# Algebraic topology, proving two spaces aren't homeomorphic

I need help on how to approach a problem of this kind, I'm given two topological spaces: $$X=\mathbb{R}^2-\{(n,0)|n\in\mathbb{N}\}\text{ and }Y=\mathbb{R}^2-\{(\frac{1}{n},0)|n\in\mathbb{N}\}$$ I want to show that they're not the same but I have no clue where to start, one way could be showing that $\pi_1(X)\neq\pi_1(Y)$, but I don't know how to compute those groups.

• Fundamental groups are pretty hard to calculate here, there are simpler ways. Just a fun fact about this question: $Y\setminus \{0\}$ is in fact homeomorphic to $X$, with the homeomorphism $x\mapsto \frac1x$
– 5xum
Nov 16 '15 at 21:46

In $Y$, there exists a point such that no neighborhood of the point is simply connected. Since $X$ does not have this property, they cannot be homeomorphic.
• The fundamental group of $Y$ is very difficult to compute explicitly (see here, for instance; that is about a different space but I believe it has the same fundamental group). It is not too difficult to show the fundamental group of $Y$ is uncountable but the fundamental group of $X$ is countable, but this is still much harder than the simple argument in hunter's answer. Nov 16 '15 at 21:45
• @EricWofsey: I agree that the Hawaiian earring and $Y$ have the same fundamental group. Specifically, if you construct the "circles" in the Hawaiian earring to have radii $2/(2n+1)$, then I think that the Hawaiian earring is a deformation retract of $Y$. Nov 16 '15 at 21:51