# If $B = B_1 \cup B_2$ where B is open and $B_1$ and $B_2$ are disjoint, then is it true that $B_1$ and $B_2$ are both open?

Suppose that $X = \mathbb{R}$ and it equip $X$ with the standard topology on $\mathbb{R}$. If $B = B_1 \cup B_2$ where $B$ is open and $B_1$ and $B_2$ are disjoint, then is it true that $B_1$ and $B_2$ are both open ?

• Hint: Try the union of sets of the form $(a,b)$, $[c,d]$ $(e,f]$ and $[g,h)$. Aug 5, 2016 at 1:16
• Furthermore if one of them is open it does not imply that the other one is open. This would mean open and closed sets were the same thing, which is only true in some boring topologies. Aug 6, 2016 at 14:42

the reals are open. the rationals and irrationals are not open.

You shouldn't even need a counter-example for this question. For any open set $B$, take ANY subset of $B$, call the subset $B_1$. Then put $B_2 = B \setminus B_1$. From what you are saying, in particular $B_1$ would be open.

So, rephrasing your question: Is it true that ALL subsets of an open set are themselves open? THIS you can answer yourself.

Take $B_1=(-1,0)$ and $B_2=[0,1)$. Then $B_1\cup B_2=(-1,1)$ is open, and $B_1$ and $B_2$ are disjoint. But $B_2$ is not open in the standard topology.

More general answer (without condition $X=\mathbb R$ equipped with standard topology).

Statement: If $B_1\cup B_2$ is open and $B_1\cap B_2=\varnothing$ then $B_1,B_2$ are both open.

If $X$ is equipped with discrete topology then this is evidently true.

Conversely observe that $X$ (the whole space) is open by definition. If $B\subseteq X$ then $B\cup B^c=X$ and $B\cap B^c=\varnothing$.

So if it is true then we can conclude that any arbitrary subset of $X$ is open.

Final conclusion:

The statement is true if and only if $X$ is equipped with the discrete topology (see the comment of copper.hat).