# property of Euler characteristic

I read the statement that the Euler characteristic is always additive with respect to closed-closed union, which means that $\chi(X\sqcup Y) = \chi(X)+\chi(Y)$ if $X$ and $Y$ are closed.

And I read that this is not true with respect to closed-open union. Can someone give me a counterexample showing that this is not additive?

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Are you sure that $\sqcup$ isn't a disjoint union? Otherwise, I don't believe the formula. – Dylan Moreland Jul 18 '11 at 17:31

E.g. $[0,1]=\{0\}\sqcup (0,1]$ and the Euler characteristics of each of the three spaces is $1$.
1) is $(0,1]$ open in $\mathbb R$, the point 1 does not have any neighborhood? i thaught it is neither open nor closed. 2)It is stille true that $[0,1]=\{0\}\sqcup[0,1]$ and so closed closed union is also not additive? !!! – palio Jul 18 '11 at 17:21
@palio: 1) $(0,1]$ is open in $[0,1]$, not in $\mathbb R$. 2) Those sets are not disjoint. Of course additivity doesn't hold in general for nondisjoint unions. E.g., $X=X\cup X$ is always true, but $\chi(X)=2\chi(X)$ isn't. – Jonas Meyer Jul 18 '11 at 17:30
what is the reason for the property being true when $X$ and $Y$ are closed? – palio Aug 26 '12 at 16:07