What is the fewest number of points you must delete from $\mathbb{R}^3$ to make it not simply connected?

Question

This question concerns a set-theoretic aspect that I found interesting in the recent question asked by user Nick R., namely, Is $\mathbb{R}^3\setminus\mathbb{Q}^3$ simply connected? He had asked whether $\mathbb{R}^3$ remains simply connected after deleting a countable set of points, such as the collection of rational points $\mathbb{Q}^3$.

That question was answered affirmatively by Martin M. W. My question is, can we do better? Specifically, I want to understand, in a general context where the continuum may be very large, exactly how many points we may freely delete from $\mathbb{R}^3$, whilst remaining simply connected. What is the fewest number of points that we must delete from $\mathbb{R}^3$ in order to make it no longer simply connected?

Let us define the simply connected deletion number, $\delta$, to be the smallest cardinality of a subset $A\subset\mathbb{R}^3$, such that the complement $\mathbb{R}^3\setminus A$ is no longer simply connected.

Martin's answer to the earlier question shows that deleting any countable number of points preserves the simply connected property, and so the simply connected deletion number is definitely uncountable, at least $\omega_1$. And since it is clearly at most the continuum, the question is settled if the continuum hypothesis holds. Like all other cardinal characteristics of the continuum, this number is more interesting when the continuum hypothesis fails.

In a comment, I had suggested that Martin's argument suggested that the simply connected deletion number should be at least as large as $\mathbf{cov}(\mathcal{M})$, the covering number of the meager ideal, which is the fewest number of meager sets whose union is the whole space. My reason for suggesting this was that as far as I understand Martin's answer (which I admit is imperfectly), he is proposing that for any one point $x$, there is a comeager set of homotopies that avoids $x$. So in order to avoid all the points in a set $P$, we need to know that the intersection of $|P|$ many comeager sets in his space of homotopies is nonempty. This is the same as knowing that the unions of $|P|$ many meager sets (the complements) is not the whole space of homotopies, in order that there is at least one desired homotopy that avoids every point in $P$.

If this is right, then we would deduce that the simply connected deletion number is at least $\mathbf{cov}(\mathcal{M})$, provided that the covering number for meager sets in his space was the same as for our other more familiar spaces. (If someone could explain and confirm this inequality in greater detail, please post an answer! I would want to see more details than Martin had provided about the space of homotopies.)

Question. What is the simply connected deletion number exactly? Is it consistent that this number is strictly less than the continuum? Is it necessarily the continuum? How does it relate to the other standard cardinal characteristics of the continuum? What is the value under Martin's axiom? Is it equal to $\mathbf{cov}(\mathcal{M}$)? Can it be strictly larger than $\mathbf{cov}(\mathcal{M})$?

Answer 1

Clearly wlog the base point is $0$.

Take a loop $a: [0,1] \to \mathbb R^3$ with $a(0)=a(1)=0$.

Fix a single homotopy $b: [0,1] \times [0,1] \to \mathbb R^3$ with $b(t,0)=a(t)$, $b(t,1)=b(0,u)=b(1,u)=0$. Also choose it to be real analytic on $(0,1) \times (0,1)$. I believe it's no problem doing this by choosing it to be harmonic or something.

Now let's consider a homotopy $b'(t,u)= b(t,u) + c u(1-u)$ for suitably chosen $c \in \mathbb R^3$. We want to choose a $c$ that avoids a set of points $p$. That means $c$ must fail to equal:

$$ \frac{ p - b(t,u)}{u(1-u)} $$

This is a real analytic parameterized surface in $\mathbb R^3$. So the new question is - how many real analytic parameterized surfaces in $\mathbb R^3$ suffice to cover $\mathbb R^3$?

I think the answer is continuum. Their intersection with a line is countable unless they contain that line, and a real analytic surface contains at most countably many lines on a plane unless it is that plane - so if you have fewer than continuum real analytic surfaces, choose one plane that is none of your surfaces, then choose a line in it that is contained in none of your surfaces, then win.

Answer 2

The simply connected deletion number equals the continuum. This follows from the fact that for any dense subset $A$ in the real line the subset $A_3=(A\times \mathbb R\times \mathbb R)\cup (\mathbb R\times A\times\mathbb R)\cup (\mathbb R\times\mathbb R\times A)$ is 2-dense in the following sense: for any continuous ap $f:[0,1]^2\to\mathbb R^3$ and any $\varepsilon>0$ there is a continuous map $g:[0,1]^2\to \mathbb R^3$ such that $g$ coincides with $f$ on the boundary of $f$ and $g((0,1)^2)\subset A_3$, and $g$ is $\varepsilon$-near to $f$. The proof of this fact is a bit more complicated than I wrote in the preceding version of this answer, so we can proceed differently.

Given a subset $S\subset\mathbb R^3$ of cardinality $|S|<\mathfrak c$, construct by induction a dense countable set $C$ in $\mathbb R^3$ such that any 3-element subset $B$ of $С$ is affinely independent and its affine hull $aff(B)$ does not intersect $S$. (To construct such a set $C$ fix any countable base $(U_n)$ of the topology and in $n$th set $U_n$ choose a point $c_n$ such that any 3-element subset $B\subset\{c_0,\dots,c_n\}$ is affinely independent and its affine hull misses $S$).

Then the union $\Delta=\bigcup\{aff(B):B\subset C,\;|B|=3\}$ is 2-dense in $\mathbb R^3$. To prove the 2-density of $\Delta$, fix any function $f:[0,1]^2\to\mathbb R^3$. We need to construct a map $g:[0,1]^2\to\mathbb R^2$ such that $g$ coincides with $f$ on the boundary of $[0,1]^2$, $g((0,1)^2)\subset\Delta$ and $g$ is near to $f$. On $(0,1)^2$ the map $g$ can be defined by the formula $g(x)=\sum_{U\in\mathcal U}\lambda_U(x)c_U$ where

• $\mathcal U$ is a cover of $(0,1)^2$ by open sets whose diameters tend to zero as $U$ tends to the boundary of the square,

• $\mathcal U$ has order $\le 3$ in the sense that each point of $(0,1)^2$ is contained in at most 3 sets of the cover $\mathcal U$ (such choice of $\mathcal U$ is possible as $dim(0,1)^2=2$);

• $(\lambda_U)_{U\in\mathcal U}$ is a partition of the unity subordinated to the cover $\mathcal U$, and

• for every $U\in\mathcal U$ the point $c_U$ belongs to $C$ and is sufficiently near to a point in the set $f(U)$.