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Do there exist subsets with internal closures $A$ of $\mathbb R$ such that $A$ , $\bar A$ , $A^\circ$ , $(\bar A)^\circ$ , $\overline{A^\circ}$ are pairwise distinct?

I found an example from a book that such a set exists. For example, consider $$A=[0,1)\cup (1,2]\cup(\mathbb Q\cap [3,4])\cup\{5\}$$

My question on this example is here:

$$\bar A=[0,1]\cup[1,2]\cup[3,4]\cup\{5\}$$

$$ A^\circ\subset(0,1)\cup(1,2) $$

$$ \overline{A^\circ}\subset[0,1]\cup[1,2]=[0,2] $$

$$ (\bar A)^\circ\subset (0,1)\cup(1,2)\cup (3,4) $$

But, from these how can I say they are all pairwise distinct?

Otherwise cite an another examples regarding these conditions?

Can anyone help me ?

Empty
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    $A$ and $\bar A$ will never be disjoint. Although, $A^o$ and $\partial A$ are disjoint. – Matthew Levy Mar 31 '15 at 05:30
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    They are pairwise distinct. I.e. no two sets are the same. What they are not is pairwise disjoint, but as Matthew points out that can't happen no matter what. $A$ and $\bar{A}$ will never be disjoint unless $A=\emptyset$. – DRF Mar 31 '15 at 05:43
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    Look carefully, it is "distinct", not "disjoint". @Matthew Levy – adember Mar 31 '15 at 05:44
  • How can you say that they are pairwise distinct? please explain it.@DRF – adember Mar 31 '15 at 05:48
  • The sets $ A_1,A_2,\dots,A_n$ are pairwise disjoint means $A_i \cap A_j =\emptyset \ \forall \ i ,j \leq n \ & \ i \neq j.$ For any $A \subset \mathbb R ,\ A^0 \subset A \subset \overline{A}$ – Mirunalini_UML Mar 31 '15 at 05:48
  • Oh, I'm sorry. I should have read more carefully. – Matthew Levy Mar 31 '15 at 05:55
  • You’ll find this post useful. – Berrick Caleb Fillmore Apr 25 '15 at 02:55
  • I don't think you need the ${\bf Q}\cap [3,4]$ part is needed. Take just $A=[0,1)\cup (1,2]\cup{5}$ and draw pictures of all these sets. – tomasz Apr 25 '15 at 02:58
  • @ tomasz):That may be..But my question is that how we find that they are distinct ? As $A^0\subset.....$ not $=$ – Empty Apr 25 '15 at 03:08
  • @ Berrick Fillmore ): In your suggested question I found the example which I already know but NOT the explanation in the answer .. – Empty Apr 25 '15 at 03:12
  • Pairwise DISTINCT means they are UNEQUAL. The subset relation in each of the last 3 lines should be an equals-sign.So there are indeed 5 different sets..... BTW, the most number of distinct sets you can get starting with some subset X of a space S, by iterations of "complement" and "closure" is 14, including X. (Hint : Int (Cl(X))=Int(Cl(Int(Cl(X)))) and Cl(Int(X)) =Cl(Int(Cl(Int(X)))). You can get 14 for some subset of reals but I dk how. – DanielWainfleet Aug 04 '15 at 07:20

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