From Hartshorne:

If $Y$ is an irreducible subset of $X$, then its closure $\overline{Y}$ in $X$ is also irreducible.

By irreducible they mean that $Y$ cannot be written $Y_1 \cup Y_2$ for two proper subsets $Y_i$ of $Y$ that are closed in the subspace topology of $Y$.

My attempt

Suppose that $\overline{Y} = Y_1 \cup Y_2$, both $Y_1, Y_2$ closed in $\overline {Y}$. Since the closed subsets of $\overline{Y}$ are precisely the closed subsets of $X$ intersected with $\overline{Y}$, we have that $Y_i = \overline{Y} \cap Y_i'$ for some closed $Y_i'$ in $X$. In other words, each $Y_i$ is closed in the whole space $X$ as well as in $\overline{Y}$.

By a previous statement, If $U \subset Y$ is an open subset, then when $Y$ is irreducible, $U$ is both irreducible (in $U$) and dense (in $Y$).

So we have that $U_i = \overline{Y} \setminus Y_i$ is irreducible in itself and dense in $\overline{Y}$, meaning $\overline{U}_i = \overline{Y} \setminus \text{Int} (Y_i) = \overline{Y}$.

Where to next?


If $\overline{Y} = Y_1 \cup Y_2$, then $Y = (Y_1 \cap Y) \cup (Y_2 \cap Y)$


Thanks @T.S.L.

If $\overline{Y} = Y_1 \cup Y_2$ then $Y = (Y_1 \cap Y) \cup(Y_2 \cap Y)$. Proof: $Y = Y \cap \overline{Y} = (Y \cap Y_1) \cup \dots$

So if $\overline{Y}$ is reducible, then since $Y\cap Y_i$ is closed in $Y$ we have that $Y$ is reducible, a contradiction as long as $Y\cap Y_i, i = 1,2$ is a proper subset of $Y$. If it were not then $Y = Y \cap Y_i \implies Y \subset Y_i$ which is closed in $X$ so that $\overline{Y} \subset Y_i \subset \overline{Y}$ and thuse $\overline{Y} = Y_i$ for some $i$ contradicting our assumption that $Y_i$ was proper.


Note: If $Z$ is irreducible, every nonempty open subset of $Z$ is dense and irreducible.

Let $Y$ be an irreducible subset of $X$.

To show: $\overline{Y}$ is irreducible.

Suppose $\overline{Y}$ is not irreducible. Then $\overline{Y} = C_{1} \cup C_{2}$, where $C_{1}$ and $C_{2}$ are proper non empty closed subsets of $\overline{Y}$.

So, $ Y = (C_{1} \cap Y) \cup (C_{2} \cap Y)$ .

Let $G_{1} = C_{1} \cap Y $ and $G_{2} = C_{2} \cap Y $. Both are closed in $Y$.

Let $ U = Y \setminus G_{1}$ and $V = Y \setminus G_{2} $. Both are nonempty open in $Y$ and their intersection is empty. This implies $Y$ has a nonempty open set that is not dense in $Y$, which contradicts the above note.


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