# Connected Spaces

1. Apr 9, 2010

### Mikemaths

Can a disconnected space be a disjoint union of two infinite sets?
Must the disjoint subspaces be finite?

2. Apr 9, 2010

### dx

No, the connected components need not be finite, nor do they have to be compact.

3. Apr 9, 2010

### Landau

Sure. E.g.$$T=[0,1]\cup [2,3]\subset\mathbb{R}$$.

Well, just take any two disjoint infinite open subsets of some topological space, then their union is by definition disconnected.

4. Apr 10, 2010

### Bacle

Landau wrote, in part:

" Well, just take any two disjoint infinite open subsets of some topological space, then their union is by definition disconnected."

I think we also need that the intersection of their closures is empty, e.g.,

(-oo,1)\/(1,oo) is not a disconnection of R, since 1 belongs to both their closures,

IOW, I think we need that the open sets have no limit point in common.

5. Apr 10, 2010

### Office_Shredder

Staff Emeritus
The reason it's not a disconnection of R is because neither has 1 in it, not because 1 is in both of their closures.

But this isn't really relevant, because $$(-\infty, 1)$$ and $$(1, \infty)$$ is a disjoint cover of the topological space $$\mathbb{R}-{1}$$

6. Apr 10, 2010

### Bacle

O.K, my bad: the actual statement should be that X is disconnected iff (def.)

X=A\/B , with ClA /\B =empty =A/\ClB , with \/ =union, /\ intersection, Cl=closure

and A, B subsets of X. And Landau was right.

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