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The following is taken from "Essential Mathematics for Undergraduates: A Guided Approach to Algebra, Geometry, Topology and Analysis" by Simon G. Chiossi

$\color{Green}{Background:}$

$\textbf{(1) Definition:}$ Let $S\neq\emptyset$ be a set. A $\textbf{Moore (closure) operator}$ is a function $K:P(S)\to P(S)$ such that

$(i)$ $X\subset K(X)$

$(ii)$ $K$ is increasing: $X\subset Y$ implies $K(X)\subset K(Y)$

$(iii)$ $K$ is idempotent: $K(K(X))=K(X)$ for all $X,Y\in P(S).$

$\textbf{(2) Exercise:}$ Show that

$i)$ $K$ is increasing if and only if $K(X\cup Y)\subset K(X)\cup K(Y)$

enter image description here

$\textit{Note:}$ From @DaveL.Renfro's comments, it there is a misprint in ii) of the exercise above.

$\color{Red}{Questions:}$

For the exercise above, I think I have to show

$X\subset Y$ implies $K(X)\subset K(Y)$ if and only if $K(X\cup Y)\subset K(X)\cup K(Y)$

I am having trouble showing both directions. In the forward direction where $X\subset Y$ implies $K(X)\subset K(Y)$ is assumed. I am having trouble showing the conclusion.

I know that from property (i) letting $X\cup Y$ be $X$, I get $X\cup Y\subset K(X\cup Y)$

and plugging in $X\cup Y$ for $Y$ in (ii), I get $X\subset X\cup Y$ implies $K(X)\cup K(X\cup Y)$ Doing the same by plugging in $Y$ for $X$ and $X\cup Y$ for $Y$ in (ii), I get $K(Y)\subset K(X\cup Y).$ I don't know what to do from here.

For the backward direction, if I assume $K(X\cup Y)\subset K(X)\cup K(Y),$ how do I even get $X\subset Y,$ since I am not sure if I can assume that $X\subset Y\subset K(X\cup Y).$

Thank you in advance.

Seth
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  • What are the assumptions on $K$ in the exercise? Presumably $K$ is a function from $P(S)$ into $P(S).$ Are we allowed to assume $K$ satisfies (i) and/or (iii)? – Dave L. Renfro Sep 19 '23 at 17:28
  • @DaveL.Renfro $K$ is assume to satisfy all three properties in the definition. Also, there is a similar question that have been asked on here in the context of closure relating to point set topology. But I don't think I can just imitate that solution outright. – Seth Sep 19 '23 at 17:33
  • I think you have the inclusion in the exercise backwards, and with this change the if and only if exercise only requires that $K$ be a function from $P(S)$ to $P(S).$ Assume the inclusion I mentioned is reversed. For the only if half, note that $X \subseteq X \cup Y$ and $Y \subseteq X \cup Y$ IMPLIES $K(X) \subseteq K(X \cup Y)$ and $K(Y) \subseteq K(X \cup Y),$ which in turn IMPLIES $K(X) \cup K(Y) \subseteq K(X \cup Y).$ For the if half, assuming $X \subseteq Y$ we have $K(X) = K(X) \cup \emptyset \subseteq K(X) \cup K(Y-X) \subseteq K(X \cup (Y-X)) = K(Y).$ (continued) – Dave L. Renfro Sep 19 '23 at 19:15
  • FYI, each of these is also equivalent to the property $K(X \cap Y) \subseteq K(X) \cap K(Y).$ Incidentally, this result is stated as an unproved Lemma at the beginning of "3. Monotone Closure Operators [satisfy (semi-K1)]" in the 1st part of this 3-part mathoverflow answer, which I mention in case you're interested in some of the mathematical applications of these ideas. If no one else posts and answer, and you can confirm the incorrect inclusion in the exercise, then at some later time I'll expand my comments a bit to an answer. – Dave L. Renfro Sep 19 '23 at 19:15
  • @DaveL.Renfro thank you so much for your answers. I made the post because I was looking into galois connections and how they relation to left and right adjoint functors. I found out that Galois connections are basically closure operators. – Seth Sep 19 '23 at 20:25
  • @DaveL.Renfro I posted a screenshot of where the exercise and definition came from. From your answers, I think there is a misprint in the exercise. – Seth Sep 19 '23 at 23:32

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