$R_1$ paracompact spaces are normal How to prove that an $R_1$ and paracompact space is a normal space?
To include definitions, let $X$ be a topological space.


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*We say that $X$ is paracompact if every open cover has an open refinement that is locally finite.

*We say that $X$ is R1 if any two topologically distinguishable points can be separated by disjoint open neighbourhoods. 


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*We say that $x,y \in X$ are topologically indistinguishable if they have exactly the same neighborhoods. That is, if $x$ and $y$ are points in $X$, and $A$ is the set of all neighborhoods that contain $x$, and $B$ is the set of all neighborhoods that contain $y$, then $x$ and $y$ are "topologically indistinguishable" if and only if $A = B$.
Otherwise we call $x,y \in X$ topologically distinguishable. This means there is an open set containing precisely one of the two points.


*We say that $X$ is a normal space if given any disjoint closed sets $E$ and $F$, there are open neighbourhoods $U$ of $E$ and $V$ of $F$ that are also disjoint.

There is a construction called the Kolmogorov quotient of a space $X$. This is the quotient space of $X$ by the relation $x\sim y$ whenever $x,y$ are topologically indistinguishable. The quotient map $q:X\to K(X)$ preserves almost all topological properties, which is not surprising since it induces a bijection $f_*:\tau_X\to\tau_{K(X)}, U\mapsto f(U)$, between the topologies, with inverse $f^*:U\mapsto f^{-1}(U)$.
In particular, if $X$ is paracompact, then so is $K(X)$: Given an open cover $\cal U$ of $K(X)$, this pulls back to an open cover of $X$, which has a locally finite open refinement cover $\cal V$, whose image is an open refinement cover for $\cal U$. It remains to show that $f_*(\cal V)$ is locally finite. For any point $y\in K(X)$, pick a point $x\in f^{-1}(y)$. This $x$ has an open neighborhood $W$ intersecting only a finite number of sets in $\cal V$. But then $f(W)$ intersects only finitely many images since $f_*$ preserves intersections (which follows from the fact that its inverse $f^*$ does).
Now $R_1$-spaces can be characterized as those spaces having a Hausdorff Kolmogorov quotient. So if $X$ is a paracompact $R_1$-space, $K(X)$ is a paracompact Hausdorff space, and therefore it is normal. This implies that $X$ is normal as well.
