A topological space is normal when every two disjoint closed subsets have disjoint open neighbourhoods.
A space is hereditarily normal when every subspace is a normal topological space. This is stronger than normality: arbitrary subspaces need not be closed. Every perfectly normal space is hereditarily normal.
A T1 space is perfectly normal when it is normal and every closed subset is a G-delta set. Equivalently every closed subset is the zero set of a continuous function into , using the closed G-delta set as a zero set theorem. If disjoint closed sets are the exact zero sets of , then has those sets as its exact zero and one levels. Every subspace is normal, although the converse fails.
For disjoint closed sets in a subspace , their ambient closures satisfy . In a perfectly normal space choose continuous functions with exact zero sets . The function is continuous on , which contains , and separates . Disjoint inverse images of endpoint neighborhoods give normality of .
If is closed, is open, and in a normal topological space, separate from the closed complement by disjoint open sets . Then . In particular, first separate two disjoint closed sets by disjoint open sets and then shrink each neighborhood this way: their closures are disjoint. No additional metric or compactness assumption is needed.
Let open sets be indexed by a dense subset of , with for . Taking the infimum of membership indices, with value one outside their union, gives a continuous function. The strict sublevel sets are unions of ; strict superlevel sets are unions of complements of . These identities prove continuity.
Every continuous real-valued function on a closed subspace of a normal space extends continuously to the whole space. If its values lie in a closed interval, the extension can be kept in that interval.
Suppose is closed in a normal space and on for a continuous real-valued on . The closed set misses . Urysohn's lemma supplies equal to zero on and one on . Then agrees with on and has range in . This permits the inverse transform in the proof of unbounded Tietze extension.
A continuous function on a compact subset of a Tychonoff space extends to the whole space. Embed the space into a cube of closed unit intervals. The Tychonoff theorem makes the cube compact Hausdorff and thus normal. The compact subset is closed in that cube, so the Tietze extension theorem extends its bounded function to the cube. Restrict this extension to the embedded original space. Normality of the original space is unnecessary.
Let be closed in a normal topological space and continuous. Separate its closed level sets and by Urysohn's lemma to obtain taking the corresponding endpoint values. Checking the outer and middle ranges gives the displayed bounds. Iteration on residuals gives a uniformly convergent series with total supremum norm at most . The uniform limit theorem gives the bounded Tietze extension theorem.
For a real function in the bounded continuous functions of norm on a closed subset of a metric space, separate its closed level sets and by . The distance to a set functions make continuous with the displayed bounds. Empty level sets are handled by the constants , or . Iterating on the residual yields a uniformly convergent continuous extension with no increase in the supremum norm.
A closed subset of a normal space is a countable intersection of open sets exactly when it is the zero set of some continuous function .
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In topology, a normal space is a specific type of topological space that satisfies certain separation properties. A topological space \( X \) is called **normal** if it meets the following criteria: 1. **It is a T1 space**: This means that for any two distinct points in the space, there exist open sets that contain one point but not the other. In other words, points can be separated by neighborhoods.