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Engelking states a theorem(2.1.7) in the book General Topology as follows:

For every $T_1$ spaces the following are equivalent:

(1) Every subspace of $X$ is normal.

(2) Every open subspace of $X$ is normal.

(3) Two separated sets have disjoint open neighborhoods.

I was asked to show that (1) and (3) are equivalent, however, I don’t know where the $T_1$ condition is used in the proof. My proof goes as follows:

If $X$ satisfies (3), then let $Y$ be a subspace and $A,B\subset Y$ disjoint closed sets in the subspace $Y$. Denote the closure operator in $X$ by $Cl$ and in $Y$ by $Cl^*$. We claim that $A, B$ are separated in $X$: $$Cl(A)\cap B= Cl(A)\cap Cl^*(B) = Cl(A)\cap Cl(B)\cap Y = Cl^*(A)\cap Cl^*(B)$$ But this is empty by the fact that $A, B$ are closed. Then we can take disjoint open neighborhood of $A,B$ in $X$, and intersect both by $Y$.

If $X$ satisfies (1), then take two separated sets $A, B$ in $X$. The open subspace $Y=X-(Cl(A)\cap Cl(B))$ contains each of $A $ and $B$ by the fact that they are separated. The closure of them in $Y$ is empty: $Cl^*(A)\cap Cl^*(B)= Cl(A)\cap Cl(B)\cap Y$ is empty because $Y$ does not contain $Cl(A)\cap Cl(B)$. Take disjoint open neighborhood of $Cl^*(A)$ and $Cl^*(B)$ in $Y$, then by the fact that $Y$ is open, we are done with the proof.

So where exactly did I use $T_1$ condition in my proof?

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    $\begingroup$ Does author require normal spaces to be Hausdorff? $\endgroup$ – William Elliot Dec 14 '18 at 3:31
  • $\begingroup$ @WilliamElliot The author requires normal spaces to be both $T_1$ and $T_4$. I understand it now-I use this book just for a reference so I was not aware it has an alternate definition for normal space. Thank you for the comment. $\endgroup$ – William Sun Dec 14 '18 at 3:35
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Be aware that Engelking has the tendency to assume extra separation axioms in some definitions: compact/paracompact includes Hausdorff, normal and regular includes $T_1$, perfectly normal includes normal (which includes $T_1$) etc.

So in order to have a space all of whose subspaces are normal, we can only consider $T_1$ spaces to begin with. It's the price of admission, as it were.

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