Does not being a subset of a set mean that you are a subset of the complement set? If I have three sets $A, B$ and $C$, where $ A   \cup  B$  is not a subset of $C$,
would it be correct to say that $A\cup B$ is a subset of $C^c$?
I'm working on a proof question, and I am not sure if I can use the two interchangeably.
 A: Everyone is giving generalized arguments for the most part, but to disprove a conjecture, all you must do is find one counterexample, and this is what Ittay was hinting at in a comment. Thus, I'll give you one example of many that invalidates your claim. Consider the following sets:


*

*$A=\{1,2\}$

*$B=\{2,3,4\}$

*$C=\{1,4,5\}$

*$U=\{1,2,3,4,5\}$

*$C^C=\{2,3\}$

*$A\cup B=\{1,2,3,4\}$


From the above, it is clear that $A\cup B$ is not a subset of $C$, and it is also clear that $A\cup B$ is not a subset of $C^C$, thus disproving your conjecture. 
A: $A=\{1\}, B=\{2\}, C=\{2,3\}$. Then $A\cup B\not\subseteq C^c$.
A: HINT: Consider the case where $A\subseteq C$ and $B\cap C=\varnothing$. What happens then? 
General piece of advice, if you have an hypothesis just try and prove it. If you failed, try to use this failure to create a counterexample; if you succeed go over your proof carefully several times, show it to someone else, and if you're finally sure it works, then you are done. 
A: The claim is false as currently stated. To see this consider the real number line with subsets $A = [0,1], B=[1,2]=C$. 
Instead of being concerned on whether or not $A\cup B \subset C$, the key thing is that $(A \cup B) \cap C = \emptyset$. If that is true, then $A\cup B \subseteq C^c$. Otherwise if $(A \cup B) \cap C \neq \emptyset$, then $A\cup B \not\subseteq C^c$
A: No. If $A,B$ are sets, then $A$ need not be a subset of $B$ or $B^c$. For example, take $A=B\cup B^c$ where $B$ and $B^c$ are both nonempty.
