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?



Homotopy type of the plane minus a sequence with no limit points
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](http://en.wikipedia.org/wiki/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](https://mathoverflow.net/questions/18454/fundamental-groups-of-noncompact-surfaces)). 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.


