# Lindelöf'ize a space?

Although much weaker, Lindelöf is in the same spirit of compactness. It occurs to me whether there is such a thing like "Lindelöf'ization", since there are various kinds of compactification process for a space, that is, to embed a non-Lindelöf space as a subspace of a Lindelöf space.

This question might be silly and I admit that I do not have really solid background in topology. Actually, I do not even know a non-Lindelöf space other than ones in Munkres's book.

Anyway, if someone has looked into this, I would be greatly appreciated!

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What have you tried? Have you tried starting with the Alexandroff one-point compactification and replacing compact with Lindelof to see what happens? – Willie Wong Mar 14 '12 at 15:36

In general there is little interest in ‘Lindelöfizations’ per se: they don’t as a class have nice enough properties. One very specific exception (or better, class of exceptions) is the one-point Lindelöfization of an uncountable discrete space. Suppose that $D$ is an uncountable discrete space, and let $p$ be a point not in $D$. Let $X=D\cup\{p\}$, and say that $V\subseteq X$ is open iff either $p\notin V$, or $p\in V$ and $X\setminus V$ is countable. Since the countable subsets of $D$ are precisely its Lindelöf subsets, the nbhds of $p$ are precisely the complements in $X$ of the Lindelöf subsets of $D$, and the construction is analogous to that of the Alexandroff one-point compactification of a locally compact space. These spaces have proved somewhat useful, especially in constructing (counter)examples.
This construction can of course be generalized. If $Y$ is a $T_3$-space in which every point has a nbhd with Lindelöf closure, we can form a Lindelöf Hausdorff space $X$ containing $Y$ as a dense open subspace in the following way. Let $p$ be a point not in $Y$, let $X=Y\cup\{p\}$, make $Y$ an open subspace of $X$ with its original topology, and let the open nbhds of $p$ be the sets of the form $X\setminus F$ for closed, Lindelöf subsets $F$ of $Y$. It’s not hard to check that $X$ has the desired properties. I can’t offhand recall seeing it used, however.