# Euler characteristic of $\mathbb{R}$ and $\mathbb{R}^2$?

I´m trying to understand the value of the Euler characteristic of the real line and the real plane.

I don´t know if it is defined, I think that it is for any topological space.

So this could be right?

If we separate $$\mathbb{R} = (-\infty,x_0] \cup [x_0,x_1] \cup [x_1,+\infty)$$

$$\mathcal{X}(\mathbb{R}) = V - E + F = 2 - 3 + 0 = -1$$

Analogously, considering a "triangle" in the plane,

$$\mathcal{X}(\mathbb{R}^2) = V - E + F = 3 - 3 + 2 = 2$$

Thanks!

NOTE: I´ve seen the related topic:

Is the Euler characteristic of $\mathbb{R}^n$ $1$ or $(-1)^n$?

But I didn´t understand it.

• What is your definition of Euler characteristic? – dcolazin May 17 at 16:52
• The problem of the "triangle" $T$ calculation is that $\mathbb{R}\setminus T$ is not a triangle – dcolazin May 17 at 17:00
• The Euler characteristic is defined for these spaces, but it can't be calculated using the $V-E+F$ definition. You have to use the ranks of homology groups. If you do assume that it is well-defined, it is an invariant of homotopy equivalence. So both $\mathbb R^2$ and $\mathbb R$ should both have $\chi=1$. – Cheerful Parsnip May 17 at 17:00
• @CheerfulParsnip you can use the CW-complexes definition of $V-E+F$ – dcolazin May 17 at 17:02
• @dcolazin, but there are infinitely many cells in each dimension. – Cheerful Parsnip May 17 at 17:03

Any reasonable definition of the Euler characteristic should satisfy the product property $$\chi(N\times M)=\chi(N)\chi(M).$$ This is violated here in your calculation for $$N=M=\Bbb R$$.