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

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 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 cov$(\cal{M})$, 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 cov$(\cal{M})$, 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 cov($\cal{M}$)? Can it be strictly larger than cov$(\cal{M})$?

• Is the analogous question about connectedness in dimension 2 (as opposed to simple-connectedness in dim 3) easy to answer? – Yaakov Baruch Aug 31 '15 at 21:45
• @YaakovBaruch The answer to that question is known, and it is the continuum. If you delete fewer than continuum many points in the plane, the resulting space remains path connected, because for any two points, there is a foliation of continuum many disjoint paths between them, and most of these paths avoid the deleted points. Does this help with the simply connected case in three dimensions? – Joel David Hamkins Aug 31 '15 at 21:49
• Let $A\subset \mathbb{R}^3$ with less than continuum points, and let $x, y$ be two points in $\mathbb{R}^3\setminus A$. Can we find a pair of parallel lines, both disjoint from $A$, passing resp. for x and for y? If so, we can work on the 2 dimensional orthogonal projection and apply the 2 dimensional result w.r.to the 2-dim projection of A. – Pietro Majer Aug 31 '15 at 22:09
• Certainly we can find lines like that, since for any line through $x$ there is a parallel line through $y$, and fewer than continuum many of the line pairs are bad. We can even find such lines in any desired parallel planes containing $x$ and $y$. – Joel David Hamkins Aug 31 '15 at 22:14
• @PietroMajer For path connectedness, you can use the same foliation idea in $\mathbb{R}^3$ that I mentioned above: just pick continuum many disjoint paths from $x$ to $y$ (disjoint except for $x$ and $y$), and most of them survive the deletion. – Joel David Hamkins Sep 1 '15 at 6:47

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.

• Great! Can we view this argument as implementing the (vague) suggestion made in my second comment on the question? – Joel David Hamkins Aug 31 '15 at 22:31
• @JoelDavidHamkins Sort of. I don't think you can get nonintersecting surfaces. Take a space-filling curve, and try to contract it in two ways that don't intersect. But you could ask for a weaker condition - like that each point is only contained in countably many surfaces. I think you can achieve that by a variant of my construction. Take an analytic disc with boundary your curve. If it contains uncountably many lines, they form a single analytic family, so you can always choose a direction such that the disc contains no lines in that direction. Project onto the perpendicular direction. – Will Sawin Sep 1 '15 at 1:55
• @JoelDavidHamkins The fibers of that map are certainly countable. Now add $u(1-u)$ times that direction times a variable. Each removed point can only remove countably many values of $u$, and you win. – Will Sawin Sep 1 '15 at 1:56
• By the way, the smallest number of tame curves that cover $\mathbb R^n$ equals $cov(\mathcal M)$, so can be consistently smaller than the continuum. A curve $C$ in $\mathbb R^n$ is tame if there exists a homeomorphism $h$ of $\mathcal R^n$ such that $h(C)$ is contained in a line. It would be interesting to know the answer for $C^k$-curves with $k>0$. This coverning number is still $cov(\mathcal M)$ or already continuum? – Taras Banakh Sep 1 '15 at 16:35

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 $$f(U)$$.

• I guess you mean the 3-element subset $B$ of $C$, not of $S$. – Alex Ravsky Jul 3 at 1:13
• @AlexRavsky Yes, you are right. $B$ should be a subset of $C$. – Taras Banakh Jul 8 at 16:52