Need a counterexample related to Interior and Closure Let "cl" denotes closure, "int" denotes interior.I'm looking for a Single example of a subset $A$ of some topological space $X$ where all the following sets are unequal:
$1.$A 
$2.$ int(A),
$3.$cl(A),
$4.$ int(cl(A)),
$5.$ Cl(int(cl(A))),
$6.$ cl(int(A)),
$7.$ int(cl(int(A)))
It's clear that such a set $A$ is neither open nor closed.Moreover,in $\mathbb R$ with usual topology It's easy to construct some example using intervals and $\mathbb Q$ where some of the above sets are unequal.But I'm finding it a bit hard to construct a example where all the above $7$ sets are unequal.Any hints/ideas?
 A: HINT: Do it in pieces. Let $A_1=[0,1)\cup(1,2)\cup\{3\}$; then $\operatorname{cl}A_1=[0,2]\cup\{3\}$, $\operatorname{int}\operatorname{cl}A_1=(0,2)$, and $\operatorname{cl}\operatorname{int}\operatorname{cl}A_1=[0,2]$. Moreover, $\operatorname{int}A_1=(0,1)\cup(1,2)$, so we’ve taken care of everything except $\operatorname{cl}\operatorname{int}A$ and $\operatorname{int}\operatorname{cl}\operatorname{int}A$. 
Now let $A_2=(-1,4)\setminus A_1=(-1,0)\cup\{1\}\cup[2,3)\cup(3,4)$. Show that $A_2$, $\operatorname{cl}A_2$, $\operatorname{int}A_2$, $\operatorname{cl}\operatorname{int}A_2$, and $\operatorname{int}\operatorname{cl}\operatorname{int}A_2$ are all distinct.
Now what if you had a set $A$ that was like a discrete union of $A_1$ and $A_2$? And how can you make such a set?
A: Try $A = ([0,2] \cap \mathbb{Q}) \cup \{ 3 \} \cup (4,5) \cup (5,6) \subseteq \mathbb{R}$, with the usual topology.
Edit: after looking at the "possible duplicate" above, the comments there show an almost identical solution.
