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Let $S\subseteq \mathbb R^3$ be homeomorphic to $\mathbb R$. Prove that $\mathbb R^3 \setminus S$ is connected.

I haven't been able to solve this, although my topology skills are pretty weak. My friend told me he managed to prove this using results from his "dimension topology" class. Although I think this should be solvable using more mainstream stuff like homology or something. But I'm not sure.

Regards.

ruadath
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Asinomás
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  • The space is homotopic to a cylinder, don't? – iam_agf Nov 23 '16 at 06:05
  • why? I don't think so. For example, in a torus, removing a loop will leave different spaces depending on the loop that is removed. – Asinomás Nov 23 '16 at 06:11
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    If $S$ is closed this is the same as asking why the complement of $S^1 \hookrightarrow S^3$ is connected, which you can see with Alexander duality. But if $S$ is not closed I'm not sure how to approach this. –  Nov 23 '16 at 06:55
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    The following is the only reasonable approach I can think of. 1) Show that, given any embedded copy of the unit interval in $\Bbb R^3$, the space of paths which do not intersect this arc is dense in the space of all paths (with fixed endpoints). I have no idea how to do this, but I think I believe it's true. A hearty literature search is encouraged. 2) Apply the Baire Category Theorem to the space of paths with fixed endpoints, where your $U_n$ are the space of paths that do not intersect $[n, n+1]$ in your embedding of $\Bbb R$. The intersection over all these sets is nonempty, as desired. –  Nov 23 '16 at 21:51

5 Answers5

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First: it is not reasonable to prove this with homology. Let's start with a story.

Suppose $X$ is a subset of the sphere homeomorphic to $D^k$. Then Hatcher proves in proposition 2B.1 that $S^n \setminus X$ has trivial reduced homology. (He also proves that if $X \cong S^k$, then $S^n \setminus X$ has the same reduced homology as $S^{n-k-1}$. This won't be as important to us.) The way the proof goes, roughly, is to use Mayer-Vietoris on the complement of $I^{k-1} \times [0,1/2]$ and $I^{k-1} \times [1/2, 1]$. If the homology isn't what we said above, we show the same is true for one of these sub-intervals, keep making the sub-intervals smaller, and then show that the homology of these complements ultimately reduces to the homology of $I^{k-1} \times \{p\}$ (all, again, a sketch - see the actual proof.)

It was essential that we used $I$ here, or else we wouldn't have access to the Mayer-Vietoris sequence - the complement of these sets is open! The same is not true for a copy of $\Bbb R$. If the $\Bbb R$ had closed image, you'd be able to access the problem with similar techniques (most easily by taking the one-point-compactification and thinking of this as the complement of an $S^1$ in $S^3$ and using the above technique). You might want to say "OK, well there they took a nested intersection of intervals and looked at their complement. We know the result works for embeddings of $[0,1]$. Why don't we try to take a nested union instead?" This does not work - just think about the union of discs on the $xy$-plane of increasing radius! Clearly their complements are path-connected but their increasing union is not. So there's something special going on about the 1-dimensionality of the line. (Note, by the way, that FINITE unions are easy to deal with via the Mayer-Vietoris sequence. Indeed it's not hard to show using the above facts that the complement of any $n-2$-dimensional finite simplicial complex embedded in $\Bbb R^n$ is path-connected.) Even the most souped-up duality theorems, like Alexander duality in Cech cohomology, need to assume compactness (even if they can drop local contractibility!) After all, consider the embedding of a copy of $\Bbb R$ inside the 2-sphere that winds around an annulus, with limit set $|z| = 1$ and $|z| = 3$. Then it has three path-components, which no form of duality could expect.

OK, so how do we deal with this? We, like your friend said, use dimension theory. I don't know this stuff very well, so I'm going to cite it away. (I don't have access to Hurewicz-Wallman right now, but when I do, I'll offer a more precise reference.) This MO comment says that the space of paths from $x$ to $y$ not in some compact 1-dimensional subspace of $\Bbb R^3$ is open and dense in the space of paths from $x$ to $y$, and this answer implies that the image of any unit interval in $\Bbb R^3$ is 1-dimensional. Now apply the Baire Category theorem to the subset $U_n \subset \mathcal P(x,y)$, where $U_n$ is the set of maps that miss $[n,n+1]$ in your embedding of $\Bbb R$. Because $\mathcal P(x,y)$ is a complete metric space with the supremum metric, it is a Baire space, and hence $\bigcap_{n \in \Bbb Z} U_n$ is nonempty (and, in fact, dense). Thus we conclude.

