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A topological space is called connected if any presentation of $X$ as $X = V_1 \uplus V_2$ by disjoint open subsets implies that one of them is trivial ($V_1 = X$ or $V_2 = X$). By taking complement one can replace the word "open" by "close".

A topological space is called irreducible if any presentation of $X$ as $X = X_1 \cup X_2$ by two closed subsets implies that one of them is trivial ($X_1 = X$ or $X_2 = X$).

Clearly every irreducible space is connected. The converse is not always true but:

Proposition: Let $X$ be a connected topological space which has an open covering by irreducible subspaces. Then $X$ is irreducible.

I want to prove this proposition.

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I assume that the two closed subsets $X_1, X_2$ in the definition of irreducible are supposed to be disjoint? –  Tom Nov 12 '13 at 13:01
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No, there can be $X_1 \cap X_2 \ne \emptyset$. For example, consider the Zariski topology on $\mathbb{A}^2(\mathbb{C})$ and the variety $X = \{ (x,y) \mid xy = 0 \}$. It is reducible and it is covered by $\{ (x,y) \mid x = 0 \}$ and $\{ (x,y) \mid y = 0 \}$ which are closed in the Zariski topology. –  LinAlgMan Nov 12 '13 at 13:14
    
Ah, I see; I misinterpreted. So $\mathbb{R}$ with the usual topology is another example of a reducible (and connected) space since $\mathbb{R} = (-\infty,0]\cup[0,\infty)$. I also suppose then that $\mathbb{R}$ with the cofinite topology is irreducible. Interesting! –  Tom Nov 12 '13 at 13:30
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@Tom: One way to think of irreducible spaces that I find helpful is that they are the spaces in which every non-empty open set is dense. –  Brian M. Scott Nov 12 '13 at 14:41
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1 Answer 1

up vote 4 down vote accepted

HINT: It’s easily checked that a space $X$ is irreducible if and only if $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open subsets of $X$. Let $X$ be connected, and let $\mathscr{U}$ be an open cover of $X$ by irreducible subspaces. Suppose that $V$ and $W$ are disjoint open subsets of $X$; you want to show that one of them is empty.

  • Show that each $U\in\mathscr{U}$ intersects at most one of $V$ and $W$.

Let

$$\begin{align*} &\mathscr{U}_V=\{U\in\mathscr{U}:U\cap V\ne\varnothing\}\;,\\ &\mathscr{U}_W=\{U\in\mathscr{U}:U\cap W\ne\varnothing\}\;,\text{ and}\\ &\mathscr{U}_0=\mathscr{U}\setminus(\mathscr{U}_V\cup\mathscr{U}_W)\;. \end{align*}$$

Consider the open sets $\bigcup\mathscr{U}_V,\bigcup\mathscr{U}_W$, and $\bigcup\mathscr{U}_0$, bearing in mind that $X$ is connected.

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Thank you. The hint helped me. –  LinAlgMan Nov 12 '13 at 15:57
    
@LinAlgMan: You’re welcome; glad it helped. –  Brian M. Scott Nov 12 '13 at 16:03
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