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Humphreys page 60, Linear algebraic groups.

Let $G$ act morphically on the nonempty variety $X$. Say $Y=G\cdot y$ is the orbit of $y\in X$. As the image of $G$ under the orbit map, $Y$ is constructible, hence contains an open dense subset of $\bar{Y}$.

This seems false. Take $Y=\{0,1,2\}\subset\Bbb A^1$, this is the union of three locally closed sets (intersect each point with the entire space and union), and contains no open subset of $Y=\bar{Y}$

What is wrong here?

Earth Cracks
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  • Or they would perhaps mean that it is open with a subspace topology on $\bar{Y}$ – Earth Cracks Jun 21 '17 at 07:41
  • Isn't $Y=\overline Y$ in this case? – lisyarus Jun 21 '17 at 07:41
  • Could you provide more context to those of us who do not have the text handy? It's hard to know whether there were previous assumptions on $Y$ or $\overline{Y}$ which might, when combined with the constructability of $Y$, yield the desired statement – KReiser Jun 21 '17 at 07:42
  • Right they mean locally open don't they. And in that case I can just take the complement of any of these three points in $\Bbb A^1$ and intersect it with $Y$ to obtain any pair is open, or the triple, or a point – Earth Cracks Jun 21 '17 at 07:42
  • @KReiser Sure thing, one second – Earth Cracks Jun 21 '17 at 07:42

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$\overline{Y}$ is an open subset of itself, and since $Y = \overline{Y}$, $Y$ contains an open sense subset of $\overline{Y}$: itself. (In fact, as $Y = \overline{Y}$, it contains all open subsets of $\overline{Y}$.) The statement indeed refers to the subspace topology on $\overline{Y}$; a similar question was asked here, and lemma 2.1 in this paper has a proof of the statement.

Stahl
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