Community
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It is sufficient to show that $X=\Bbb R^3\setminus S$ is path-connected. Now for any couple of points $p,q$ in $X$ let $\overline{pq}$ be the segment joining them, then this is a path that. If $\overline{pq}\cap S=\emptyset$, there is no problem. In the case that $\overline{pq}\cap S\neq \emptyset$, take a third point $r$, different to the previous and such that $\overline{pr}\cap S=\emptyset$ and $\overline{rq}\cap S=\emptyset$, and then $\overline{pr}\cup\overline{rq}$ is a path joining $p$ with $q$.

InsideOut
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    I don't see why we can suppose $S$ is the $z$-axis. The potential for $S$ to be some kind of knot is what make this question difficult. Perhaps you can modify this to construct a path from $p$ to $q$ guaranteed to avoid the curve $S$. – 2'5 9'2 Nov 23 '16 at 07:58
  • I edit my post! thanks for the remark. – InsideOut Nov 23 '16 at 09:41
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    It is by no means clear that you can choose such point $r$. How does the fact that $S$ is homeomorphic to $\Bbb{R}$ enter your argument? For instance, how does your proof distinguish our case from the non-example $\Bbb{R}^3 \setminus \Bbb{R}^2$? – Sangchul Lee Nov 23 '16 at 09:49
  • If pq intersect $S$, let $l$ the straight line perpendicular to the segment passing through the intersection point. Maybe this line intersect $S$ many times, but you can choose the point $r$ judiciously such that pr don't intersect the line. I use the fact that $S$ is a line because if the segment interest it then I turn around $S$. – InsideOut Nov 23 '16 at 09:55
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    Consider this example: Using the space-filling curve, you can create a surjective continuous map $\omega : [1, 2] \to S^2$ such that $\omega(1) = (0,0,-1)$ and $\omega(2) = (0,0,1)$. Now define $\gamma : \Bbb{R} \to \Bbb{R}^3$ by $$ \gamma(t) = \begin{cases} t \omega(t), & 1 \leq t \leq 2 \ (0,0,t-2), & t < 1 \ (0,0, t), &t > 2 \end{cases}. $$ It is not hard to check that $\gamma(\Bbb{R})$ is homeomorphic to $\Bbb{R}$. But any half-line emanating from $p = (0,0,0)$ intersects $\gamma(\Bbb{R})$. For me, your claim is far from triviality for this example. – Sangchul Lee Nov 23 '16 at 10:12
  • Mmmh, interesting example, I have to think about this! – InsideOut Nov 23 '16 at 10:14
  • If your that curve is surjective to $S^2$ then it divide the space on two connected components, so the statement in general is false, right? – InsideOut Nov 23 '16 at 10:17
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    Notice that my $\gamma$ restricted on $[1, 2]$ is not the same as $\omega$, but rather $t$ is multiplied to adjust the distance from the origin. So it may not block every half-line emanating from a point other than the origin. The point is that you may need to produce a very complicated path to join two points in $\Bbb{R}^3\setminus S$. – Sangchul Lee Nov 23 '16 at 10:21
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Here's a nice hint: The zero-th homology $H_0(\mathbb{R}^3 - S)$ must be a direct sum of $n$ copies of $\mathbb{Z}$, where $n$ denotes the number of connected components of $\mathbb{R}^3 - S$.

Tom Gannon
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Assume that $\mathbb{R}^3\setminus S$ is disconnected. So there are nontrivial open, disjoint sets $U,V$ with

$$\mathbb{R}^3\setminus S=U\sqcup V$$ $$\mathbb{R}^3=U\sqcup V\sqcup S$$

Since $U,V$ are nontrivial, $\partial U,\partial V\neq\emptyset$.

Since $U,V$ are open, $U\cap\partial U=V\cap\partial V=\emptyset$.

