# Possible cardinalities of the remainders of compactifications of $\Bbb R$

With the usual topology on $$\Bbb R$$, a compactification $$\mathrm{id}_{\Bbb R}:\Bbb R\to v\Bbb R$$ can have a remainder $$v\Bbb R \setminus \Bbb R$$ of cardinality $$1,2, 2^{\aleph_0}=\mathfrak c,$$ or $$2^{\mathfrak c}.$$ The only possibilities less than $$\mathfrak c$$ are $$1,2.$$

Suppose $$\mathfrak c^+<2^{\mathfrak c}.$$ What possible cardinals between $$\mathfrak c$$ and $$2^{\mathfrak c}$$ can be the cardinals of such remainders?

Is there perhaps a Forcing argument that can answer or partly answer this?

Every connected compact Hausdorff space of weight $$\aleph_1$$ is the remainder $$v \mathbb R \setminus \mathbb R$$ of some compactification of $$\mathbb R$$. In particular, $$[0,1]^{\aleph_1}$$ is the remainder of a compactification of $$\mathbb R$$, and therefore $$\mathbb R$$ has a compactification with remainder of cardinality $$2^{\aleph_1}$$.

Using forcing, one can show that it is consistent to have $$\mathfrak{c} < 2^{\aleph_1} < 2^{\mathfrak{c}}$$. (For example, Easton's Theorem immediately implies that we may get a model where $$2^{\aleph_0} = \aleph_2$$, $$2^{\aleph_1} = \aleph_3$$, and $$2^{\aleph_2} = \aleph_4$$, although Easton's Theorem is a bit overkill for this.) Thus it is consistent that $$\mathbb R$$ has a compactification with cardinality in $$[\mathfrak{c}^+,2^{\mathfrak{c}})$$.

The result about weight-$$\aleph_1$$ continua is proved by Dow and Hart in this paper. But the special case of $$[0,1]^{\aleph_1}$$ is actually much easier to prove, using the fact that $$[0,1]^{\aleph_1}$$ is separable. Let $$\{d_1,d_2,d_3,\dots\}$$ be a countable dense subset of $$[0,1]^{\aleph_1}$$. Map $$\mathbb R$$ into $$[0,1] \times [0,1]^{\aleph_1}$$ as follows. First map $$\mathbb R$$ onto the ray $$[1,\infty)$$, and then map $$[1,\infty)$$ into $$[0,1] \times [0,1]^{\aleph_1}$$ by linearly mapping each interval $$[n,n+1]$$ to the line segment connecting $$(\frac{1}{n},d_n)$$ to $$(\frac{1}{n+1},d_{n+1})$$ in $$[0,1] \times [0,1]^{\aleph_1}$$. This mapping embeds the ray $$[1,\infty)$$ in $$[0,1] \times [0,1]^{\aleph_1}$$, and its boundary in this embedding is precisely the set $$\{0\} \times [0,1]^{\aleph_1} \approx [0,1]^{\aleph_1}$$.

Edit: It is also possible to find a compactification $$v \mathbb R$$ of $$\mathbb R$$ such that $$|v \mathbb R \setminus \mathbb R|$$ has countable cofinality. In fact, I claim that the set $$T = \{|v\mathbb R \setminus \mathbb R| \,:\, v\mathbb R \text{ is a compactification of } \mathbb R \}$$ includes all cardinals of the form $$2^\kappa$$, where $$\aleph_0 \leq \kappa \leq \mathfrak{c}$$, and all countable limits of such cardinals. So, for example, in a model of set theory where $$2^{\aleph_n} = \aleph_{\omega+n+1}$$ for all $$n$$ (which is consistent, by Easton's Theorem), there is a compactification $$v \mathbb R$$ of $$\mathbb R$$ such that $$|v \mathbb R \setminus \mathbb R| = \aleph_{\omega+\omega}$$.

Lemma: Suppose $$X$$ is a connected compact Hausdorff space, and $$X$$ has a dense subspace $$D$$ that is both separable and path connected. Then there is a compactification of $$\mathbb R$$ whose remainder is (homeomorphic to) $$X$$.

Proof: The main ideas are already present in the third paragraph above. Let $$\{d_1,d_2,d_3,\dots\}$$ be a countable dense subset of $$D$$. Map $$\mathbb R$$ into $$[0,1] \times X$$ as follows. First map $$\mathbb R$$ onto the ray $$[1,\infty)$$, and then map $$[1,\infty)$$ into $$[0,1] \times X$$ by linearly mapping each interval $$[n,n+1]$$ to some path connecting $$(\frac{1}{n},d_n)$$ to $$(\frac{1}{n+1},d_{n+1})$$ in $$[0,1] \times D$$. This mapping embeds the ray $$[1,\infty)$$ in $$[0,1] \times X$$, and its boundary in this embedding is precisely the set $$\{0\} \times X \approx X$$.

My claim above follows almost immediately from this lemma. Each of the spaces $$[0,1]^\kappa$$, where $$\aleph_0 \leq \kappa \leq \mathfrak{c}$$, is separable and path connected, and so $$|[0,1]^\kappa| = 2^\kappa \in T$$ by the lemma.

To get countable limits of such cardinals, fix some infinite cardinals $$\kappa_1,\kappa_2,\kappa_3,\dots \leq \mathfrak{c}$$. Let $$Y$$ be the space obtained by gluing the endpoints of an interval to some (any) point of $$[0,1]^{\kappa_1}$$ and some (any) point of $$[0,1]^{\kappa_2}$$, gluing the endpoints of another interval to $$[0,1]^{\kappa_2}$$ and $$[0,1]^{\kappa_3}$$, gluing the endpoints of another interval to $$[0,1]^{\kappa_3}$$ and $$[0,1]^{\kappa_4}$$, and so on. (In other words, $$Y$$ is obtained by stringing together the $$[0,1]^{\kappa_n}$$ like beads on a necklace.) Finally, let $$X$$ be the one point compactification of $$Y$$. Then $$X$$ satisfies the hypotheses of the lemma, and $$|X| = \sup_n 2^{\kappa_n}$$.

Interestingly, this method seems hopeless for getting a compactification of $$\mathbb R$$ with a remainder of size $$\aleph_\omega$$. I wonder if this is possible?

• I originally asked this on MSE as math.stackexchange.com/questions/3674921/… . Asaf Karagila conjectured that $cf (|v\Bbb R$ \ $\Bbb R |)\ne \omega.$ Jul 9, 2020 at 7:51
• @DanielWainfleet: Hmm, that's an interesting idea. I'll have to think about it. Jul 9, 2020 at 12:28
• @DanielWainfleet: OK, I think we can get cofinality $\omega$. The short version: By the argument in my last paragraph, we can get $[0,1]^\kappa$ for any $\kappa \leq \mathfrak{c}$ as a remainder, and by modifying the argument a bit, we can get a connected space that contains any countably many such spaces. I'll post some details this afternoon when I have time. Jul 9, 2020 at 12:40
• Neat! That is pretty cool. Jul 9, 2020 at 20:25