In an exercise, I had to prove, that the following statement is true:
$A\setminus B = A \cap \bar{B}$
The task itself was not hard to solve, but I immediately had the question, if this statement would still hold for the empty set $\emptyset$
One can rewrite:
$A\setminus B = \{x|x\in A \wedge x\notin B\}$
My reasoning was, that if x is in the empty set and not in B, but the empty set is contained in every set, x must also be contained in B. But my reasoning also tells me that, if A is empty, $x\in A$ should be false.
However, when I try to input $x\in A$ into Wolfram Alpha, I get the result:
undetermined (can be true or false)
I asked my professor and we both concluded that this was a mistake on WolframAlphas side, specifically because the Axiom of empty set has to hold
A set $M$ is the empty, when the proposition $x \in M$ holds false for each x.
Out of curiosity though, I want to know where the problem is. Is my input wrong or do I have to use another expression in WolframAlpha?
We think that WolframAlpha is interpreting the empty set as an character (or constant). However, even with the input $x \in \{\}$ I get the same result.
Please note that this is my first term in university, so I might not understand every answer well.
Edit: Again, what I typed into WolframAlpha was "x element of empty set" and "x element of {}" respectively.