Equivalent Definition for Irreducible space
拓扑里关于不可约空间有好几个等价的定义, 这篇文章对此作了一个整理.
Proposition
(不可约空间的等价定义)
Let $X$ be a topological space. The following are equivalent
- If $X = Z_1 \cup Z_2$ with $Z_i$ closed, then either $X = Z_1$ or $X = Z_2$. In other words, $X$ cannot be written as a finite union of closed subsets strictly contained in $X$.
- For every non-empty open sets $U$ and $V$, there intersection $U \cap V \ne \varnothing$.
- Any non-empty open subset is dense in $X$.
- Any open subset is connected.
$X$ is said to be irreducible if satisfying one of the condition above.
Proof.
(2) $\Rightarrow$ (3): If $U$ is an open subset which is not dense,
then one can take $y \in U^c$ and $V$ its open neighborhood
such that $U \cap V = \varnothing$.
(3) $\Rightarrow$ (2): Suppose $U \cap V = \varnothing$ for some non-empty open subset $U$ and $V$. Then for all $x \in V, x \notin \overline{U}$, this shows that $U$ is not dense.
(2) $\Rightarrow$ (1): Suppose $X = Z_1 \cup Z_2$ with $Z_i$ closed, then $\varnothing = X^c = Z_1^c \cap Z_2^c$. Hence one of $Z_1^c$ and $Z_2^c$ is empty, i.e. one of $Z_1$ and $Z_2$ has to be $X$.
(1) $\Rightarrow$ (4): Let $U$ be an open subset. If $U$ is not connected, then $U = U_1 \sqcup U_2$ with $U_i$ open and strictly contained in $U$. Now $X = U_1^c \cup U_2^c$.
(4) $\Rightarrow$ (2): Let $U, V$ be open subset. If $U \cap V = \varnothing$, then $U \cup V$ is open and both $U$ and $V$ are open in $U \cup V$, which means that $U \cup V$ is not connected.
Done. $\square$
(3) $\Rightarrow$ (2): Suppose $U \cap V = \varnothing$ for some non-empty open subset $U$ and $V$. Then for all $x \in V, x \notin \overline{U}$, this shows that $U$ is not dense.
(2) $\Rightarrow$ (1): Suppose $X = Z_1 \cup Z_2$ with $Z_i$ closed, then $\varnothing = X^c = Z_1^c \cap Z_2^c$. Hence one of $Z_1^c$ and $Z_2^c$ is empty, i.e. one of $Z_1$ and $Z_2$ has to be $X$.
(1) $\Rightarrow$ (4): Let $U$ be an open subset. If $U$ is not connected, then $U = U_1 \sqcup U_2$ with $U_i$ open and strictly contained in $U$. Now $X = U_1^c \cup U_2^c$.
(4) $\Rightarrow$ (2): Let $U, V$ be open subset. If $U \cap V = \varnothing$, then $U \cup V$ is open and both $U$ and $V$ are open in $U \cup V$, which means that $U \cup V$ is not connected.
Done. $\square$
Corollary
- An irreducible space is necessarily connected.
- Any non-empty open subset of an irreducible space is again irreducible and dense.
- If $Y$ is an irreducible subset of a space $X$, then its closure $\overline{Y}$ is also irreducible.
Proof.
(1) 因为全空间 $X$ 就是一个开集, 再利用 part (4) of Proposition (b31a71b)
(2) 设 $Y$ 是不可约空间 $X$ 的一个开子集. $Y$ is dense by part (3) of Proposition (b31a71b). 注意到对任意 $U \subseteq Y$, 我们有 $U$ is open in $Y$ if and only if $U$ is open in $X$, so $Y$ is irreducible by part (4) of Proposition (b31a71b).
(3) Immediate by part (1) of Proposition (b31a71b).
See also [1, Chapter I Example 1.1.3 and 1.1.4] . $\square$
(2) 设 $Y$ 是不可约空间 $X$ 的一个开子集. $Y$ is dense by part (3) of Proposition (b31a71b). 注意到对任意 $U \subseteq Y$, 我们有 $U$ is open in $Y$ if and only if $U$ is open in $X$, so $Y$ is irreducible by part (4) of Proposition (b31a71b).
(3) Immediate by part (1) of Proposition (b31a71b).
See also [1, Chapter I Example 1.1.3 and 1.1.4] . $\square$
References
- 1
- Robin Hartshorne. Algebraic Geometry. GTM 52.