Set Theory question about equalities of sets

Question

Consider the following sets:

$A=\{3,4\},$ $B=\{4,3\}\bigcup \emptyset, C=\{4,3\}\bigcup\{\emptyset\}$

Which pairs of sets are equal?

My attempt: We obviously have $C=\{4,3,\emptyset\}$ We also have(?) $B=\{4,3\}$ as by the Null set Axiom, the empty set is the unique set having no elements.

Now by the extensionality axiom, $A$ and $B$ are equal since they have the same elements. However, they are not equal to $C$ as they don't contain the empty set as an element.

Is that true? I'm still unsure about the bit where I look at the empty set as an element of $C$ since technically, no element of $C$ is not an element of $A$ or vice versa, hence we can use ZF1 again to conclude that they are equal??

Thanks in advance

Answer 1

You are correct. Except that to properly conclude that $A\ne C$ (and likewise $B\ne C$) you need to show that $\emptyset\notin A$, i.e., that $3\ne\emptyset$ and $4\ne\emptyset$.

Answer 2

If we have two sets $A$ and $B$ such that these two sets contain the same number of elements (the same cardinal number or cardinality), then both these sets are equivalent. The order of listing is not important, e.g. $$\{a, b, c\} = \{b, a, c\}.$$ Nevertheless, they are not necessarily equal. A multiple of sets are equal if every element in one set is also a member of every other set. Therefore, every pair of sets that are equal are also equivalent. A multiple of sets that have the same number of elements, but not the same elements, are equivalent sets. We arrive at the conclusion that every equal set is equivalent, but not every equivalent set is equal, and that the two sets shown in the example above are equal. In comparison, every prime number greater than $2$ is odd, but not every odd number greater than $2$ is prime. Below is an example of equivalent but not equal sets. $$\{a, b, c\} \leftrightarrow \{1, 2, 3\}.$$ Sometimes the equivalence operation $\leftrightarrow$ is replaced with $\sim$, but I will use the former. The above expression is also known as an equivalence relation, and since these sets are equivalent, but not equal, they are referred to as showing a one-to-one correspondence. $$\underbrace{n(A) = n(B)\Rightarrow A\operatorname*{\longleftrightarrow}^{\text{strictly}} B\Leftrightarrow \exists x\in A\ \land \ \nexists x\in B.}_{\text{Equivalence Axiom.}}$$

Thus, if we have three sets $A =\{3, 4\}$, $B =\{4, 3\}\cup \varnothing$ and $C = \{4, 3\}\cup\{\varnothing\}$ then we first note that all sets are equivalent except $C$ because it has three elements whilst the other sets $A$ and $B$ have only two elements. $$A\leftrightarrow B\not\leftrightarrow C.$$ Since $A$ and $B$ contain the same elements, they are equal. $$A = B.$$ The order of listing does not matter, so $\{3, 4\} = \{4, 3\}$ and so the above equation remains valid, and these sets both show a one-to-one correspondence.

If confused about the union of empty sets, go here.