1
$\begingroup$

I'm given a list of test-prep problems, which involve unstated theorems which we are to guess the statement of and then prove. I'm able to work out what most expect me to show, however the following is unclear.

Problem. Suppose that $A$ is a proper, non-empty subset of $\mathbb{R}$. What can be proven about $A$? Prove it. Hint: note that $A \subseteq \bar{A}$, and $A^c \subseteq \bar{A^c}$.

(Here, $A^c$ denotes the complement of $A$ with respect to the reals, and $\bar{A}$ denotes the closure of $A$ (that is, the union of $A$ and all its limit points))


Is there some obvious property of $A$, $\bar{A}$, or $A^c$ that can be proven here? My best guess is that I am to prove some set relationship (given the hints), however it is unclear to me what that might be (perhaps some subset relationship between $\bar{A}$ and $A^c$?).

I suspect that this is some "standard" problem, however I'm unable to turn anything up with simple Google/Stack Exchange searches.

$\endgroup$
6
  • $\begingroup$ Besides the fact that $~A \cup A^c~ = \Bbb{R},~$ you have a problem expressing $~\overline{A}~$ (i.e. the closure of $~A~$ in terms of either $~A~$ or $~A^c.~$ If $~A~$ is closed then $~\overline{A} = A.~$ However, in the stated problem, you are not given any information about whether $~A~$ or $~A^c~$ is closed. For example, it could be that neither $~A~$ nor $~A^c~$ is closed. $\endgroup$ Commented Mar 26, 2023 at 23:13
  • $\begingroup$ What you can say is that $~A \subseteq \overline{A}.$ This implies that $~\left[ ~\overline{A} ~\right]^c \subseteq A^c.$ $\endgroup$ Commented Mar 26, 2023 at 23:14
  • $\begingroup$ @user2661923 Sorry, could you clarify the meaning of the bracket notation ( I assume that $[\bar{A}]^c \not = \bar{A}^c$)? $\endgroup$ Commented Mar 27, 2023 at 0:08
  • $\begingroup$ I am only using the bracket notation to represent order of operations. You start with the set $~A.~$ Then you compute the closure of $~A.~$ Then you compute the complement of (the closure of $~A$). $\endgroup$ Commented Mar 27, 2023 at 0:12
  • $\begingroup$ There are (literally) infinitely many theorems that can be proven about $A$. For a trivial example, you can prove "$1 \in A\text{ or } 1\notin A$". Without further guidance, the question is unanswerable. And calling a necessary part of the question a "hint" beggars the meaning of that word. So presumably that further guidance should have come from the context of the problem that you haven't given here. If you could indicate the various things that were defined in the material being covered by the test, then we could probably figure out what they want, but as is, no. We can only guess. $\endgroup$ Commented Mar 27, 2023 at 18:11

1 Answer 1

0
$\begingroup$

Although the question is open-ended, I would argue that the following identity is the natural 'answer': $$ \overline{\left(\overline{\left(\overline{\left(\overline{A}\right)^c}\right)^c}\right)^c} = \overline{\left(\overline{A}\right)^c} $$ It is universal in the sense that not only is it true for all $A\subseteq \mathbb{R}$, but any other equality of sets (obtained through successive closure and complement of a subset $A\subseteq \mathbb{R}$) that is true for all $A\subseteq \mathbb{R}$, is a consequence of this identity (in a sense that we make formal below).

Formally, let $L$ denote the closure operation on subsets of the reals, and let $X$ denote the complement operation on subsets of the reals. Then $X,L$ generate a monoid $M$, which is partially ordered by: $U\leq V$ if and only if $U(A)\subseteq V(A)$ for all $A\subseteq \mathbb{R}$.

It is easy to see that the following hold in $M$:\begin{eqnarray}&1)& X^2=1\\&2)&L^2=L\\&3)& L\geq1 \\&4)& \forall U,V,W \in M, \,\, U\leq V \implies UW\leq VW \\&5)& \forall U,V \in M, \,\, U\leq V \implies LU\leq LV \\&6)& \forall U,V \in M, \,\, U\leq V \implies XU\geq XV \end{eqnarray}

Using just $(1)-(6)$ above, one can deduce the identity $$LXLXLXL=LXL\qquad (*)$$

Further $M$ is the monoid freely generated by $X,L$ subject to the relations $$*,\qquad L^2=L,\qquad X^2=1.$$

To see this, note that any element of $M$ is a word in $X,L$, and from the second and third relation above, we may choose that word to consist of alternating $X$'s and $L$'s. From $(*)$ we may choose that word to not contain $LXLXLXL$. This only leaves $14$ words.

Thus any identity between words in $X,L$ that is not implied by the three relations above, must equate two of these $14$ elements of $M$. However this is not possible, as we may find a single set $A\subseteq\mathbb{R}$ such that applying these $14$ elements of $M$ to $A$ results in fourteen distinct sets. Thus $$|M|=14.$$

$\endgroup$
1
  • $\begingroup$ I left the proof of $(*)$ and the construction of $A$ (such that $|MA|=14$) as exercises, as it looked like you wanted the statements to prove yourself. If you want any help with the proof\construction just let me know. $\endgroup$
    – tkf
    Commented Mar 29, 2023 at 19:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .