Definition talk:T5 Space/Definition 2

We have a flag:

This is exactly how the condition is worded in : $\text{I}: \ \S 2$.

Is it actually a "mistake" as such, then? Or is it indeed accurate despite the a priori condition? --prime mover (talk) 10:10, 25 November 2012 (UTC)


 * I have written out what it means, and it seems that definition 2 is implied by, but does not imply definition 1. The latter is seen as follows:


 * Given $A^- \cap B = A \cap B^- = \varnothing$, the only reasonable thing to do is to take $Y = B^c$, yielding the premises of def 2. Thus we find $A'$ closed with $A \subseteq A' \subseteq Y$, and juggling with complements yields $B \subseteq (A')^c, A \subseteq Y^\circ$ and so $(A')^c, Y^\circ$ as the only natural $U,V$. It remains to verify they are disjoint, but $(A')^c \cap Y^\circ = \varnothing \iff Y^\circ \subseteq A'$. This condition cannot be derived (take e.g. $A = A' = \varnothing$) and so def 2 doesn't imply def 1.


 * I haven't managed to adjust def 2 to be equivalent to def 1. The intuitive repairing condition (demanding $Y^\circ \subseteq A'$ possible) comes down to asking, for all $Y$, $(Y^\circ)^- \subseteq Y$, but I haven't checked this against def 1; it may be too strong. --Lord_Farin (talk) 13:29, 25 November 2012 (UTC)


 * That def 2 couldn't be right is immediate from the fact that its content is vacuously true... --Lord_Farin (talk) 13:30, 25 November 2012 (UTC)


 * Suppose $A \subseteq Y$. In order for $A^-$ to be a neighborhood of $A$, there needs to exist $U \in \tau$ such that $A \subseteq U \subseteq A^-$.


 * Let $\R$ be the real number line under the usual topology. Let $A = \left({0, 1}\right]$. Then $A^- = \left[{0, 1}\right]$ but there is no open interval $U$ such that $A \subseteq U \subseteq A^-$. So $A^-$ is not, in this context, a closed nbhd of $A$. So if $Y = \left[{0, 1}\right]$ then while it is the case that $A^- \subseteq Y$, it is not the case that $Y$ a priori contains a closed nbhd of $A$.


 * Or am I missing something? --prime mover (talk) 15:06, 25 November 2012 (UTC)

Definition 2 does imply definition 1. Let $B = Y^{\complement}$. The conditions that $A \subseteq Y^\circ$ and $A^- \subseteq Y$ mean that $A$ and $B$ are separated. If $N$ is a closed neighborhood of $A$ that is contained in $Y$, then $N^{\complement}$ is an open neighborhood of $B$. We have $N \cap N^{\complement} = \varnothing$. Anything wrong? --abcxyz (talk) 17:41, 25 November 2012 (UTC)