If $A\subseteq B$, then $A'\subseteq B'$

Question

Proof or counterexample: If $A\subseteq B$, then $A'\subseteq B'$. I have no idea where to start. Only thing I know is the definition of limit points.

$A'$ is the set of all limit points of $A$.

Answer 1

If $x\in A'$, then for every open neighborhood $U$ of $x$, there exists point $y_U \in U$ such that $y_U\in A\setminus\{x\}$. As $A\subseteq B$, $y_U\in B$. What conclusion can you make?

Therefore, for every open neighborhood $U$ of $x$, the point $y_U\in U$ is a point distinct from $x$ that is also in $B$. By the definition of limit points, $x$ must be a limit point of $B$, whence $A'\subseteq B'$.

Answer 2

Hint: argue by the contrapositive: that is, show that if $x \not \in B'$, then $x \not \in A'$.

Take $x \not \in B'$. Therefore, there is an open neighbourhood $U$ of $x$ with $U \setminus \{x\} \cap B = \emptyset$. Hence $U \setminus \{x\} \cap A \subset U \setminus \{x\} \cap B = \emptyset$, and so $U \setminus \{x\} \cap A = \emptyset$, or equivalently $x \not \in A'$.