$A,B\subset \mathbb R^2$ s.t $A\cap B=\Phi$ and $A\cup B$ is open in $\mathbb R^2.$ Given that $A$ is open and $A\cup B$ is connected then is $B$ closed or open ?

Now if , $B$ is open too then $A\cup B$ cannot be connected so $B$ must be closed.

Now one possibility of this is that $A$ is any open set and $B=A^c$ the closed set.So the union is the whole set.

If $A$ is an open disc and $B$ is just any closed disc then the union is not open.

So my question is whether there is such a thing that this case is meaningful iff $A\cup B$ is allowed to be the whole set and otherwise not.Can I prove that union of an open set A and a closed set B can never be open in $\mathbb R^2$ unless $B=A^c$ and $A\cup B=\mathbb R\ ?$ Thanks.


First of all, note that "open" and "closed" aren't the only types of set there are!

For example, let $A=\{(x, y): x^2+y^2<1\}$, and let $B=\{(x, y): 1\le x^2+y^2<2\}$. Then

  • $A$ is open and connected,

  • $A\cap B=\emptyset$,

  • $A\cup B$ is open and connected,

  • but $B$ is neither open nor closed.

Your second question is: if $A$ is open and connected, $A\cap B=\emptyset$, $B$ is closed, and $A\cup B$ is open and connected, must $B=A^c$?

The answer to this is no: take $A=\{(x, y): 0<x^2+y^2<1\}$ and $B=\{(0, 0)\}$. (Thanks to Eric Wofsey for pointing out how silly my previous answer was - I was using simple connectedness!)

  • 1
    $\begingroup$ @EricWofsey Derp. Fixed, thanks! $\endgroup$ – Noah Schweber Sep 28 '16 at 22:18

Let $A=(0,1)\times(0,1)$ and $B=(0,1)\times[1,2)$; then $A\cup B=(0,1)\times(0,2)$ is open and connected, as is $A$, and $A\cap B=\varnothing$, but $B$ is neither closed nor open.


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