I have this problem:

"Show that a topological space X$\neq\emptyset$ is irreducible $\Leftrightarrow$ $\forall$ U open subset of X, U is connected. "

I can easily prove the $\Rightarrow$ part, using that every open subset of an irreducible space is itself irreducible (therefore connected).

But what about the $\Leftarrow$ part? I have no idea how to prove it, and to be fair I'm not even sure it's true.

Thanks everybody!

  • $\begingroup$ Welcome to math stack exchange! $\endgroup$ – Peter Mar 30 '18 at 14:38

Suppose $U$ is non-empty open and disconnected so that $U =U_1 \cup U_2$, where $U_1$ and $U_2$ are non-empty disjoint and open in $U$ (and so also open in $X$).

Then $X = (X\setminus U_1) \cup (X\setminus U_2)$ (as $U_1$ and $U_2$ are disjoint) and both sets are closed and proper (as both $U_i$ are non-empty). This contradicts that $X$ is irredicible.

So all non-empty open $U$ are connected.

The other direction is similar. Suppose $X$ has no open disconnected subsets and suppose that $X = F_1 \cup F_2$ where $F_1$ and $F_2$ are proper non-empty closed sets. Then (by de Morgan) $$\emptyset = (X\setminus F_1) \cap (X\setminus F_2)$$

so $$U = (X\setminus F_1) \cup (X\setminus F_2)$$ is an open set that is disconnected (as both sets in this union are non-empty proper open subsets of $U$). This contradiction shows that $X$ is irreducible.


Let $A$, $B$ be any two non-empty open sets in $X$. Then, $A\cup B$ is open and non-empty, and thus connected. So $A\cap B\ne\emptyset$.

  • $\begingroup$ Does that mean, that the other direction is valid as well ? $\endgroup$ – Peter Mar 30 '18 at 14:35
  • 1
    $\begingroup$ @Peter Indeed. Irreducible $\Leftrightarrow$ all open are connected $\Leftrightarrow$ all non-empty open are dense. $\endgroup$ – user228113 Mar 30 '18 at 14:36
  • $\begingroup$ OK, I have overseen the "for all" $\endgroup$ – Peter Mar 30 '18 at 14:37

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