How 'normal' are normal spaces?

Question

Is normality a property that is 'easily' satisfy by a given space? I mean by this: are non-normal spaces hard to construct/unnatural compared to normal spaces?

Answer 1

Normality is one of the separation axioms in topology. The higher you climb up the ladder of separation axioms, the more the topology behaves 'like' one expects, and thus, in a sense, the more tame the topology is (or least, the more we recognize the topological behaviour to agree with our more day-to-day expectations).

Normal spaces are in abundance, among other reasons since every metrizable space is normal, and metric spaces are in abundance. However, there are lots of normal spaces which are not metrizable. In that sense, normality is quite easy to assure.

At the same time, it is also easy to construct spaces that are not normal. Any non Hausdorff space, for instance, is not normal and there are plenty of such spaces (all finite topological spaces, other than the discrete ones, are non-Hausdorff and thus not normal). It is somewhat more difficult to construct a regular space that is not normal.

So, there are lots of examples of normal spaces and, at the same time, it is easy to construct topological spaces that are not normal. This would show that the dichotomy you anticipated regarding the concept of normality is not there.

Whether or not non-normal spaces are natural is a question of context. In a sense when the context is strongly geometric, for instance in Hilbert or Banach space theory, most (perhaps all...) spaces or interest are normal. In contrast, in digital geometry, where most spaces of interest are finite, normality is almost never met.

The issue of course is that topology is extremely broad. While topology was born in the realm of analysis it had spread to practically all other realms. In some sense, it is simply too easy to manufacture a topology, including very 'weird' ones from an analysis point-of-view. This is not at all bad of course, but we just have to keep in mind how rich topology is.

Answer 2

In a way, normal spaces are everywhere, since all metric spaces are normal and sticking a metric to something is very common. This means a large part of topology will deal with normal spaces, and yes, non-normal spaces are much less common and harder to find (or construct).

Answer 3

As Ittay Weiss implies in his answer, whether normal spaces are "normal" really depends on how you approach topology. In some areas, normality of the topological spaces under consideration may be almost a preliminary assumption (all second-countable manifolds are normal, for example), but for those that consider more general spaces, the non-normality of examples may be required.

Below, I'll take normal to mean "every pair of disjoint closed subsets can be separated by disjoint open neighbourhoods" (so, in particular, a normal space need not be Hausdorff). I will refer to normal Hausdorff (equivalently normal T₁) spaces as T₄.

There are some fairly modest assumptions that guarantee that a space is normal.

• Any finite T₁-space is discrete, and is therefore normal (and T₄).

• All Lindelöf regular spaces are normal (and so, in particular, all countable regular spaces, and all second-countable regular spaces are normal).

• All compact Hausdorff spaces are normal (hence T₄).

Normality (together with other topological properties) can also impose strong properties on space. For instance, we have Urysohn's Metrization Theorem:

• All second-countable T₄-spaces are metrizable.

At the same time, normality is not very robust topological property, and is not preserved by the simple topological constructions:

• A subspace of a normal (T₄) space need not be normal.

  Example. Suppose that $X$ and $Y$ are infinite sets, with $X$ uncountable, and pick $x_* \in X$, $y_* \in Y$. Define a topologies on $X$ and $Y$, respectively, by

  • $U \subseteq X$ is open iff $x_* \in U$ or $X \setminus U$ is finite.
  • $V \subseteq Y$ is open iff $y_* \in V$ or $Y \setminus V$ is finite.

  Then the product $X \times Y$ is T₄, however the subspace $( X \times Y ) \setminus \{ \langle x_* , y_* \rangle \}$ is not normal. (It is not too difficult to show that $\{ \langle x_* , y \rangle : y \in Y \setminus \{ y_* \} \}$ and $\{ \langle x , y_* \rangle : X \in X \setminus \{ x_* \} \}$ are disjoint closed subsets of this subspace which cannot be separated by disjoint open neighbourhoods.)

• The product of two T₄ spaces need not be normal.

  Example. The Sorgenfrey line $\mathbb{S}$ ($\mathbb{R}$ with the lower-limit topology) is a T₄ space, but the square $\mathbb{S} \times \mathbb{S}$ is not normal. (Using a Baire Category argument one can show that $\{ \langle q , -q \rangle : q \in \mathbb{Q} \}$ and $\{ \langle x , -x \rangle : x \in \mathbb{R} \setminus \mathbb{Q} \}$ are disjoint closed sets which cannot be separated by disjoint open neighbourhoods.)

• The product of a T₄-space with a compact metric space need not be normal.

  A T₄ space $X$ such that $X \times [0,1]$ is not normal is called a Dowker space. There are known examples of Dowker spaces, the first of which was constructed by M.E. Rudin, but they are not very easy to describe.

• The quotient of a T₄ space need not be normal.

  Example. The real line $\mathbb{R}$ is clearly T₄, however the quotient space $\mathbb{R} / \mathbb{Q}$ (identifying all rational numbers) is neither normal nor Hausdorff (as all nonempty open sets contain the equivalence class of $\mathbb{Q}$).

  However, the quotient of a T₄ space by a closed equivalence relation is T₄.