If $\hat A= A \cup \left\{\right.$ connected components of $X-A$ which are relatively compact in $X\left.\right\}$, then for every $A \subseteq X$ (Here, $B$ is relatively compact means the closure of $B$ is compact.)

  
*
  
*$\hat A$ is compact.
  
*$\hat A=\hat {\hat A}$.
  
*$\hat A$ is connected.
  
*$\hat A=X$.

I try to eliminate the options by using an example.
Consider $X=\Bbb R - \{1,2,3\}$ with metric topology and let $A=(-\infty,1)$.
Then $\hat A=(-\infty,1) \cup (1,2) \cup (2,3)$.
Hence options 1,3,4 are false.
So I select option 2 as an answer.
Is my method correct?
 A: I think $\hat A=A$ for your example. The closure in your $X$ of $(1,2)$ equals $(1,2)$, which is not compact, and similarly for $(2,3)$. But you can still mark off 1 and 4 because of this.
How about $X=\{1\}\cup \{2\}$ and $A=\{1\}$. Then $\hat A=X$, which is not connected.  So you can rule out option 3.
Thus option 2 is correct (you may want to try a simple proof of this).
A: For (1) let $X=R$ and $A=(0,1)$. The connected components of $X-A$ are $(-\infty,0]$ and $[1,\infty)$ which are closed in $X$ but not compact.So $\hat A=A$, which is not compact.For (3) let $X=R^2$ and let $A$ contain exactly 2 points.Then $X-A$ is connected but $\overline {X-A}=X$ is not compact. So $\hat A=A$, which is not connected. Either of these will take care of (4)..... I have also seen the term "pre-compact" as a synonym for "relatively compact."
A: For (2), $X-\hat{A}$ is the union of non-relatively compact connected components of $X-A$ by definition. Let $O$ be any connected component of $X-A$ which is not relatively compact, then $O$ is a non-relatively compact connected component of $X-\hat{A}$ since $X-\hat{A}\subset X-A$. Therefore, $X-\hat{A}$ has no relatively compact connected components, which shows that $\hat{\hat{A}}=A$.
