An open subset of an irreducible set is dense.

I'm trying to understand this example in Hartshorne's algebraic geometry book In order to prove the irreducible part, suppose $Y$ is an irreducible space and $Y'$ a open subset of $Y$ with $Y'=Y'_1\cup Y'_2$ with $Y'_1,Y'_2$ proper closed subsets. Then $Y=(Y-Y')\cup (Y'_1\cup Y'_2)$ contradiction because $Y$ is irreducible.

Am I right? I need help also in the density part, I'm really stuck I don't know even how to begin.

Thanks a lot.

• Hint for density: Being irreducible is equivalent to "any two non-empty open subsets intersect non-trivially" (by taking complements in the usual definition). – Tobias Kildetoft Aug 5 '13 at 7:17
• @TobiasKildetoft Is my solution in the irreducible part right? In the density part, do I need to use the definition of Zariski topology or this is true in general? – user42912 Aug 5 '13 at 7:42
• In your argument, why should $Y_1\cup Y_2$ be closed in $Y$? (Note that this is $Y'$ which is assumed open in $Y$. It would probably be easier also for that part to rewrite the definition in the form I mentioned above. And no, this is all for arbitrary irreducible spaces, not just those coming from the Zariski topology. – Tobias Kildetoft Aug 5 '13 at 7:48
• @TobiasKildetoft because I suppose by contradiction that $Y'$ is irreducible, then by definition of irreducible $Y'$ can be written as a union of closed proper subsets, say $Y'=Y'_1\cup Y'_2$, $Y'_1, Y'_2$ closed subsets in $Y'$, so they are closed in the whole space "Y" also. – user42912 Aug 5 '13 at 8:35
• No, closed subsets of open subset are not necessarily closed in the larger space. – Tobias Kildetoft Aug 5 '13 at 8:38

Your proof for the irreducible part is wrong. For example what do you mean by $Y' = Y_1 \cup Y_2$ with $Y_1,Y_2$ closed? The key question is closed in which space? First I would prove that an open subset of an irreducible space is dense: If $U$ open and non-empty is such that $\overline{U} \neq Y$ then immediately we can write $Y= U^c \cup \overline{U}$ contradicting $Y$ being irreducible.
Then now you can use this result to prove that any open subset of an irreducible space is irreducible: If $U = A \cup B$ with $A,B$ closed in $U$ then taking closures in $Y$ we get that $Y = \overline{U} = \overline{A} \cup \overline{B}$ and so this forces $\overline{A} = Y$ say. But now the closure of $A$ in $U$ which is equal to $\overline{A} \cap U$ is also equal to $A$ because $A$ was closed in $U$ by assumption. Thus $A = \overline{A} \cap U = Y \cap U = U$. It follows $U$ is irreducible in the subspace topology on it.
• I don't understand how $Y$ is equal to $U^{\complement} \cup \overline{U}$ and not merely a subset, since $U$ is open and thus $U \neq \overline{U}$. – G. Chiusole Nov 2 '18 at 16:13
• @G.Chiusole But $U \subsetneq \overline{U}$ – Ping Wan Mar 4 at 2:17
For a slightly different proof not using the density, consider that if an open subset $U \subset Y$ is reducible we can write $U = (U \setminus V) \cup (U \setminus W)$ for $V, W$ open non-empty subsets of $U$. Since $U$ is itself open, $V$ and $W$ will also be open subsets of $Y$. Hence we can write $Y = (Y \setminus V) \cup (Y \setminus W)$, so $Y$ is reducible.