compact Hausdorff spaces are normal



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

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In fact a stronger statement hods: paracompact Hausdorff spaces are normal.

To prove this, consider the following lemma:


(separation by neighbourhoods of points from compact subspaces in Hausdorff spaces)


  1. (X,τ)(X,\tau) be a Hausdorff topological space;

  2. YXY \subset X a compact subspace.

Then for every xX\Yx \in X \backslash Y there exists

  1. an open neighbourhood U x{x}U_x \supset \{x\};

  2. an open neighbourhood U YYU_Y \supset Y

such that

  • they are still disjoint: U xU Y=U_x \cap U_Y = \emptyset.

of lemma

By the assumption that (X,τ)(X,\tau) is Hausdorff, we find for every point yYy \in Y disjoint open neighbourhoods U x,y{x}U_{x,y} \supset \{x\} and U y{y}U_y \supset \{y\}. By the nature of the subspace topology of YY, the restriction of all the U yU_y to YY is an open cover of YY:

{(U yY)Y} yY. \left\{ (U_y \cap Y) \subset Y \right\}_{y \in Y} \,.

Now by the assumption that YY is compact, there exists a finite subcover, hence a finite set SYS \subset Y such that

{(U yY)Y} ySY \left\{ (U_y \cap Y) \subset Y \right\}_{y \in S \subset Y}

is still a cover.

But the finite intersection

U xsSYU x,s U_x \coloneqq \underset{s \in S \subset Y}{\cap} U_{x,s}

of the corresponding open neighbourhoods of xx is still open, and by construction it is disjoint from all the U sU_{s}, hence in particular from their union

U YsSYU s. U_Y \coloneqq \underset{s \in S \subset Y}{\cup} U_s \,.

Therefore U xU_x and U YU_Y are two open subsets as required.


of prop.

First we claim that (X,τ)(X,\tau) is regular. To show this, we need to find for each point xXx \in X and each disjoint closed subset YXY \in X dijoint open neighbourhoods U x{x}U_x \supset \{x\} and U YYU_Y \supset Y. But since closed subspaces of compact spaces are compact, the subset YY is in fact compact, and hence this is in fact the statement of lemma .

Next to show that (X,τ)(X,\tau) is indeed normal, we apply the idea of the proof of lemma once more:

Let Y 1,Y 2XY_1, Y_2 \subset X be two disjoint closed subspaces. By the previous statement then for every point y 1Yy_1 \in Y we find disjoint open neighbourhoods U y 1{y 1}U_{y_1} \subset \{y_1\} and U Y 2,y 1Y 2U_{Y_2,y_1} \supset Y_2. The union of the U y 1U_{y_1} is a cover of Y 1Y_1, and by compactness of Y 1Y_1 there is a finite subset SYS \subset Y such that

U Y 1sSY 1U y 1 U_{Y_1} \coloneqq \underset{s \in S \subset Y_1}{\cup} U_{y_1}

is an open neighbourhood of Y 1Y_1 and

U Y 2sSYU Y 2,s U_{Y_2} \coloneqq \underset{s \in S \subset Y}{\cap} U_{Y_2,s}

is an open neighbourhood of Y 2Y_2, and both are disjoint.


Last revised on April 28, 2017 at 07:36:08. See the history of this page for a list of all contributions to it.