Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

E.g. $[0,1]=\{0\}\sqcup (0,1]$ and the Euler characteristics of each of the three spaces is $1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.