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In p. 126 of Hatcher's Algebraic Topology, there is a discussion after theorem 2.26 about local homology groups. In particular, he says that the local homology groups of $X$ at a point $x \in X$ are the groups $H_n(X, X - \{x \})$. Until here everything is clear.

But then he uses excision theorem to claim that for every open neighborhood $U$ of $x$, $H_n(X, X - \{x\}) \equiv H_n(U, U- \{x \})$. What I don't understand is why he is not assuming that $x$ is a closed point of $X$. If it were closed, then we just use excision theorem with the open sets $U$ and $X - \{x\}$, but without this assumption I don't understand how can he use the excision theorem.

So the question is: do we need to assume that $x$ is closed in $X$?

Thank you!

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    $\begingroup$ In my version of that book, in the last paragraph on page 126, he writes "...assuming points are closed in $X$, ..." $\endgroup$ – Stefan Hamcke Feb 25 '15 at 15:47
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    $\begingroup$ Actually this is specified also in the pdf linked in the question... $\endgroup$ – Dario Feb 25 '15 at 16:06
  • $\begingroup$ Thank you! In my printed version it is not mentioned, I should have checked more carefully the online version :) $\endgroup$ – Pedro A. Castillejo Feb 25 '15 at 16:41
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Hatcher makes separation axiom assumptions way beyond points being closed. Mostly he assumes the spaces are CW complexes, which are normal spaces in which points are closed. In algebraic topology the spaces generally are normal or at least completely regular.

I would imagine these assumptions are mentioned in chapter 0.

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