I was using Van Kampen's Theorem and induction to show that $\mathbb{R}^n$ with finitely many points removed is simply connected for $n\geq 3$. However, this made me wonder whether we can remove countably many points and get the same conclusion. I'm not sure how I would go about proving it, nor can I think of a counterexample. I'm not even sure whether $\mathbb{R}^n\setminus\mathbb{Q}^n$ is simply connected.
$\mathbb{R}^n$ with countably many points removed is simply connected
algebraic-topologyhomotopy-theory
Related Solutions
Let $\{Y_i\}_i$ be the collection of path components of $Y$, so then $X*Y=\bigcup_i X*Y_i$. Let $Z$ be the portion of $X*Y$ associated with $[0,1/2)$, and let $A_i=Z\cup (X*Y_i)$. The intersection of any two of these $A_i$ is $Z$, which is path connected (deformation retracting onto $X$), so if we prove that $X*Y_i$ is simply connected, then $X*Y$ is simply connected by the van Kampen theorem. While it is true that the $A_i$ are not open in $X*Y$, this is fine because the surjectivity part of the theorem only relies on $f^{-1}(A_i)$ being open in $[0,1]$ for any path $f:[0,1]\to X*Y$. We defer showing openness to the appendix.
Because of this, let us assume $Y$ is path connected as well. The space $X*Y$ decomposes into two open subsets, one associated with $[0,2/3)$ and the other associated with $(1/3,1]$. Call these $Z_1$ and $Z_2$. The first deformation retracts onto $X$, the second deformation retracts onto $Y$, and their intersection $Z_1\cap Z_2$ deformation retracts onto $X\times Y$. The inclusion maps $X\times Y\to Z_1$ and $X\times Y\to Z_2$ induce projections on $\pi_1$, so the van Kampen theorem gives $\pi_1(X*Y)$ to be the free product $\pi_1(X)*\pi_1(Y)$ modulo the subgroup generated by all $i_{1*}(x,y)i_{2*}(x,y)^{-1}$, which are all $xy^{-1}$. Since $x,y$ may be arbitrary, the resulting $\pi_1(X*Y)$ is trivial, so $X*Y$ is simply connected.
Appendix. First, $Z$ is open in $X*Y$, so $f^{-1}(Z)$ is open. Take a point $f(s)=(x,y,t)$ in $X*Y_i$ with $t>0$. By continuity of $f$ there is a $\delta$ such that $f((s-\delta,s+\delta))\subset X\times Y\times (0,1]$. Since the image of this interval must lie in a path component, $f((s-\delta,s+\delta))\subset A_i$.
After posting this question I came across this blog post http://wildtopology.wordpress.com/2014/06/28/the-griffiths-twin-cone/ and now feel like I can answer my own question.
As mentioned in the comments the space I describe looks like this 
but it is homeomorphic to a space that looks like this 
Using the second of these two spaces and two applications of the Seifert–van Kampen theorem first with $U$ and $V_\text{odd}$ and then with $U \cup V_\text{odd}$ and $V_{\text{even}}$ we can see that the inclusion map identifying the Hawaiian earring with the intersection of the space with the $xy$ plane induces a surjection of fundamental groups.
The kernel of this map is the conjugate closure of the union of subgroups that contain loops only going around either even or odd circles but not both. This is not the entire fundamental group because the loop that goes once around every circle from biggest to smallest is not contained in this union.
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Best Answer
Yes, it is simply connected. See the proof given here: It will apply to any countable subset of any simply-connected topological manifold of dimension $\ge 3$. A related MSE question is here.