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I am working on a problem from a class-note that is rife with typos, and I think there is also a bug here in this problem for I could not make any sense out of it:

If $\tau $ is a set of open sets in $X$ then the map $\chi : \mathscr P(X) \to \mathscr P(X)$ defined by
$\chi (A) = \{ x \in X :$ for every $U \in \tau$ if $x \in U$ then $U \cap A $ then $\neq \emptyset \}$
is an operation of closure.

Note that I copied it exactly as it is without any revision, with the definitions are for open set and operation of closure as follow:

Open Set: $\tau \subseteq \mathscr P (X)$ is a set of open sets in $X$ if $\tau$ satisfies the following:
$(\mathscr O_1)$ : $\emptyset \in \mathscr \tau$
$(\mathscr O_2)$ : $X \in \tau$
$(\mathscr O_3)$ : If $\{ U_i \}_{i \in I} \in \tau$, then $\cup _{i \in I} \in \tau$
$(\mathscr O_4)$ : If $U, V \in \tau$, then $U \cap V \in \tau$

Operation of Closure:
$(\mathscr C_1) : \kappa (\emptyset ) = \emptyset $
$(\mathscr C_2) : $ For $\forall A \in \mathscr P(X), A \subseteq \kappa (A)$
$(\mathscr C_3) : $ For $\forall A \in \mathscr P(X), \kappa (\kappa (A)) = \kappa (A)$
$(\mathscr C_4) : $ For $\forall A, B \in \mathscr P(X), \kappa (A \cup B) = \kappa (A) \cup \kappa (B)$.

My question is very simple: Do you see any typo here? Can you make any sense out of it? Thank you very much for your time and help. Happy holidays!

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  • $\begingroup$ @BrianM.Scott : I am so sorry - I have just made the correction. Thanks again. $\endgroup$ – Amanda.M Dec 15 '14 at 8:45
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The only direct problem I see is that $$\chi (A) = \{ x \in X :\text{for every }U \in \tau\text{ if }x \in U\text{ then }U \cap A \text{ then}\neq \emptyset \}$$ should be $$\chi (A) = \{ x \in X :\text{for every }U \in \tau,\text{ if }x \in U\text{ then }U \cap A \neq \emptyset \}$$ that is, without the second "then".

Also, the book really ought to use the standard term "topology" instead of "set of open sets", but that is probably on purpose ...

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  • $\begingroup$ Thank you! I will tackle the problem again after getting convinced that there is no typo. Thanks again. $\endgroup$ – Amanda.M Dec 15 '14 at 9:44

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