Completely normality is equivalent to the requirement that every subspace is normal. If we think of completely normality as a restriction of normality to obtain the normality of subspaces, then there should be a stricter condition $P$ than complete normality satisfying the following property:

$P(X)$ if and only if all subspaces in $X$ are completely normal.

Now if we keep this process going, should we ultimately find a property $P_n$ such that every subspaces in a space $X$ satisfying $P_n$ also satisfy $P_n$?

  • $\begingroup$ I doubt it. The reason a subspace of a completely normal space can fail to be normal is that there are sub spaces of the sub space that are separated in the subspace topology but not in the ambient space. My guess is that if every subspace of a space is completely normal it’s topology is very trivial. $\endgroup$ – Charlie Frohman Dec 13 '18 at 2:37
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    $\begingroup$ No, as I explain below, complete normality is already hereditary almost by definition. $\endgroup$ – Henno Brandsma Dec 13 '18 at 4:45
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    $\begingroup$ A subspace of a completely normal space is completely normal! Your premise is wrong. $\endgroup$ – Henno Brandsma Dec 13 '18 at 4:46

A subspace $Y$ of a completely normal space $X$ is again completely normal: if $Z$ is a subspace of $Y$, then $Z$ is also a subspace of $X$ and so $Z$ is normal.

Another name for this property is "hereditarily normal". In general we can define, when we have a topological property $P$, the (probably) stronger property "hereditarily P": all subspaces of $X$ have property $P$.

For $P$ = "normal" this indeed is a stronger property than $P$. Hereditarily Lindelöf spaces are also a standard property, as are hereditarily separable spaces. In all these cases there are spaces with property $P$ that are not hereditarily $P$.

The process (if it can be called that) stops here though: a space that is hereditarily $P$ is also "hereditarily hereditarily $P$", using the same remark that a subspace of a subspace is still a subspace of the original space (transitivity of subspace topologies).

  • $\begingroup$ Nice answer! I’ll slightly edit the description of my question to prevent it from misleading. $\endgroup$ – William Sun Dec 13 '18 at 4:54

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