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We just learned about various properties of Lindelöf subsets and theorems concernining when $Y \subset X$ is $C^{*}$-embedded in $X$. This means that every bounded continuous real-valued function on $Y$ has a continuous extension over $X$.

Is it true that every Lindelöf subset of $\beta \omega$ is $C^{*}$-embedded?

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Yes, this is true. In fact even more is true:

A space $X$ is called weakly Lindelöf iff every open cover of $X$ has a countable subfamily whose union is dense in $X$. It's clear that every Lindelöf space is weakly Lindelöf and every separable space is too (just pick one member from the open cover for every member of a countable dense subset of $X$).

In this paper by Comfort, Hindman and Negrepontis it is shown in corollary 1.7 that every weakly Lindelöf subspace of an $F'$-space is $C^{\ast}$-embedded in its closure.

Now, $\beta\omega$ is a normal $F$-space, which makes it an $F'$-space (check the paper for the definitions, which I will not repeat here) and being normal implies that closed subsets are $C^{\ast}$-embedded in it, so: all weakly Lindelöf subspaces (in particular all Lindelöf and separable ones) are $C^{\ast}$-embedded in it.

This is quite remarkable: note that $\mathbb{Q} \setminus \{ 0 \}$ is separable metrizable, but is not $C^{\ast}$-embedded in $\beta\mathbb{Q}$, (while $\mathbb{Q}$ as a subspace is, and is homeomorphic to it) e.g., so $\beta\omega$ being an $F$-space is quite important. Also this shows, in this case, that an intrinsic property of a subspace implies something about how it lies into a larger space ($C^{\ast}$-embedding).

Remark: under CH (the continuum hypothesis), Woods showed that being weakly Lindelöf and being $C^{\ast}$-embedded are equivalent for subsets of $\beta\omega$, while later Dow and Woods showed that this equivalence for subsets of $\beta\omega$ was itself equivalent to CH, completing the circle.

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