N-point compactification of $\mathbb{R}$.

Question

Let $X = \mathbb{R} \cup \{\infty_1,..., \infty_n\} $ be a topological space in which the topology is: the neighborhood of $\infty_i$ are $A \cup \infty_i$ with $A^c$ a compact set in $\mathbb{R}$. The problem asks to find the fundamental group of X. I have no idea how to do this. Thanks in advance for every hint\answer.

Answer 1

Actually $\mathbb R$ does have an obvious 2-point Hausdorff compactification, (not for any larger n though), but the compactification defined in your problem is different and definitely not Hausdorff.

Think of it as bending $\mathbb R$ round into a circle with one point missing, but then putting n points into the gap. They are distinct but have the same neighbourhoods within $\mathbb R$. A path going round the circle can pass through different infinities, but only one on each circuit. A loop which goes through one infinity but then comes back through another will not be homotopic to the null loop so the fundamental group is not the same as that of a circle. See if you can work it out from there.

Answer 2

Hint: This is actually completely straightforward with van Kampen's theorem. What are some simple open sets you could use to cover $X$, which let you ignore the non-Hausdorff weirdness when looking at just one of the open sets at a time?

A stronger hint is hidden below.

Cover $X$ with the open sets $U_i=\mathbb{R}\cup\{\infty_i\}$ for $i=1,\dots,n$. Each $U_i$ is just an ordinary 1-point compactification of $\mathbb{R}$, i.e. a circle.

Answer 3

$\mathbb{R}$ only has a one-point and two-point compactfication (homeomorphic to $S^1$ resp. $[0,1]$)

$\mathbb{R}^n$ has just a one-point compactification, not an $m$-point one for $m>1$: consider the intersection of the complements of the open balls $B(0,n)$ in such a compactification: it's an intersection of compact and connected (!) sets hence compact and connected and so cannot be a finite $m$-point set.