Homotopy type of the plane minus a sequence with no limit points

Question

It is well known that the plane minus $n$ points is homotopy equivalent to a wedge of circles and hence its fundamental group is free on $n$ letters.

Question: Is the plane minus an infinite sequence of points having no limit point homotopy equivalent to an infinite wedge circles?

I'm pretty sure that this could fail if the sequence had a limit point, since then the space in question might be something more like the Hawaiian earing. But is the sequence having a limit point the only thing that could break down?

Answer 1

Yes. More generally, if $X$ is a proper closed subset of $\mathbb{R}^2$, then every path component $M$ of $\mathbb{R}^2 \setminus X$ is homotopy equivalent to a wedge of circles (observe that $X$ might be something like a Cantor set, which makes this a little more surprising). First, $M$ is a noncompact $2$-manifold, so its universal cover is homeomorphic to a disc (eg this follows from the uniformization theorem and the fact that every surface can be made into a Riemann surface). Second, the fundamental group of $M$ is free (see the answers to this question). It follows that $M$ is an Eilenberg-MacLane space for a free group. Free groups also have wedges of circles for their Eilenberg-MacLane spaces, so by the uniqueness of Eilenberg-MacLane spaces $M$ is homotopy equivalent to a wedge of circles.

Answer 2

Here is a more naive solution, as least if the sequence is countable. Let $\Bbb N \subset \Bbb R^2$ be the embedding defined by the sequence. Then there is an isotopy from this embedding to the standard inclusion into the $x$-axis (inductively move the points one by one through embeddings to each integer point on the $x$-axis. The homotopy type of the complement does not change through an isotopy. Furthermore, the homotopy type of the complement of the set of integer points on the $x$-axis clearly has the homotopy type of a wedge of circles (this can be seen, e.g., as follows: (1). the complement of the integer points in $\Bbb R$ has the homotopy type of an infinite wedge of zero-spheres, (2). passing from $\Bbb R^1$ to $\Bbb R^2$ has the effect of suspending the complement).

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Addendum: maybe the following is a better way to see the answer. Let $X = \lbrace x_n \rbrace$ be the sequence.

We can find a sequence of spaces $D_1 \subset D_2 \subset \cdots $ exhausting $\Bbb R^2$ such that $D_k$ is homeomorphic to a closed disk, $D_k$ is embedded in the interior of $D_{k+1}$ and $X$ meets each $D_k$ in its interior. Let $X_k = X \cap D_k$, and let $C_k$ be its complement in $D_k$. Then $C_k$ is a finite wedge of circles up to homotopy, $C_k \subset C_{k+1}$ and and $C:= \cup_k C_k$ is the complement of $X$ in $\Bbb R^2$.

Furthermore, the inclusion $C_k \subset C_{k+1}$ is a cofibration and admits a retraction, so we can write $$C_{k+1} \simeq C_k \vee E_k$$ and $E_k$ is a finite wedge of circles.

Then $C$, which is a colimit of the $C_k$, coincides with the homotopy colimit of the $C_k$, and with respect to the displayed identification we see that the homotopy colimit is a countable wedge of circles.

Answer 3

Expanding on John's answer a bit: If $X = \{x_1,x_2,\ldots\} \subset R^2$ is a countable discrete set, then there is a diffeomorphism $f:R^2 \to R^2$ with $f(x_i)=i$. Proof: by Sards theorem, we can assume that $X \cap N =0$ (translate a bit if necessary).

Claim 1: we can find paths $p_n: [0,1] \to R^2$ which are embeddings, $p_n (0)=x_n$, $p_n (1)= n$ and such that the images of all the $p_n$ are all disjoint. Proof by induction on $n$. Suppose that $p_1,\ldots ,p_{n-1}$ have been constructed. By homological duality theory, the complememt $R^2 \setminus \cup_{i=1}^{n-1} p_i ([0,1])$ is path-connected. We can then find a path from $x-n$ to $n$, avoiding $\cup_{i=1}^{n-1} p_i ([0,1])$ and also $X \cup N$.

If we have found these paths, we can find neighborhoods $U_n$ of the images, $U_i \cap U_j=\emptyset$ for $i \neq j$. There exist diffeomorphisms $f_n:U_n \to U_n$ with compact support and $f_i(x_i)=i$. These glue together to a global diffeomorphism $f$ with the desired property.