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Proposition 1.10. If $Y$ is a quasi-affine variety, then $\dim Y=\dim \overline{Y}$.

The first line in this proof says

If $Z_{0}\subset Z_{1}\subset \cdots \subset Z_{n}$ is a sequence of distinct closed irreducible subsets of $Y$, then $\overline{Z_{0}}\subset \overline{Z_{1}}\subset\cdots \subset \overline{Z_{n}}$ is a sequence of distinct closed irreducible subsets of $\overline{Y}$ (1.1.4), ...

My question is: why are the sets $\overline{Z_{0}},\overline{Z_{1}},\cdots ,\overline{Z_{n}}$ all distinct?

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  • $\begingroup$ The bar means projective closure ? $\endgroup$ – Rene Schipperus Dec 4 '17 at 3:37
  • $\begingroup$ @ReneSchipperus no, closure inside the variety that $Y$ is an open subset of. For Hartshorne at this point, a quasi-affine variety is an open set of some closed subset of $\Bbb A^n_k$ for some fixed algebraically closed field $k$. $\endgroup$ – KReiser Dec 4 '17 at 3:39
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$Z_i=Y\cap \overline{Z_i}$, so if any two $\overline{Z_i}$ were the same, then two $Z_i$ would be the same, which is a contradiction.

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    $\begingroup$ Thank you. I didn't know that $Z_{i}=Y\cap \overline{Z_{i}}$, then I looked at Munkres, theorem 17.4 proves that. $\endgroup$ – Delong Dec 4 '17 at 4:09

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