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I have been considering the set $$V = \{(x,y) \in \mathbb{R}^2 | 1+x^2+xy \neq 0\},$$ and I have proven earlier that this set is open. I want to prove the disconnectedness of this set, but I am having issues with this given my instructor's definition of a disconnected set.

Most definitions of a disconnected essentially boil down to the existence of two open sets $A,B \subset V$ such that $A \cap B = \emptyset, A \cup B = V, A \cap V \neq \emptyset, B \cap V \neq \emptyset .$ Under these conditions, it is pretty apparent that in our case, the sets $$A = \{(x,y) \in \mathbb{R}^2 | 1+x^2+xy < 0\}$$ and $$B = \{(x,y) \in \mathbb{R}^2 | 1+x^2+xy > 0\}$$ satisfy these conditions.

However, my instructor recently gave me the definition that a set $U$ is disconnected if there exists two disjoint open set $A$ and $B$ such that $U \subset A \cup B, U \cap A \neq \emptyset, U \cap B \neq \emptyset .$ Note that this case is slightly altered given the fact that $A \cup B$ must strictly contain $U$. Any recommendations on how to alter my current solution to satisfy this particular definition?

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    $\begingroup$ It's difficult to obtain strict inclusion (as in $\subsetneq$) if you were to ask, for instance, about the connectedness of a whole topological space, not just a subset of one. What else would they contain? I don't think strict inclusion is meant. $\endgroup$ – Arthur May 31 '16 at 22:34
  • $\begingroup$ If $V = A\cup B,$ then $V \subset A\cup B$ $\endgroup$ – Doug M May 31 '16 at 22:51
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    $\begingroup$ The symbol $\subset$ doesn't always mean strict inclusion. Many authors use it for ordinary inclusion, and that's almost certainly what's intended here. $\endgroup$ – Jack Lee May 31 '16 at 23:20
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Both definitions of disconnectedness are equivalent. If we have two disjoint nonempty open sets $A,B \subseteq V$ under the subspace topology such that $A \cup B = V$, then as Doug M pointed out in the comments, $V \subseteq A \cup B$. Further, since $A$ and $B$ are nonempty and are contained in $V$, we have $A \cap V \neq \emptyset$ and $V \cap B \neq \emptyset$.

Conversely, if there exist disjoint open sets $A$ and $B$ such that $V \subseteq A \cup B$, $V \cap A \neq \emptyset$, and $V \cap B \neq \emptyset$, then the sets $V \cap A$ and $V \cap B$ are open in the subspace topology, are disjoint, and $V= (V \cap A) \cup (V \cap B)$.

In short, there is no need to modify your answer.

I should point out that the sets $A$ and $B$ in the first definition are only required to be open in the subspace topology. This is typically the result of connectedness being defined in terms of the ambient space $X$, and then defining it for subsets of $X$ using the subspace topology.

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