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All spaces are assumed Hausdorff. We call a topological space compact if every open cover has a finite subcover. We call it Lindelöf if every open cover has a countable subcover, and hereditarily Lindelöf if moreover every subspace is Lindelöf.

It is obvious that every compact space is Lindelöf. We have that every closed subset of a compact space is compact, i.e. it is Lindelöf and 'so much more'. This leads to the natural question, is every open set Lindelöf, too? (this would suffice to make the space hereditarily Lindelöf; see for example here).

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The OP's own answer, the one-point compactification of an uncountable discrete space, is a classical one. Another is $\{0,1\}^X$, where $X$ is uncountable, where the set of all points with exactly one $1$ is uncountable and discrete (and so not Lindelöf too).

Of course, every closed subspace of a Lindelöf space is also Lindelöf, so by the same argument, every Lindelöf space should be hereditarily so...

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    $\begingroup$ The remark about ordered spaces baffles me. Surely $\omega_1 + 1$ is Lindelöf, but $\omega_1$ is not. $\endgroup$ – Niels J. Diepeveen Apr 6 '16 at 22:50
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    $\begingroup$ As Niels says, the statement about ordered spaces is simply false, $\omega_1+1$ and its subspace $\omega_1$ actually being one of the classic counterexamples. $\endgroup$ – Brian M. Scott Apr 7 '16 at 3:25
  • $\begingroup$ @BrianM.Scott you're right of course (edited). I was confused with $c=hL $ for ordered spaces. $\endgroup$ – Henno Brandsma Apr 7 '16 at 4:30
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    $\begingroup$ Since every Tychonoff space has a Hausdorff compactification, asking "is every compact space hereditarily Lindelöf" is the same as asking "is every Tychonoff space Lindelöf". Counterexamples abound. $\endgroup$ – bof Apr 7 '16 at 4:37
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Unfortunately, this is not true. Take for example, any uncountable set $X$ with the discrete topology (so that singletons are open). It is obviously not Lindelöf.

Now, 'create' the space $Y$ by adding a point $*$ to $X$ with neighbourhoods of the form $U=\{*\}\cup C$, where $C$ is a co-finite subset of $X$. To be clear, $Y=X\cup\{*\}$, with all points in $X$ being open, and the above-mentioned neighbourhoods of $*$.

Then $Y$ is obviously compact, and $X$ is obviously an open subset of $Y$, but $X$, being uncountable, is pretty far from being Lindelöf. To be specific, the open cover $\mathcal{U}=\{\{x\}:x\in X\}$ has no countable subcover.

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  • $\begingroup$ What is $F$, maybe $X\setminus C$? $\endgroup$ – sqtrat Apr 6 '16 at 14:44
  • $\begingroup$ Sorry, initially I had called the co-finite set '$F$', but then decided this can be confusing since '$F$' is usually used for finite, not co-finite, sets. When changing to '$C$', I omitted to change all occurrences. $\endgroup$ – Simon_Peterson Apr 6 '16 at 15:27

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