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paracompact Hausdorff spaces are normal

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Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

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topological homotopy theory

Contents

Statement

Proposition

(Dieudonné‘s theorem)

Every paracompact Hausdorff space is normal.

In particular compact Hausdorff spaces are normal.

Proof

Let (X,τ)(X,\tau) be a paracompact Hausdorff space

We first show that it is regular: To that end, let xXx \in X be a point, and let CXC \subset X be a closed subset not containing xx. We need to find disjoint open neighbourhoods U x{x}U_x \supset \{x\} and U CCU_C \supset C.

First of all, by the Hausdorff property there exists for each cCc \in C disjoint open neighbourhods U x,c{x}U_{x,c} \supset \{x\} and U c{c}U_c \supset \{c\}. As cc ranges, the latter clearly form an open cover {U cX} cC\{U_c \subset X\}_{c \in C} of CC, and so the union

{U cX} cCX\C \{U_c \subset X\}_{c \in C} \,\cup\, X \backslash C

is an open cover of XX. By paracompactness of (X,τ)(X,\tau), there exists a locally finite refinement, and by this lemma we may assume its elements to share the original index set and be contained in the original elements of the same index. Hence

{V cU cX} cC \{V_c \subset U_c \subset X\}_{c \in C}

is a locally finite collection of subsets, such that

U CcCV c U_C \coloneqq \underset{c \in C}{\cup} V_c

is an open neighbourhood of CC.

Now by definition of local finiteness there exists an open neighbourhood W x{x}W_x \supset \{x\} and a finite subset KCK \subset C such that

cC\K(W xV c=). \underset{c \in C \backslash K}{\forall}( W_x \cap V_c = \emptyset ) \,.

Consider then

U xW x(kK(U x,k)). U_x \;\coloneqq\; W_x \cap \left( \underset{k \in K}{\cap} \left( U_{x,k} \right) \right) \,.

which is an open neighbourhood of xx, by the finiteness of KK.

It thus only remains to see that

U xU C=. U_x \cap U_C = \emptyset \,.

But this holds because the only V cV_{c} that intersect W xW_x are the V kU kV_{k} \subset U_{k} for kKk \in K and each of these is by construction disjoint from U x,kU_{x,k} and hence from U xU_x.

This establishes that (X,τ)(X,\tau) is regular. Now we prove that it is normal. For this we use the same approach as before:

Let C,DXC,D \subset X be two disjoint closed subsets. We need to produce disjoint open neighbourhoods for these.

By the previous statement of regularity, we may find for each cCc \in C disjoint open neighbourhoods U c{c}U_c \supset \{c\} and U D,cDU_{D,c} \supset D. Hence the union

{U cX} cCX\C \left\{ U_c \subset X \right\}_{c \in C} \cup X \backslash C

is an open cover of XX, and thus by paracompactness has a locally finite refinement, whose elements we may, again by this lemma, assume to have the same index set as before and be contained in the previous elements with the same index. Hence we obtain a locally finite collection of subsets

{V cU cX} cC \{ V_c \subset U_c \subset X \}_{c \in C}

such that

U CcCV c U_{C} \coloneqq \underset{c \in C}{\cup} V_c

is an open neighbourhood of CC.

It is now sufficient to see that every point dDd \in D has an open neighbourhood U dU_d not intersecting U CU_C, for then

U DdDU d U_D \coloneqq \underset{d \in D}{\cup} U_d

is the required open neighbourhood of DD not intersecting U CU_C.

Now by local finiteness of {V cX} cC\{V_c \subset X\}_{c \in C}, every dDd \in D has an open neighbourhood W dW_d such that there is a finite set K dCK_d \subset C so that

cC\K d(V cW d=). \underset{c \in C \backslash K_d}{\forall} \left( V_c \cap W_d = \emptyset \right) \,.

Accordingly the intersection

U dW d(cK dCU D,c) U_d \coloneqq W_d \cap \left( \underset{c \in K_d \subset C}{\cap} U_{D,c} \right)

is still open and disjoint from the remaining V kV_k, hence disjoint from all of U CU_C.

Last revised on June 24, 2017 at 16:28:10. See the history of this page for a list of all contributions to it.