Probabilities of events being certain or impossible - implications 
Let $\Omega$ be the sample space and $A\subseteq \Omega$ be an event.Given below are two statements.

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*Statement 1: If $P(A)=0$ then $A=\phi$.


*Statement 2: If $P(A)=1$ then $A = \Omega$.
Determine whether each of the above statements as true or false.

This is a question from a paper and the answer key gives answer as both statements are wrong. I am unable to prove either of these statements to be false.
Can anyone justify this (either through a proof or using a counterexample)?
 A: Disambiguation: There are two ways that we can define a sample space and this will ultimately impact the types of examples we can use. One definition of the sample space is the set of all possible outcomes. The second is a (more relaxed definition) stating that it is a set that contains all possible outcomes. These definitions are similar, but are not the same.
Case 1 (sample space only contains all possible events): The first case is more restrictive. If every $x \in \Omega$ is a possible event, then we can look at the following example. If we are asked to pick a number in the interval $[0,1]$, then the probability we pick any single number is $0$ - e.g $P( \{0.5 \})=0$ - therefore statement $1$ is false. Similarly, if we consider interval $A :=[0,1]$ \ $\{0.5 \}$ (ie. the entire interval minus the point $0.5$), then $P(A)=1$, therefore statement $2$ is also false.
Case 2 (sample space contains all possible events): Let $\Omega = \{ 1,2,3,4,5,6,7 \}$. Now consider the outcomes of a roll of a fair six sided die. To address statement $1$, we see that the set {$7$} $\subseteq \Omega$ satisfies $P(${$7$}$) \space = 0$. Therefore, this is a counterexample for statement $1$ (since $7$ is not possible to roll on a standard $6$ sided die the probability of getting this number is $0$, however, {$7$} is not equal to $\emptyset$). Now we turn to statement $2$. For the set $\{ 1,2,3,4,5,6 \} \subseteq \Omega$, then $P(${$1,2,3,4,5,6$}$) \space=1$. This is a counterexample for statement $2$ (since $A$ is not equal to $\Omega$, but the probability of $A$ is still $1$ as it allows for every possible number to roll on the die).
Therefore, both of the statements are false (under both definitions of a sample space).
Note: it is true that $P( \emptyset) \space = 0$ and $P( \Omega) \space = 1$, however, the reason that the statements are incorrect is that we can find examples where $P(A) = 0$ or $P(A) = 1$ where $A \neq \emptyset$ and $A \neq \Omega$.