We must have that $\partial U\cap V=\emptyset$ or else there is $x\in \partial U\cap V$. This would mean an open ball $W$ exists surrounding $x$ that is contained within $V$. And there would be points in $W\setminus\{x\}$ that are also in $U$, which would contradict $U$ and $V$ being disjoint.

This implies that $\partial U\subset S$. And similarly, $\partial V\subset S$.

Suppose for a moment that $S\setminus(\partial U \cup\partial V)$ is nonempty. Let $y\in S\setminus(\partial U \cup\partial V)$. Well, there is a small enough ball $Y$ surrounding $y$ that manages to neither intersect $U$ nor $V$, since $y$ is not in their boundaries. Therefore $Y$, a ball, is contained in $S$, homeomorphic to $\mathbb{R}$. This is a contradiction.

So $S\setminus(\partial U \cup\partial V)$ must be empty, and given the containment established earlier, therefore $S=\partial U \cup\partial V$. Now use the topology on $S$ induced by its homeomorphism to $\mathbb{R}$. $\partial U$ and $\partial V$ are still closed (since they are nonempty), but they union to $S$ which is open in this topology. Therefore they must have a nontrivial intersection.

Consider $z\in\partial U\cap\partial V$ and an open ball $Z$ containing $z$, with $Z$ so small that $\partial Z\cap S$ consists of two points $\{a,b\}$ and within $Z$, $S$ is homotopy equivalent to the union of the line segments $\overline{az}$ and $\overline{zb}$ while fixing $a$, $z$, and $b$. This is possible because $S$ is homeomorphic to $\mathbb{R}$.

Now, $Z\setminus S$ is connected owing to the simple form of $Z\cap S$. ($Z\setminus S$ is homotopy equivalent to a cylinder.) And intersecting the sides of the first equation above with $Z$, we have the disjoint union

$$Z\setminus S=(Z\cap U)\sqcup (Z\cap V)$$

However $z$ is in both $\partial U$ and $\partial V$, so neither set on the right is empty. This contradicts $Z\setminus S$ being connected.

So it all falls apart. The original assumption must have been false, and so $\mathbb{R}^3\setminus S$ is connected.

