# Irreducible components of topological space

Let $X$ be a topological space. Let $\Sigma$ be the set of irreducible components of $X$. Let $X=\cup_{i\in I} X_i=\cup_{j\in J} Y_j$, $X_i,Y_j\in \Sigma$ for some index set $I,J$. $X_i$'s are distinct from each other. $Y_j$'s are distinct from each other.

I want an example such that $X$ has two distinct expression, i.e. there exist $\{X_i|i\in I\}\neq\{Y_j|j\in J\}$, such that $X=\cup_{i\in I} X_i=\cup_{j\in J} Y_j$.

Something we knew (but not helpful for this question): $X$ must not be one of the following case:

1. $X$ is Noetherian, then it can be uniquely written as a union of finite distinct irreducible components.
2. $X$ can be written as a union of finite distinct irreducible components, then all irreducible components of $X$ are in this expression, and expression is unique.
3. $X$ is a scheme, since {irreducible components} 1:1 correspond {generic points}. Thus expression is unique.

Update: 4. As stated, if $X$ is Hausdorff, then every irreducible set is a single point. (Because $E$ irr.$\Leftrightarrow$ every two nonempty opens intersect.)

Thanks.

-
I'm no expert on the topic, so I'm leaving this comment to be addressed to by greater minds; should this question be tagged under [algebraic-topology] as well/instead? – Asaf Karagila Jun 21 '11 at 15:45
I've added [algebraic-geometry], since this is the field of math in which irreducible components are important. – Jim Belk Jun 21 '11 at 15:58
It's worth mentioning that an example can't be Hausdorff. Any Hausdorff space with at least two points has the property that two points have disjoint neighborhoods, so those neighborhoods have closures which are not the entire space. So the only irreducible Hausdorff spaces are points and the irreducible decomposition of a Hausdorff space is trivially unique. – Qiaochu Yuan Jun 21 '11 at 16:09
More generally, an example can’t be a sober space. (I’m not familiar with schemes, but I suspect that I’m essentially just restating wxu’s (3).) – Brian M. Scott Dec 9 '11 at 18:37
@BrianM.Scott: yes, schemes are sober spaces. – user18119 Dec 10 '11 at 22:20

This should work. Take the square $[0,1]\times [0,1]$ with this strange topology. A subset is open if it is open in the Zariski topology (no generic points!) of every vertical segment, and in the Zariski topology of the horizontal segment $[0,1]\times\{0\}$.

Then, it's clear that the irriducible components are the vertical segments and $[0,1]\times\{0\}$, but to cover the square it's enough to use the vertical ones.

-