Prove that $\overline{A}\subset\overline{B}$ Let $A,\ B$ denote subsets of a space $X$. Prove that if $A\subset B$, then $\overline{A}\subset\overline{B}$.
My attempt:
Let $x\in\overline{A}$
$\implies x\in\bigcap\limits_{\alpha\in K}U_\alpha$ where $U_\alpha$ is a closed set of $X$ and $U_\alpha\supset A$
$\implies x\in U_\alpha\ \forall\alpha\in K$
How do I proceed?
 A: If $x$ belongs to every closed set containing $A$, then, in particular, $x$ belongs to every closed set containg $B$ (since $A\subset B$) and therefore $x\in\overline B$.
A: Suppose $A \subseteq B$.
Then always $B \subseteq \overline{B}$, and $\overline{B}$ is closed. Both are true from the definition of $\overline{B}$. 
Now, $\overline{A} = \bigcap \{C: C \text{ is closed and } A \subseteq C\}$, and
as we saw that $\overline{B}$ is closed and contains $A$ (as $A \subseteq B \subseteq \overline{B}$) the set $\overline{B}$ is just one of the sets we take the intersection of in this definition of $\overline{A}$ and as the intersection of a (non-empty) family is a subset of each set we are intersecting, by definition, we thus have $\overline{A} \subseteq \overline{B}$.
A: Let $x\in\overline{A}$ then by definition for every $\varepsilon>0$ the ball $B(x,\varepsilon)\cap A\neq\emptyset$. Since $A\subset B$ then  $$\emptyset\neq B(x,\varepsilon)\cap A=B(x,\varepsilon)\cap (A\cap B)=(B(x,\varepsilon)\cap B)\cap A\Rightarrow B(x,\varepsilon)\cap B\neq \emptyset$$
hence $x\in\overline{B}$.
$\textbf{For a general topological space:}$ Let $X$ be a topological space and $A, B\in X$ such that $A\subset B$. If $x\in \overline{A}$ then by definition for each neighborhood $N$ of $x$ we have $N\cap A\neq\emptyset$. We need to show $\overline{A}\subset\overline{B}$. From the argument above since $A\subset B$ then 
$$N\cap A=N\cap (A\cap B)=(N\cap B)\cap A\neq\emptyset$$
implies $N\cap B\neq\emptyset$ for any neighborhood $N$ of $x$. Hence $x\in\overline{B}$.
A: Let $x\in \bar A$.  Then $x$ is a limit point of $A$.  Then $\exists$ a sequence $(x_n)$ in $A$ with $x_n\to x$.  But $(x_n)$ is also a sequence in $B$ ($x_n\in A\subset B \forall n$).  Thus $x\in \bar B$.
A: Correct me if wrong.
$A \subset B.$
$\overline{A}= A\cup A'$ , where $A'$ is the set of limit points of $A.$
Need to show that $A' \subset B'$.
Let $x \in A'$. Then every open neighbourhood $K_r(x)$ contains points $y \in A,$ and  $y \not = x.$
But $A \subset B$ :
Thus every open neighbourhood $K_r(x)$ , $x \in A'$, contains points $y \in B$, and $y \not = x$, i.e. $x \in B'$.
Hence: $A' \subset B'.$
Finally: 
$\overline{A} =A \cup A' \subset B\cup B'= \overline{B}$.
A: Let $x\in \overline{A}$, then $x$ is in all closed set containing $A$.
Every closed set containing $B$ contain $A$, so contain $x$. Then $x$ is in all closed set cointaining $B$, therefore $x\in\overline{B}$