2'5 9'2
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    How do you know $U$ and $V$ are open in $\mathbb{R}^3$? A priori they are only open in $\mathbb{R}^3\setminus S$. – Eric Wofsey Nov 23 '16 at 08:55
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    Also, there need not exist a $Z$ with the properties you state, since $S$ might be "wild" at $z$. For instance, there could be infinitely many knotted segments of $S$ that accumulate at $z$, as in the example shown at https://en.wikipedia.org/wiki/Wild_knot. – Eric Wofsey Nov 23 '16 at 09:05
  • @Eric Wofsey regarding your first comment. Isn't it true that an open set minus a closed set is open? Thus $\mathbb R^3\setminus S$ is open. The rest follows since an open subset of an open subset is open in the original space. – R_D Nov 23 '16 at 12:35
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    @R_D It was never assumed that $S$ was closed. –  Nov 23 '16 at 13:34
  • $\mathbb R $ is a closed subset of $\mathbb R^3$ and S being homeomorphic to $\mathbb R $ makes S closed. Am I wrong? – R_D Nov 23 '16 at 14:25
  • @R_D Just passing by. Please excuse my ignorance (I've long forgotten most of the topology stuffs I learnt in the past): if the homeomorphism is some $f:\mathbb R^3\to \mathbb R^3$, of course it's an open map, but here we have only $f:S\to\mathbb R$. Why must $S$ be closed in $\mathbb R^3$? – user1551 Nov 23 '16 at 15:48
  • Well as I understood it, $f$ being a homeomorphism is saying that it is a topological embedding of $\mathbb R $ into $\mathbb R^3$ with $S$ as the image. Doesn't that mean $\mathbb R $ is embedded as a closed subset of $\mathbb R^3$. That is, S is a closed subset of $\mathbb R^3$. Is this reasoning wrong? – R_D Nov 23 '16 at 15:55
  • @R_D Surely you can embed $\mathbb R$ into $\mathbb R^3$, but still the homeomorphism is between $S$ and $\mathbb R\times0^2$ only. What I mean is, if $A,B$ are topological subspaces of a topological space $S$, and $f$ is merely a homeomorphism between $A$ and $B$, it can happen that $A$ is closed in $S$ but $B$ is not. You may see an example on this site. – user1551 Nov 23 '16 at 16:31
  • @EricWofsey When OP says $S$ is homeomorphic to $\mathbb{R}$, I take that to mean $S$ with its subset topology induced from $\mathbb{R}^3$. This precludes the wild knot and other such things, and implies that $U,V$ are open in $\mathbb{R}^3$. I'll clarify in the answer post in a bit. – 2'5 9'2 Nov 23 '16 at 16:32
  • @EricWofsey I give on the "$U,V$ open in $\mathbb{R}^3$" point. For instance the image might be ${(0,0,t)\mid 0<t<1}$. Note to myself: If I come back to amend this, my strategy will be to use the subset topology that $S$ inherits to justify replacing $S$ with the closure of $S$ (which will imply $U,V$ open), address the point about $Z$, and still show that $\mathbb{R^3}\setminus \bar{S}$ is connected, which will imply that $\mathbb{R^3}\setminus {S}$ is connected as long as $\bar{S}\setminus S$ behaves. – 2'5 9'2 Nov 23 '16 at 17:18
  • @alex.jordan: No, a wild arc is still homeomorphic to $\mathbb{R}$. Its subspace topology is perfectly tame; all that is wild is the way it's embedded in $\mathbb{R}^3$. Also, note that replacing $S$ with its closure is problematic, since its closure could easily contain an entire sphere (consider a path inside the unit ball that is slowly approaching the unit sphere and becomes "space-filling" in the spherical direction as it approaches, similar to the example at http://math.stackexchange.com/questions/2026842/mathbb-r3-minus-a-line-is-connected/2026985?noredirect=1#comment4160282_2026921) – Eric Wofsey Nov 23 '16 at 19:26
  • @EricWofsey This is homeomorphic to $\mathbb{R}$ if you omit the accumulation point. And if you omit that point, then the curve itself contains no "wild points" as suggested by your objection in the second comment. On the other hand if you don't omit the accumulation point, the curve is not homeomorphic to $\mathbb{R}$. (All your other objections I agree with now, just still not on the same page with this one.) – 2'5 9'2 Nov 24 '16 at 02:34
  • No, the whole point of a wild arc is that it is actually homeomorphic to $\mathbb{R}$ (wild points included) even though the local behavior of its embedding is bad. Think about what the topology at the accumulation point looks like: any interval around the accumulation point can still be realized as the intersection with some open subset of $\mathbb{R}^3$. – Eric Wofsey Nov 24 '16 at 02:38
  • When I look at the curve here, I imagine a homeomorphism with $\mathbb{R}$ where moving toward the accumulation point from the right corresponds to moving toward $\infty$, and moving toward the accumulation point from the left corresponds to moving toward $-\infty$. If the curve (with wild point included) is homeomorphic to $\mathbb{R}$, then what is the image of the accumulation point? Seems like if you are including the wild point then you have something homeomorphic to the one-point compactification of $\mathbb{R}$, not to $\mathbb{R}$. – 2'5 9'2 Nov 24 '16 at 02:49
  • @EricWofsey Forgot to tag you. – 2'5 9'2 Nov 24 '16 at 02:50
  • Retracted. If it's indeed true that the the wild point has topology surrounding it that is still homeomorphic to a segment of $\mathbb{R}$, then I could just puncture the curve at some non-wild point. I should read about the topology around the wild point. – 2'5 9'2 Nov 24 '16 at 02:53
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    Right, that's a picture of a wild knot, which is homeomorphic to $S^1$, not $\mathbb{R}$. But if you remove any single point (which doesn't have to be the wild point), you get something homeomorphic to $\mathbb{R}$). – Eric Wofsey Nov 24 '16 at 02:55
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A short answer using transversality and its properties:

Any path $\gamma$ joining two points $p$ and $q$ in $\mathbb{R}^3\setminus S$ is homotopic to a path $\gamma'$ (with fixed endings) which is tranverse to $S$. But $\gamma'\pitchfork S$ (thanks to codimension $2$ of the two submanifolds) means $\gamma'\cap S=\varnothing$.

amine
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