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Question: Let B be a fixed subset of $X$ and for each nonempty $A\subset X$, let $\bar{A}=A\cup B$, with $\bar{\emptyset}=\emptyset$. Verify that $A \rightarrow \bar{A}$ is a closure operation.

Comments: I want to verify that my thought process for fitting these requirements is legitimate. The definition of Closure and the theorem used were from Willard's, "General Topology." Thank you for any critique or comment in advance.

Definition of Closure: $\bar{E}=\bigcap\{K\subset X$: K is closed and $E\subset K\}$

Theorem: The operation $A \rightarrow \bar{A}$ in a topological space has the following properties:

K-a) $A\subset \bar{A}$
Proof:
$A\subset (A\cup B)=\bar{A}\rightarrow A\subset \bar{A}$

K-b) $\bar{(\bar{A})}=\bar{A}$
Proof: $\overline{(\bar{A})}=\overline{A\cup B}=(A\cup B)\cup(A\cup B)^{'}$. Since $A\cup B$ is closed, $A\cup B=(A\cup B)^{'}$, so $\overline{(\bar{A})}=\overline{A\cup B}=(A\cup B)\cup(A\cup B)^{'}=A\cup B=\bar{A}$

K-c) $\overline{A\cup B}=\bar{A}\cup \bar{B}$
Notes: Since $A\cup B$ is closed, $A\cup B=(A\cup B)\cup (A\cup B)^{'}$ and $\bar{A}=A\cup B\rightarrow B$ is closed $\rightarrow \bar{B}=B\rightarrow\bar{B}\subset\bar{A}$
Proof: $\bar{A}\cup\bar{B}=\bar{A}=A\cup B=(A\cup B)\cup (A\cup B)^{'}=\overline{(A\cup B)}$

k-d) $\bar{\emptyset}=\emptyset$

This condition is met by definition.

$\therefore$ $A \rightarrow \bar{A}$ is a closure operation.

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  • $\begingroup$ Your "definition of closure" looks wrong. $\bar E$ should be the intersection of all $K$s, not the set of them. $\endgroup$ – Henning Makholm Oct 6 '17 at 19:31
  • $\begingroup$ More seriously, though, it looks like you should be starting out with an abstract definition of what a "closure operation" means. You're given (modulo the above comment) a definition of the operation called "closure" in a topological space -- which is presumably a "closure operation", but is not the operation you're asked to prove is a "closure operation" here. $\endgroup$ – Henning Makholm Oct 6 '17 at 19:34
  • $\begingroup$ You aren't given a topology. You are given a set $X$ and an operation $A \mapsto \overline{A}$on the subsets of $X$. You are being asked to show that operation satisfies the axioms for an abstract closure operation. When you have done this, you will know that $A \mapsto \overline{A}$ is the closure operation for a topology on $X$ (in this case, the topology whose closed sets comprise the empty set and the sets that contain the set $B$). $\endgroup$ – Rob Arthan Oct 6 '17 at 19:48
  • $\begingroup$ Your work seems to be aconfusion of showing A union B is closure operator and showing the topological closure of E is a closure operator. $\endgroup$ – William Elliot Oct 6 '17 at 19:57
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It's simpler than you think. You just have to show that the given operation obeys the 4 axioms for a topological closure operation:

  1. $\overline{\emptyset} = \emptyset$ is given by the definition.
  2. $A \subseteq A \cup B = \overline{A}$ for all $A$.
  3. $\overline{A_1} \cup \overline{A_2} = A_1 \cup B \cup A_2 \cup B = (A_1 \cup A_2) \cup B = \overline{A_1 \cup A_2}$ for all $A_1, A_2 \subseteq X$.
  4. $\overline{\overline{A}} = \overline{A \cup B} = A \cup B \cup B = A \cup B = \overline{A}$ for all $A$.

That's it really. Just simple set theory calculations.

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  • $\begingroup$ Thank you! Did I do axiom k-3) incorrectly then? I am new to these theorems and definitions, so I am using this problem to understand them. $\endgroup$ – James Snell Oct 6 '17 at 22:55
  • $\begingroup$ @JamesSnell No, that part didn’t make any sense to me. Closedness or derived sets have nothing to do with proving that axiom. $\endgroup$ – Henno Brandsma Oct 7 '17 at 8:07

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