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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.

J. W. Tanner
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user183002
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    What is your definition of Euler characteristic? – dcolazin May 17 '19 at 16:52
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    The problem of the "triangle" $T$ calculation is that $\mathbb{R}\setminus T$ is not a triangle – dcolazin May 17 '19 at 17:00
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    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 '19 at 17:00
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    @CheerfulParsnip you can use the CW-complexes definition of $V-E+F$ – dcolazin May 17 '19 at 17:02
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    @dcolazin, but there are infinitely many cells in each dimension. – Cheerful Parsnip May 17 '19 at 17:03
  • @CheerfulParsnip how do you build $\mathbb{R}$ as a CW-complex? $0 \ 0$-cells, $1 \ 1$-cell ;) but it's not an homotopy invariant – dcolazin May 17 '19 at 17:06
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    @dcolazin Please look up the definition of CW complex. It needs at least one $0$ cell, and the boundaries of the $1$-cells have to attach to the $0$ skeleton. – Cheerful Parsnip May 17 '19 at 17:07
  • @CheerfulParsnip you are right: construction as a partition of cell, not as a CW-complex – dcolazin May 17 '19 at 17:19
  • @dcolazin: It never occurred to me calculate Euler characteristic using an arbitrary partition of a space into cells. I gather from the link in the OP that this is what some people call combinatorial Euler characteristic. – Cheerful Parsnip May 17 '19 at 19:39
  • @CheerfulParsnip exactly, and the two characteristics are not the same – dcolazin May 17 '19 at 19:42

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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 $N=M=\Bbb R$.

Dietrich Burde
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