ is a limit point $A$. This means every open set containing $x$ contains points of $A$ distinct from $x$. Since $A\subset B$, we say every point in $A$ is in $B$. Thus $x$ is a limit point of $B$. Therefore $A^{'}\subset B^{'}$.

Am I on the right track? Any suggestions or feedback?

I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.

• It's fine for me. For the interior, an alternative proof would be to say that, since $\operatorname{int} A$ is the largest open set contained in $A$, and $A\subset B$, it is an open set contained in $B$, hence in the largest open set contained in $B$. – Bernard Feb 17 '15 at 19:15

• I'm sorry, but a for a limit point $x$, the neighbourhoogs must contain a point of $A$ distinct from $x$. An isolated point of $A$ is not a limit point. – Bernard Feb 17 '15 at 19:26