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Let $H$ be a homology theory satisfying Eilenberg-Steenrod axioms and $X$ an arbitrary topological space. We can write $X$ as a disjoint union of its points $$X= \coprod_{x \in X}{\{x\}}$$ Now the additivity axiom implies that $H_n(X)$ is isomorphic to $\bigoplus_{x \in X } H_n(\{x\})$. By the dimension axiom, we have that $H_n(X)=0$ for $n\neq0$.

That generally $H_n(X)=0$ for $n\neq0$ holds is of course wrong (just look at homology groups of the 2-sphere). My question is now:

Where is the wrong argument in the first paragraph?

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    $\begingroup$ As a set, X is a disjoint union of its points, but not necessarily so as a topological space. $\endgroup$ – fixedp Feb 9 '16 at 19:54
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Eilenberg-Steenrod axioms says that, if $X= \coprod_{\alpha} X_\alpha$ the disjoint union of family of topological spaces $X_\alpha$ then $H_n(X)= \bigoplus H_n(X_\alpha)$ , so here each $X_\alpha$ is open in $X$. But the family you have defined, in that case each singleton $x\in X$ may not be an open set in $X$, and only possibility is in case of discrete space.

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  • $\begingroup$ Thanks for the quick answer. Of course, it is meant the disjoint union of topological spaces not only of sets. $\endgroup$ – bjn Feb 9 '16 at 20:10

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