# Show every irreducible subset of a topological space $X$ is contained in a maximal irreducible subset

Let $X$ be a topological space. A subset $A$ is irreducible if for every open $U,V\subseteq A$, we have $U\cap V\neq\varnothing$. Show that any irreducible subset $A\subseteq X$ is contained in a maximal irreducible set.

So here's basically what I want to do: let $A$ be an irreducible subset of $X$ and $\hat A$ be the union of all irreducible subsets containing $A$. I think this is the maximal irreducible subset I'm looking for.

To show this, let $U,V\subseteq$ be open in $A$. I want to show $U\cap V\neq\varnothing$ but I'm not sure how to do this. It's clear that an open subset of an irreducible set is irreducible, so I could show this if I knew that $U$ and $V$ were both contained in an irreducible set. But I don't know if this is even true. Any hints?

• nitpick: $U$ and $V$ must be non-empty open sets, for obvious reasons. – Henno Brandsma Jun 2 '16 at 5:06

Suppose $A_i, i \in I$ is a chain of irreducible subsets of $X$. Let's try to show that $C = \cup_{i \in I} A_i$ is irreducible. So let $U,V$ be non-empty open in $C$. So $U \cap A_i$ is non-empty for some $i \in I$, and $V \cap A_j$ is also irreducible for some $j \in I$. As we have a chain, either $A_i \subseteq A_j$ or $A_j \subseteq A_i$, say the former. Then $U \cap A_j,V \cap A_j$ are non-empty open in $A_j$ so they intersect in $A_j$, and so in $U,V$ intersect in $C$. So $\cup_i A_i$ is irreducible.
Now we can apply Zorn's lemma to the poset of all irreducible subsets of $X$ that contains some fixed $A$, ordered by inclusion, as the union is clearly an upperbound for the chain. This gives us a maximal element around every irreducible subset $A$. (As $A$ itself is in this poset, it's non-empty).
First, let me note that there might be more than one maximal irreducible set containing $A$. For instance, let $X=\{a,b,c\}$ with $\{b\}$ and $\{c\}$ as the only nontrivial open sets. Then $A=\{a\}$ is irreducible, but $\{a,b\}$ and $\{a,c\}$ are two different maximal irreducible sets containing it.
In particular, in this example, your $\hat{A}$ would be all of $X$, which is not irreducible. So your approach will not work. More generally, you should not expect there to be any canonical way to construct a maximal irreducible set, because of this non-uniqueness.